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waves Ocean combine properties of both transverse and longitudinal waves. With proper balance and timing, a surfer can capture some of the waves energy and take it for a ride.
OUTLINE
13.1 13.2 13.3
Hookes Law Elastic Potential Energy Comparing Simple Harmonic Motion with Uniform Circular Motion 13.4 Position, Velocity, and Acceleration as a Function of Time 13.5 Motion of a Pendulum 13.6 Damped Oscillations 13.7 Waves 13.8 Frequency, Amplitude, and Wavelength 13.9 The Speed of Waves on Strings 13.10 Interference of Waves 13.11 Reection of Waves
Rick Doyle/Corbis
13
CHAPTER
Vibrations and Waves
Periodic motion, from masses on springs to vibrations of atoms, is one of the most important kinds of physical behavior. In this chapter we take a more detailed look at Hookes law, where the force is proportional to the displacement, tending to restore objects to some equilibrium position. A large number of physical systems can be successfully modeled with this simple idea, including the vibrations of strings, the swinging of a pendulum, and the propagation of waves of all kinds. All these physical phenomena involve periodic motion. Periodic vibrations can cause disturbances that move through a medium in the form of waves. Many kinds of waves occur in nature, such as sound waves, water waves, seismic waves, and electromagnetic waves. These very different physical phenomena are described by common terms and concepts introduced here.
13.1 HOOKES LAW
One of the simplest types of vibrational motion is that of an object attached to a spring, previously discussed in the context of energy in Chapter 5. We assume that the object moves on a frictionless horizontal surface. If the spring is stretched or compressed a small distance x from its unstretched or equilibrium position and then released, it exerts a force on the object as shown in Active Figure 13.1. From experiment this spring force is found to obey the equation
Hookes law
Fs
kx
[13.1]
424
13.1
Hookes Law
425
where x is the displacement of the object from its equilibrium position (x 0) and k is a positive constant called the spring constant. This force law for springs was discovered by Robert Hooke in 1678 and is known as Hookes law. The value of k is a measure of the stiffness of the spring. Stiff springs have large k values, and soft springs have small k values. The negative sign in Equation 13.1 means that the force exerted by the spring is always directed opposite the displacement of the object. When the object is to the right of the equilibrium position, as in Active Figure 13.1a, x is positive and Fs is negative. This means that the force is in the negative direction, to the left. When the object is to the left of the equilibrium position, as in Active Figure 13.1c, x is negative and Fs is positive, indicating that the direction of the force is to the right. Of course, when x 0, as in Active Figure 13.1b, the spring is unstretched and Fs 0. Because the spring force always acts toward the equilibrium position, it is sometimes called a restoring force. A restoring force always pushes or pulls the object toward the equilibrium position. Suppose the object is initially pulled a distance A to the right and released from rest. The force exerted by the spring on the object pulls it back toward the equilibrium position. As the object moves toward x 0, the magnitude of the force decreases (because x decreases) and reaches zero at x 0. However, the object gains speed as it moves toward the equilibrium position, reaching its maximum speed when x 0. The momentum gained by the object causes it to overshoot the equilibrium position and compress the spring. As the object moves to the left of the equilibrium position (negative x-values), the spring force acts on it to the right, steadily increasing in strength, and the speed of the object decreases. The object A before accelerating back towards x 0 and nally comes briey to rest at x ultimately returning to the original position at x A. The process is then repeated, and the object continues to oscillate back and forth over the same path. This type of motion is called simple harmonic motion. Simple harmonic motion occurs when the net force along the direction of motion obeys Hookes law when the net force is proportional to the displacement from the equilibrium point and is always directed toward the equilibrium point. Not all periodic motions over the same path can be classied as simple harmonic motion. A ball being tossed back and forth between a parent and a child moves repetitively, but the motion isnt simple harmonic motion, because the force acting on the ball doesnt take the form of Hookes law, Equation 13.1. The motion of an object suspended from a vertical spring is also simple harmonic. In this case, the force of gravity acting on the attached object stretches the spring until equilibrium is reached and the object is suspended at rest. By denition the equilibrium position of the object is x 0. When the object is moved away from equilibrium by a distance x and released, a net force acts toward the equilibrium position. Because the net force is proportional to x, the motion is simple harmonic. The following three concepts are important in discussing any kind of periodic motion:
s
Fs (a) x Fs = 0 m x=0 m x x m x=0 (b) x
Fs (c)
x
x=0 ACTIVE FIGURE 13.1 The force exerted by a spring on an object varies with the displacement of the object from the equilibrium position, x 0. (a) When x is positive (the spring is stretched), the spring force is to the left. (b) When x is zero (the spring is unstretched), the spring force is zero. (c) When x is negative (the spring is compressed), the spring force is to the right.
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.1 to choose the spring constant and the initial position and velocity of the block and see the resulting simple harmonic motion.
s s
The amplitude A is the maximum distance of the object from its equilibrium position. In the absence of friction, an object in simple harmonic motion oscillates between the positions x A and x A. The period T is the time it takes the object to move through one complete cycle of motion, from x A to x A and back to x A. The frequency f is the number of complete cycles or vibrations per unit of time, and is the reciprocal of the period ( f 1/T).
EXAMPLE 13.1 Measuring the Spring Constant
Goal Use Newtons second law together with Hookes law to calculate a spring constant. Problem A common technique used to evaluate a spring constant is illustrated in Figure 13.2. A spring is hung vertically (Fig. 13.2a), and an object of mass m is attached to the lower end of the spring and slowly lowered a
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Vibrations and Waves
Figure 13.2 (Example 13.1) Determining the spring constant. The elongation d of the spring is due to the suspended weight mg. Because the upward spring force balances the weight when the system is in equilibrium, it follows that k mg/d. d
distance d to the equilibrium point (Fig. 13.2b). Find the value of the spring constant if the spring is displaced by 2.00 cm and the mass is 0.550 kg. Strategy This is an application of Newtons second law. The spring is stretched by a distance d from its initial position under the action of the load mg. The spring force is upward, balancing the downward force of gravity mg when the system is in equilibrium. (See Fig. 13.2c.) The suspended mass is in equilibrium, so set the sum of the forces equal to zero. Solution Apply the second law (with a spring constant k:
Fs
mg (a) (b) (c)
0) and solve for the
F k
Fg mg d
Fs
mg
kd
0 2.70 102 N/m
(0.550 kg)(9.80 m/s2) 2.00 10 2 m
Remarks In this case the spring force is positive, because its directed upward. Once the mass is pulled down from the equilibrium position and released, it oscillates around the equilibrium position, just like the horizontal spring. Exercise 13.1 A spring with constant k 475 N/m stretches 4.50 cm when an object of mass 25.0 kg is attached to the end of the spring. Find the acceleration of gravity in this location. Answer 0.855 m/s2 (The location is evidently an asteroid or small moon.)
The acceleration of an object moving with simple harmonic motion can be found by using Hookes law in the equation for Newtons second law, F ma. This gives ma
Acceleration in simple harmonic motion
F
kx k x m [13.2]
a
TIP 13.1 ConstantAcceleration Equations Dont Apply
The acceleration a of a particle in simple harmonic motion is not constant; it changes, varying with x, so we cant apply the constant acceleration kinematic equations of Chapter 2.
Equation 13.2, an example of a harmonic oscillator equation, gives the acceleration as a function of position. Because the maximum value of x is dened to be the amplitude A, the acceleration ranges over the values kA/m to kA/m. In the next section we will nd equations for velocity as a function of position and for position as a function of time.
Quick Quiz 13.1
A block on the end of a spring is pulled to position x A and released. Through what total distance does it travel in one full cycle of its motion? (a) A/2 (b) A (c) 2A (d) 4A
Quick Quiz 13.2
For a simple harmonic oscillator, which of the following pairs of vector quantities cant both point in the same direction? (The position vector is the displacement from equilibrium.) (a) position and velocity (b) velocity and acceleration (c) position and acceleration
13.2
Elastic Potential Energy
427
EXAMPLE 13.2 Simple Harmonic Motion on a Frictionless Surface
Goal Calculate forces and accelerations for a horizontal spring system. Problem A 0.350-kg object attached to a spring of force constant 1.30 102 N/m is free to move on a frictionless horizontal surface, as in Active Figure 13.1. If the object is released from rest at x 0.100 m, nd the force on it and its acceleration at x 0.100 m, x 0.050 0 m, x 0 m, x 0.050 0 m, and x 0.100 m. Strategy Substitute given quantities into Hookes law to nd the forces, then calculate the accelerations with Newtons second law. The amplitude A is the same as the point of release from rest, x 0.100 m. Solution Write Hookes force law: Substitute the value for k, and take x nding the force at that point: A 0.100 m,
Fs Fmax
kx kA 13.0 N (1.30 102 N/m)(0.100 m)
Solve Newtons second law for a and substitute to nd the acceleration at x A:
ma a
F max F max m 13.0 N 0.350 kg
Force (N) 13.0 6.50 0 6.50 13.0
37.1 m/s2
Acceleration (m/s2) 37.1 18.6 0 18.6 37.1
Repeat the same process for the other four points, assembling a table:
Position (m) 0.100 0.050 0 0.050 0.100
Remarks Notice that when the initial position is halved, the force and acceleration are also halved. Further, positive values of x give negative values of the force and acceleration, while negative values of x give positive values of the force and acceleration. As the object moves to the left and passes the equilibrium point, the spring force becomes positive (for negative values of x), slowing the object down. Exercise 13.2 For the same spring and mass system, nd the force exerted by the spring and the position x when the objects acceleration is 9.00 m/s 2. Answers 3.15 N, 2.42 cm
13.2 ELASTIC POTENTIAL ENERGY
In this section we review the material covered in Section 4 of Chapter 5. A system of interacting objects has potential energy associated with the conguration of the system. A compressed spring has potential energy that, when allowed to expand, can do work on an object, transforming spring potential energy into the objects kinetic energy. As an example, Figure 13.3 (page 428) shows a ball being projected from a spring-loaded toy gun, where the spring is compressed a distance x. As the gun is red, the compressed spring does work on the ball and imparts kinetic energy to it. Recall that the energy stored in a stretched or compressed spring or some other elastic material is called elastic potential energy, PE s , given by PEs
12 2 kx
[13.3]
Elastic potential energy
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Figure 13.3 A ball projected from a spring-loaded gun. The elastic potential energy stored in the spring is transformed into the kinetic energy of the ball.
Energy = elastic PEs
x
Energy = KE
Recall also that the law of conservation of energy, including both gravitational and spring potential energy, is given by (KE PEg PEs)i (KE PEg PEs )f [13.4]
If nonconservative forces such as friction are present, then the change in mechanical energy must equal the work done by the nonconservative forces: W nc (KE PEg PEs )f (KE PEg PEs)i [13.5]
Rotational kinetic energy must be included in both Equation 13.4 and Equation 13.5 for systems involving torques. As an example of the energy conversions that take place when a spring is included in a system, consider Figure 13.4. A block of mass m slides on a frictionless horizontal surface with constant velocity :i and collides with a coiled spring. The v description that follows is greatly simplied by assuming that the spring is very light and therefore has negligible kinetic energy. As the spring is compressed, it exerts a force to the left on the block. At maximum compression, the block comes to rest for just an instant (Fig. 13.4c). The initial total energy in the system (block plus spring) before the collision is the kinetic energy of the block. After the block
x=0 vi
1 E = mvi2 2
(a) v
Figure 13.4 A block sliding on a frictionless horizontal surface collides with a light spring. (a) Initially, the mechanical energy is entirely the kinetic energy of the block. (b) The mechanical energy at some arbitrary position is the sum of the kinetic energy of the block and the elastic potential energy stored in the spring. (c) When the block comes to rest, the mechanical energy is entirely elastic potential energy stored in the compressed spring. (d) When the block leaves the spring, the mechanical energy is equal to the blocks kinetic energy. The total energy remains constant.
(b) x
1 1 E = mv 2 + kx 2 2 2
v=0 (c) xm
1 E = mvi2 2 1 E = kx m2 2
vi
(d)
13.2
Elastic Potential Energy
429
collides with the spring and the spring is partially compressed, as in Figure 13.4b, 1 the block has kinetic energy 2mv 2 (where v vi) and the spring has potential 12 energy 2kx . When the block stops for an instant at the point of maximum compression, the kinetic energy is zero. Because the spring force is conservative and because there are no external forces that can do work on the system, the total mechanical energy of the system consisting of the block and spring remains constant. Energy is transformed from the kinetic energy of the block to the potential energy stored in the spring. As the spring expands, the block moves in the opposite direction and regains all of its initial kinetic energy, as in Figure 13.4d. When an archer pulls back on a bowstring, elastic potential energy is stored in both the bent bow and stretched bowstring (Fig. 13.5). When the arrow is released, the potential energy stored in the system is transformed into the kinetic energy of the arrow. Devices such as crossbows and slingshots work the same way.
Eric Lars Baleke/Black Star
Figure 13.5 Elastic potential energy is stored in this drawn bow.
A P P L I C AT I O N Archery
Quick Quiz 13.3
When an object moving in simple harmonic motion is at its maximum displacement from equilibrium, which of the following is at a maximum? (a) velocity, (b) acceleration, or (c) kinetic energy.
EXAMPLE 13.3 Stop That Car!
Goal Apply conservation of energy and the work energy theorem with spring and gravitational potential energy. Problem A 13 000-N car starts at rest and rolls down a hill from a height of 10.0 m (Fig. 13.6). It then moves across a level surface and collides with a light springloaded guardrail. (a) Neglecting any losses due to friction, nd the maximum distance the spring is compressed. Assume a spring constant of 1.0 106 N/m. (b) Calculate the maximum acceleration of the car after contact with the spring, assuming no frictional losses. (c) If the spring is compressed by only 0.30 m, nd the energy lost through friction.
10 m k
Figure 13.6 (Example 13.3) A car starts from rest on a hill at the position shown. When the car reaches the bottom of the hill, it collides with a spring-loaded guardrail.
Strategy Because friction losses are neglected, use conservation of energy in the form of Equation 13.4 to solve for the spring displacement in part (a). The initial and nal values of the cars kinetic energy are zero, so the initial potential energy of the car spring Earth system is completely converted to elastic potential energy in the spring at the end of the ride. In part (b), apply Newtons second law, substituting the answer to part (a) for x because the maximum compression will give the maximum acceleration. In part (c) friction is no longer neglected, so use the work energy theorem, Equation 13.5. The change in mechanical energy must equal the mechanical energy lost due to friction. Solution (a) Find the maximum spring compression, assuming no energy losses due to friction. Apply conservation of mechanical energy. Initially, there is only gravitational potential energy, and at maximum compression of the guardrail, there is only spring potential energy. Solve for x : (KE 0 PEg mgh PEs)i 0 (KE 0 0 PEg
12 2 kx
PEs)f
x
2mgh k
2(13 000 N)(10.0 m) 1.0 106 N/m
0.51 m
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(b) Calculate the maximum acceleration of the car by the spring, neglecting friction. Apply Newtons second law: Substitute values: ma kx : (1.0 a kx m kxg mg kxg w
a
106 N/m)(0.51 m)(9.8 m/s2) 13 000 N
380 m/s2 (c) If the compression of the guardrail is only 0.30 m, nd the mechanical energy lost due to friction. Use the work energy theorem: W nc (KE PEg PEs)f (KE PEg PEs)i
(0 0 1 kx 2) (0 mgh 0) 2 1 (1.0 106 N/m)(0.30)2 (13 000 N)(10.0 m) 2 Wnc 8.5 104 J
Remarks The answer to part (b) is about 40 times greater than the acceleration of gravity, so wed better be wearing our seat belts. Note that the solution didnt require calculation of the velocity of the car. Exercise 13.3 A spring-loaded gun res a 0.100-kg puck along a tabletop. The puck slides up a curved ramp and ies straight up into the air. If the spring is displaced 12.0 cm from equilibrium and the spring constant is 875 N/m, how high does the puck rise, neglecting friction? (b) If instead it only rises to a height of 5.00 m because of friction, what is the change in mechanical energy?
6.43 m (b) 1.40 J
In addition to studying the preceding example, its a good idea to review those given in Section 5.4.
Velocity as a Function of Position
x=0 A m (a) E=
1 2
v=0
kA2
x m v (b) E=
1 2
Conservation of energy provides a simple method of deriving an expression for the velocity of an object undergoing periodic motion as a function of position. The object in question is initially at its maximum extension A (Fig. 13.7a) and is then released from rest. The initial energy of the system is entirely elastic potential energy 1 stored in the spring, 2kA2. As the object moves toward the origin to some new position x (Fig. 13.7b), part of this energy is transformed into kinetic energy, and the potential energy stored in the spring is reduced to 1kx 2. Because the total energy of 2 1 the system is equal to 2kA2 (the initial energy stored in the spring), we can equate this quantity to the sum of the kinetic and potential energies at the position x:
1 2 2 kA 1 2 2 mv 12 2 kx
Solving for v, we get v
kx 2
+
1 2
mv 2
Figure 13.7 (a) An object attached to a spring on a frictionless surface is released from rest with the spring extended a distance A. Just before the object is released, the total energy is the elastic potential energy kA 2/2. (b) When the object reaches position x, it has kinetic energy mv 2/2 and the elastic potential energy has decreased to kx 2/2.
k (A2 m
x 2)
[13.6]
This expression shows that the objects speed is a maximum at x 0 and is zero at the extreme positions x A. The right side of Equation 13.6 is preceded by the sign because the square root of a number can be either positive or negative. If the object in Figure 13.7 is moving to the right, v is positive; if the object is moving to the left, v is negative.
13.2
Elastic Potential Energy
431
EXAMPLE 13.4 The Object Spring System Revisited
Goal Apply the time-independent velocity expression, Equation 13.6. Problem A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. (b) What is the velocity of the object when the displacement is 2.00 cm? (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. Strategy The total energy of the system can be found most easily at the amplitude x A, where the kinetic energy is zero. There, the potential energy alone is equal to the total energy. Conservation of energy then yields the speed at x 0. For part (b), obtain the velocity by substituting the given value of x into the time-independent velocity equation. Using this result, the kinetic energy asked for in part (c) can be found by substitution, and the potential energy from substitution into Equation 13.3. Solution (a) Calculate the total energy and maximum speed if the amplitude is 3.00 cm. Substitute x A 3.00 cm and k 20.0 N/m into the equation for the total mechanical energy E: E KE 00 9.00 (KE PEg 00 PEg PEs
1 2 (20.0
1 2 2 kA 10 3 J
N/m)(3.00
10
2
m)2
Use conservation of energy with xi A and xf compute the speed of the object at the origin:
0 to
PEs)i 1 2 2 kA
1 2 2 mv max
(KE 9.00
PEg 10
1 2 2 mv max
PEs)f 00
3
J 0.190 m/s
v max (b) Compute the velocity of the object when the displacement is 2.00 cm. Substitute known values directly into Equation 13.6: v
18.0 10 3 J 0.500 kg
k (A2 x 2) m 20.0 N/m ((0.030 0 m)2 0.500 kg
(0.020 0 m)2)
0.141 m/s (c) Compute the kinetic and potential energies when the displacement is 2.00 cm. Substitute into the equation for kinetic energy: Substitute into the equation for spring potential energy: KE PEs
1 2 2 mv 12 2 kx 1 2 (0.500 1 2 (20.0
kg)(0.141 m/s)2 10
2
4.97 m)2
10
3
J
N/m)(2.00
4.00
10
3
J
Remark With the given information, it is impossible to choose between the positive and negative solutions in part (b). Notice that the sum KE PEs in part (c) equals the total energy E found in part (a), as it should (except for a small discrepancy due to rounding). Exercise 13.4 For what values of x is the speed of the object 0.10 m/s? Answer 2.55 cm
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Lamp
13.3 COMPARING SIMPLE HARMONIC MOTION WITH UNIFORM CIRCULAR MOTION
We can better understand and visualize many aspects of simple harmonic motion along a straight line by looking at its relationship to uniform circular motion. Active Figure 13.8 is a top view of an experimental arrangement that is useful for this purpose. A ball is attached to the rim of a turntable of radius A, illuminated from the side by a lamp. We find that as the turntable rotates with constant angular speed, the shadow of the ball moves back and forth with simple harmonic motion. This fact can be understood from Equation 13.6, which says that the velocity of an object moving with simple harmonic motion is related to the displacement by v C A2 x2
Q Ball A P Turntable Screen A Shadow of ball ACTIVE FIGURE 13.8 An experimental setup for demonstrating the connection between simple harmonic motion and uniform circular motion. As the ball rotates on the turntable with constant angular speed, its shadow on the screen moves back and forth with simple harmonic motion.
where C is a constant. To see that the shadow also obeys this relation, consider Figure 13.9, which shows the ball moving with a constant speed v0 in a direction tangent to the circular path. At this instant, the velocity of the ball in the xdirection is given by v v0 sin , or sin v v0
From the larger triangle in the gure we can obtain a second expression for sin : sin
A2
A
x2
Equating the right-hand sides of the two expressions for sin , we nd the following relationship between the velocity v and the displacement x: v v0 or v v0 A
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.8 to adjust the frequency and radial position of the ball and see the resulting simple harmonic motion of the shadow.
A2
A
x2
A2
x2
C A2
x2
A P P L I C AT I O N Pistons and Drive Wheels
u v A
v0 A2 x 2 u x x-axis
The velocity of the ball in the x-direction is related to the displacement x in exactly the same way as the velocity of an object undergoing simple harmonic motion. The shadow therefore moves with simple harmonic motion. A valuable example of the relationship between simple harmonic motion and circular motion can be seen in vehicles and machines that use the back-and-forth motion of a piston to create rotational motion in a wheel. Consider the drive wheel of a locomotive. In Figure 13.10, the curved housing at the left contains a piston that moves back and forth in simple harmonic motion. The piston is connected to an arrangement of rods that transforms its back-and-forth motion into rotational motion of the wheels. A similar mechanism in an automobile engine transforms the back-and-forth motion of the pistons to rotational motion of the crankshaft.
Period and Frequency
The period T of the shadow in Active Figure 13.8, which represents the time required for one complete trip back and forth, is also the time it takes the ball to make one complete circular trip on the turntable. Because the ball moves through the distance 2 A (the circumference of the circle) in the time T, the speed v0 of the ball around the circular path is v0 2A T
Figure 13.9 The ball rotates with constant speed v 0. The x-component of the balls velocity equals the projection of :0 on the x-axis. v
13.3
Comparing Simple Harmonic Motion with Uniform Circular Motion
433
and the period is
Link/Visuals Unlimited
T
2A v0
[13.7]
Imagine that the ball moves from P to Q , a quarter of a revolution, in Active Figure 13.8. The motion of the shadow is equivalent to the horizontal motion of an object on the end of a spring. For this reason, the radius A of the circular motion is the same as the amplitude A of the simple harmonic motion of the shadow. During the quarter of a cycle shown, the shadow moves from a point where the energy of the system (ball and spring) is solely elastic potential energy to a point where the energy is solely kinetic energy. By conservation of energy, we have
1 2 2 kA 1 2 2 mv 0
Figure 13.10 The drive wheel mechanism of an old locomotive.
which can be solved for A/v 0 : A v0
2
m k
Substituting this expression for A/v 0 in Equation 13.7, we nd that the period is T
m k
[13.8]
The period of an object spring system moving with simple harmonic motion
Equation 13.8 represents the time required for an object of mass m attached to a spring with spring constant k to complete one cycle of its motion. The square root of the mass is in the numerator, so a large mass will mean a large period, in agreement with intuition. The square root of the spring constant k is in the denominator, so a large spring constant will yield a small period, again agreeing with intuition. Its also interesting that the period doesnt depend on the amplitude A. The inverse of the period is the frequency of the motion: f 1 T 1 2 [13.9]
Therefore, the frequency of the periodic motion of a mass on a spring is f
k m
1),
[13.10] or hertz (Hz). The angular fre-
Frequency of an object spring system
The units of frequency are cycles per second (s quency is 2f
k m
[13.11]
Angular frequency of an object spring system
The frequency and angular frequency are actually closely related concepts. The unit of frequency is cycles per second, where a cycle may be thought of as a unit of angular measure corresponding to 2 radians, or 360 . Viewed in this way, angular frequency is just a unit conversion of frequency. Radian measure is used for angles mainly because it provides a convenient and natural link between linear and angular quantities. Although an ideal mass spring system has a period proportional to the square root of the objects mass m, experiments show that a graph of T 2 versus m doesnt pass through the origin. This is because the spring itself has a mass. The coils of the spring oscillate just like the object, except the amplitudes are smaller for all coils but the last. For a cylindrical spring, energy arguments can be used to show that the effective additional mass of a light spring is one-third the mass of the spring. The square of the period is proportional to the total oscillating mass, so a graph of T 2 versus total mass (the mass hung on the spring plus the effective oscillating mass of the spring) would pass through the origin.
TIP 13.2
Twin Frequencies
The frequency gives the number of cycles per second, while the angular frequency gives the number of radians per second. These two physical concepts are nearly identical linked by the conversion factor 2 rad/cycle.
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Quick Quiz 13.4
An object of mass m is attached to a horizontal spring, stretched to a displacement A from equilibrium and released, undergoing harmonic oscillations on a frictionless surface with period T0. The experiment is then repeated with a mass of 4m. Whats the new period of oscillation? (a) 2T0 (b) T0 (c) T0/2 (d) T0/4.
Quick Quiz 13.5
Consider the situation in Quick Quiz 13.4. The subsequent total mechanical energy of the object with mass 4m is (a) greater than, (b) less than, or (c) equal to the original total mechanical energy.
Applying Physics 13.1 Bungee Jumping
A bungee cord can be modeled as a spring. If you go bungee jumping, you will bounce up and down at the end of the elastic cord after your dive off a bridge (Fig. 13.11). Suppose you perform a dive and measure the frequency of your bouncing. You then move to another bridge, but nd that the bungee cord is too long for dives off this bridge. What possible solutions might be applied? In terms of the original frequency, what is the frequency of vibration associated with the solution? Explanation There are two possible solutions: Make the bungee cord smaller or fold it in half. The latter would be the safer of the two choices, as well see. The force exerted by the bungee cord, modeled as a spring, is proportional to the separation of the coils as the spring is extended. First, we extend the spring by a given distance and measure the distance between coils. We then cut the spring in half. If one of the halfsprings is now extended by the same distance, the coils will be twice as far apart as they were for the complete spring. Therefore, it takes twice as much force to stretch the half-spring through the same displacement, so the half-spring has a spring constant twice that of the complete spring. The folded bungee cord can then be modeled as two half-springs in parallel. Each half has a spring constant that is twice the original spring constant of the bungee cord. In addition, an object hanging on the folded bungee cord will experience two forces one from each half-spring. As a result, the required force for a given extension will be
Figure 13.11 (Applying Physics 13.1) Bungee jumping from a bridge.
four times as much as for the original bungee cord. The effective spring constant of the folded bungee cord is therefore four times as large as the original spring constant. Because the frequency of oscillation is proportional to the square root of the spring constant, your bouncing frequency on the folded cord will be twice what it was on the original cord. This discussion neglects the fact that the coils of a spring have an initial separation. Its also important to remember that a shorter coil may lose elasticity more readily, possibly even going beyond the elastic limit for the material, with disastrous results. Bungee jumping is dangerous discretion is advised!
EXAMPLE 13.5 That Car Needs Shock Absorbers!
Goal Understand the relationships between period, frequency, and angular frequency. Problem A 1.30 103-kg car is constructed on a frame supported by four springs. Each spring has a spring constant of 2.00 104 N/m. If two people riding in the car have a combined mass of 1.60 102 kg, nd the frequency of vibration of the car when it is driven over a pothole in the road. Find also the period and the angular frequency. Assume the weight is evenly distributed.
Telegraph Colour Library/FPG International/Getty Images
13.4
Position, Velocity, and Acceleration as a Function of Time
435
Strategy Because the weight is evenly distributed, each spring supports one-fourth of the mass. Substitute this value and the spring constant into Equation 13.10 to get the frequency. The reciprocal is the period, and multiplying the frequency by 2 gives the angular frequency. Solution Compute one-quarter of the total mass: m
1 4 (m car
m pass)
1 4 (1.30
103 kg
1.60
102 kg)
365 kg Substitute into Equation 13.10 to nd the frequency: f 1 2 1 f 2f
k m
1 2
2.00
104 N/m 365 kg
1.18 Hz
Invert the frequency to get the period: Multiply the frequency by 2 to get the angular frequency:
T
1 1.18 Hz
0.847 s
2 (1.18 Hz)
7.41 rad/s
Remark Solving this problem didnt require any knowledge of the size of the pothole, because the frequency doesnt depend on the amplitude of the motion. Exercise 13.5 A 45.0-kg boy jumps on a 5.00-kg pogo stick with spring constant 3 650 N/m. Find (a) the angular frequency, (b) the frequency, and (c) the period of the boys motion. Answers (a) 8.54 rad/s (b) 1.36 Hz (c) 0.735 s
13.4 POSITION, VELOCITY, AND ACCELERATION AS A FUNCTION OF TIME
We can obtain an expression for the position of an object moving with simple harmonic motion as a function of time by returning to the relationship between simple harmonic motion and uniform circular motion. Again, consider a ball on the rim of a rotating turntable of radius A, as in Active Figure 13.12. We refer to the circle made by the ball as the reference circle for the motion. We assume that the turntable revolves at a constant angular speed . As the ball rotates on the reference circle, the angle made by the line OP with the x-axis changes with time. Meanwhile, the projection of P on the x-axis, labeled point Q, moves back and forth along the axis with simple harmonic motion. From the right triangle OPQ , we see that cos x/A. Therefore, the x-coordinate of the ball is x A cos t (see [13.12]
ACTIVE FIGURE 13.12 A reference circle. As the ball at P rotates in a circle with uniform angular speed, its projection Q along the x-axis moves with simple harmonic motion.
y P A O u xQ
v
x
Because the ball rotates with constant angular speed, it follows that Chapter 7), so we have x A cos( t)
In one complete revolution, the ball rotates through an angle of 2 rad in a time equal to the period T. In other words, the motion repeats itself every T seconds. Therefore, t 2 T 2f [13.13]
where f is the frequency of the motion. The angular speed of the ball as it moves around the reference circle is the same as the angular frequency of the projected simple harmonic motion. Consequently, Equation 13.12 can be written
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.12 to compare the oscillations of two blocks starting from different initial positions and to verify that the frequency is independent of the amplitude.
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ACTIVE FIGURE 13.13 (a) Displacement, (b) velocity, and (c) acceleration versus time for an object moving with simple harmonic motion under the initial conditions x 0 A and v 0 0 at t 0.
x A (a) O A v T 2 O T
x = A cos vt 3T 2 t
x
A cos(2 ft)
[13.14a]
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.13 to adjust the graphical representation and see the resulting simple harmonic motion of the block.
v = vA sin vt t T 3T 2
(b) T 2 a
O
(c) T 2
O
t T 3T 2 a = v2A cos vt
This cosine function represents the position of an object moving with simple harmonic motion as a function of time, and is graphed in Active Figure 13.13a. Because the cosine function varies between 1 and 1, x varies between A and A. The shape of the graph is called sinusoidal. Active Figures 13.13b and 13.13c represent curves for velocity and acceleration as a function of time. To nd the equation for the velocity, use Equations 13.6 and 13.14a together with the identity cos2 sin2 1, obtaining v A sin(2 ft) [13.14b]
where we have used the fact that k/m . The sign is no longer needed, because sine can take both positive and negative values. Deriving an expression for the acceleration involves substituting Equation 13.14a into Equation 13.2, Newtons second law for springs: a A
2
cos(2 ft)
[13.14c]
m
Motion of paper
Figure 13.14 An experimental apparatus for demonstrating simple harmonic motion. A pen attached to the oscillating object traces out a sinusoidal wave on the moving chart paper.
The detailed steps of these derivations are left as an exercise for the student. Notice that when the displacement x is at a maximum, at x A or x A, the velocity is zero, and when x is zero, the magnitude of the velocity is a maximum. Further, when x A, its most positive value, the acceleration is a maximum but in the negative x-direction, and when x is at its most negative position, x A, the acceleration has its maximum value in the positive x-direction. These facts are consistent with our earlier discussion of the points at which v and a reach their maximum, minimum, and zero values. The maximum values of the position, velocity, and acceleration are always equal to the magnitude of the expression in front of the trigonometric function in each equation, because the largest value of either cosine or sine is 1. Figure 13.14 illustrates one experimental arrangement that demonstrates the sinusoidal nature of simple harmonic motion. An object connected to a spring has a marking pen attached to it. While the object vibrates vertically, a sheet of paper is moved horizontally with constant speed. The pen traces out a sinusoidal pattern.
Quick Quiz 13.6
If the amplitude of a system moving in simple harmonic motion is doubled, which of the following quantities doesnt change? (a) total energy, (b) maximum speed, (c) maximum acceleration, (d) period.
EXAMPLE 13.6 The Vibrating Object Spring System
Goal Identify the physical parameters of a harmonic oscillator from its mathematical description. Problem (a) Find the amplitude, frequency, and period of motion for an object vibrating at the end of a horizontal spring if the equation for its position as a function of time is x (0.250 m)cos 8.00 t
(b) Find the maximum magnitude of the velocity and acceleration. (c) What is the position, velocity, and acceleration of the object after 1.00 s has elapsed?
13.4
Position, Velocity, and Acceleration as a Function of Time
437
Strategy In part (a), the amplitude and frequency can be found by comparing the given equation with the standard form in Equation 13.14a, matching up the numerical values with the corresponding terms in the standard form. (b) The maximum speed will occur when the sine function in Equation 13.14b equals 1 or 1, the extreme values of the sine function (and similarly for the acceleration and the cosine function). In each case, nd the magnitude of the expression in front of the trigonometric function. Part (c) is just a matter of substituting values into Equations 13.14a, b, and c. Solution (a) Find the amplitude, frequency, and period. Write the standard form given by Equation 13.14a, and underneath it write the given equation: x x Equate the factors in front of the cosine functions to nd the amplitude: The angular frequency is the factor in front of t in equations (1) and (2). Equate these factors: Divide by 2 to get the frequency f : f A A cos(2 ft) (0.250 m) cos 0.250 m 8.00 t (1) (2)
2f
8.00
rad/s
0.393 rad/s
2 1 f
0.062 5 Hz
The period T is the reciprocal of the frequency: (b) Find the maximum magnitudes of the velocity and the acceleration. Calculate the maximum speed from the factor in front of the sine function in Equation 13.14b: Calculate the maximum acceleration from the factor in front of the cosine function in Equation 13.14c: (c) Find the position, velocity, and acceleration of the object after 1.00 s. Substitute t 1.00 s in the given equation:
T
16.0 s
v max
A
(0.250 m)(0.393 rad/s)
0.098 3 m/s
a max
A
2
(0.250 m)(0.393 rad/s)2
0.038 6 m/s2
x v v
(0.250 m) cos (0.393 rad)
0.231 m
Substitute values into the velocity equation:
A sin( t) (0.250 m)(0.393 rad/s) sin (0.393 rad/s 1.00 s) 0.037 6 m/s A 2 cos( t) (0.250 m)(0.393 rad/s2)2 cos (0.393 rad/s 1.00 s) 0.035 7 m/s2
Substitute values into the acceleration equation:
a a
Remarks In evaluating the cosine function, the angle is in radians, so you should either set your calculator to evaluate trigonometric functions based on radian measure or convert from radians to degrees. Exercise 13.6 If the object spring system is described by x (0.330 m) cos (1.50t), nd (a) the amplitude, the angular frequency, the frequency, and the period, (b) the maximum magnitudes of the velocity and acceleration, and (c) the position, velocity, and acceleration when t 0.250 s. 1.50 rad/s, f 0.239 Hz , T Answers (a) A 0.330 m, (c) x 0.307 m, v 0.181 m/s, a 0.691 m/s2 4.19 s; (b) v max 0.495 m/s, a max 0.743 m/s2;
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13.5 MOTION OF A PENDULUM
u L s mg sin u u mg cos u mg ACTIVE FIGURE 13.15 A simple pendulum consists of a bob of mass m suspended by a light string of length L. (L is the distance from the pivot to the center of mass of the bob.) The restoring force that causes the pendulum to undergo simple harmonic motion is the component of gravitational force tangent to the path of motion, mg sin . T m
A simple pendulum is another mechanical system that exhibits periodic motion. It consists of a small bob of mass m suspended by a light string of length L xed at its upper end, as in Active Figure 13.15. (By a light string, we mean that the strings mass is assumed to be very small compared with the mass of the bob and hence can be ignored.) When released, the bob swings to and fro over the same path; but is its motion simple harmonic? Answering this question requires examining the restoring force the force of gravity that acts on the pendulum. The pendulum bob moves along a circular arc, rather than back and forth in a straight line. When the oscillations are small, however, the motion of the bob is nearly straight, so Hookes law may apply approximately. In Active Figure 13.15, s is the displacement of the bob from equilibrium along the arc. Hookes law is F kx, so we are looking for a similar expression involving s, Ft ks, where Ft is the force acting in a direction tangent to the circular arc. From the gure, the restoring force is Ft Since s mg sin
L , the equation for Ft can be written as Ft mg sin s L
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.15 to adjust the mass of the bob, the length of the string, and the initial angle and see the resulting oscillation of the pendulum. The period is slightly larger for larger initial angles.
This expression isnt of the form Ft ks, so in general, the motion of a pendulum is not simple harmonic. For small angles less than about 15 degrees, however, the angle measured in radians and the sine of the angle are approximately equal. For example, 10.0 0.175 rad, and sin(10.0 ) 0.174. Therefore, if we restrict the motion to small angles, the approximation sin is valid, and the restoring force can be written Ft Substituting s/L, we obtain Ft mg L s mg sin mg
TIP 13.3 Pendulum Motion is not Harmonic
Remember that the pendulum does not exhibit true simple harmonic motion for any angle. If the angle is less than about 15 , the motion can be modeled as approximately simple harmonic.
This equation follows the general form of Hookes force law Ft ks, with k mg/L. We are justied in saying that a pendulum undergoes simple harmonic motion only when it swings back and forth at small amplitudes (or, in this case, small values of , so that sin ). Recall that for the object spring system, the angular frequency is given by Equation 13.11: 2f
k m
Substituting the expression of k for a pendulum, we obtain
The period of a simple pendulum depends only on L and g
mg/L m
L g
g L
This angular frequency can be substituted into Equation 13.12, which then mathematically describes the motion of a pendulum. The frequency is just the angular frequency divided by 2 , while the period is the reciprocal of the frequency, or T 2
[13.15]
This equation reveals the somewhat surprising result that the period of a simple pendulum doesnt depend on the mass, but only on the pendulums length and on the free-fall acceleration. Furthermore, the amplitude of the motion isnt a
13.5
Motion of a Pendulum
439
a max
ACTIVE FIGURE 13.16 Simple harmonic motion for an object spring system, and its analogy, the motion of a simple pendulum.
umax Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.16 to set the initial position of the block and see the block spring motion and the analogous pendulum motion.
vmax
a max
umax
vmax
a max
umax
x A 0 A
factor as long as its relatively small. The analogy between the motion of a simple pendulum and the object spring system is illustrated in Active Figure 13.16. Galileo rst noted that the period of a pendulum was independent of its amplitude. He supposedly observed this while attending church services at the cathedral in Pisa. The pendulum he studied was a swinging chandelier that was set in motion when someone bumped it while lighting candles. Galileo was able to measure its period by timing the swings with his pulse. The dependence of the period of a pendulum on its length and on the free-fall acceleration allows us to use a pendulum as a timekeeper for a clock. A number of clock designs employ a pendulum, with the length adjusted so that its period serves as the basis for the rate at which the clocks hands turn. Of course, these clocks are used at different locations on the Earth, so there will be some variation of the free-fall acceleration. To compensate for this variation, the pendulum of a clock should have some movable mass so that the effective length can be adjusted. Geologists often make use of the simple pendulum and Equation 13.15 when prospecting for oil or minerals. Deposits beneath the Earths surface can produce irregularities in the free-fall acceleration over the region being studied. A specially designed pendulum of known length is used to measure the period, which in turn is used to calculate g. Although such a measurement in itself is inconclusive, its an important tool for geological surveys.
A P P L I C AT I O N Pendulum Clocks
A P P L I C AT I O N Use of Pendulum in Prospecting
Quick Quiz 13.7
A simple pendulum is suspended from the ceiling of a stationary elevator, and the period is measured. If the elevator moves with constant velocity, does the period (a) increase, (b) decrease, or (c) remain the same? If the elevator accelerates upward, does the period (a) increase, (b) decrease, or (c) remain the same?
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Quick Quiz 13.8
A pendulum clock depends on the period of a pendulum to keep correct time. Suppose a pendulum clock is keeping correct time and then Dennis the Menace slides the bob of the pendulum downward on the oscillating rod. Does the clock run (a) slow, (b) fast, or (c) correctly?
Quick Quiz 13.9
The period of a simple pendulum is measured to be T on Earth. If the same pendulum were set in motion on the Moon, would its period be (a) less than T, (b) greater than T, or (c) equal to T ?
EXAMPLE 13.7 Measuring the Value of g
Goal Determine g from pendulum motion. Problem Using a small pendulum of length 0.171 m, a geologist counts 72.0 complete swings in a time of 60.0 s. What is the value of g in this location? Strategy First calculate the period of the pendulum by dividing the total time by the number of complete swings. Solve Equation 13.15 for g and substitute values. Solution Calculate the period by dividing the total elapsed time by the number of complete oscillations: Solve Equation 13.15 for g and substitute values: time # of oscillations 2 4 60.0 s 72.0 4
2
T
0.833 s L g 9.73 m/s2
T g
2L
L g
:
T2
T2
(39.5)(0.171 m) (0.833 s)2
Exercise 13.7 What would be the period of the same pendulum on the Moon, where the acceleration of gravity is 1.62 m/s2 ? Answer 2.04 s
The Physical Pendulum
The simple pendulum discussed thus far consists of a mass attached to a string. A pendulum, however, can be made from an object of any shape. The general case is called the physical pendulum. In Figure 13.17, a rigid object is pivoted at point O, which is a distance L from the objects center of mass. The center of mass oscillates along a circular arc, just like the simple pendulum. The period of a physical pendulum is given by
L
Pivot
O
T
L sin CM
2
I mgL
[13.16]
where I is the objects moment of inertia and m is the objects mass. As a check, notice that in the special case of a simple pendulum with an arm of length L and negligible mass, the moment of inertia is I mL2. Substituting into Equation 13.16 results in
mg Figure 13.17 A physical pendulum pivoted at O.
T
2
mL2 mgL
2
L g
which is the correct period for a simple pendulum.
13.7
Waves
441
Oil or other viscous fluid Coil spring
Shock absorber
Figure 13.18 (a) A shock absorber consists of a piston oscillating in a chamber lled with oil. As the piston oscillates, the oil is squeezed through holes between the piston and the chamber, causing a damping of the pistons oscillations. (b) One type of automotive suspension system, in which a shock absorber is placed inside a coil spring at each wheel.
Piston with holes
(a)
(b)
13.6 DAMPED OSCILLATIONS
The vibrating motions we have discussed so far have taken place in ideal systems that oscillate indenitely under the action of a linear restoring force. In all real mechanical systems, forces of friction retard the motion, so the systems dont oscillate indenitely. The friction reduces the mechanical energy of the system as time passes, and the motion is said to be damped. Shock absorbers in automobiles (Fig. 13.18) are one practical application of damped motion. A shock absorber consists of a piston moving through a liquid such as oil. The upper part of the shock absorber is rmly attached to the body of the car. When the car travels over a bump in the road, holes in the piston allow it to move up and down in the uid in a damped fashion. Damped motion varies with the uid used. For example, if the uid has a relatively low viscosity, the vibrating motion is preserved but the amplitude of vibration decreases in time and the motion ultimately ceases. This is known as underdamped oscillation. The position vs. time curve for an object undergoing such oscillation appears in Active Figure 13.19. Figure 13.20 compares three types of damped motion, with curve (a) representing underdamped oscillation. If the uid viscosity is increased, the object returns rapidly to equilibrium after its released and doesnt oscillate. In this case, the system is said to be critically damped, and is shown as curve (b) in Figure 13.20. The piston returns to the equilibrium position in the shortest time possible without once overshooting the equilibrium position. If the viscosity is made greater still, the system is said to be overdamped. In this case, the piston returns to equilibrium without ever passing through the equilibrium point, but the time required to reach equilibrium is greater than in critical damping, as illustrated by curve (c) in Figure 13.20. To make automobiles more comfortable to ride in, shock absorbers are designed to be slightly underdamped. This can be demonstrated by a sharp downward push on the hood of a car: After the applied force is removed, the body of the car oscillates a few times about the equilibrium position before returning to its xed position. A P P L I C AT I O N Shock Absorbers
x A
0
t
ACTIVE FIGURE 13.19 A graph of displacement versus time for an underdamped oscillator. Note the decrease in amplitude with time.
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.19 to adjust the spring constant, the mass of the object, and the damping constant and see the resulting damped oscillation of the object. x
13.7
WAVES
b a
c t
The world is full of waves: sound waves, waves on a string, seismic waves, and electromagnetic waves, such as visible light, radio waves, television signals, and x-rays. All of these waves have as their source a vibrating object, so we can apply the concepts of simple harmonic motion in describing them. In the case of sound waves, the vibrations that produce waves arise from sources such as a persons vocal chords or a plucked guitar string. The vibrations of electrons in an antenna produce radio or television waves, and the simple up-and-down motion
Figure 13.20 Plots of displacement versus time for (a) an underdamped oscillator, (b) a critically damped oscillator, and (c) an overdamped oscillator.
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of a hand can produce a wave on a string. Certain concepts are common to all waves, regardless of their nature. In the remainder of this chapter, we focus our attention on the general properties of waves. In later chapters we will study specic types of waves, such as sound waves and electromagnetic waves.
What Is a Wave?
When you drop a pebble into a pool of water, the disturbance produces water waves, which move away from the point where the pebble entered the water. A leaf oating near the disturbance moves up and down and back and forth about its original position, but doesnt undergo any net displacement attributable to the disturbance. This means that the water wave (or disturbance) moves from one place to another, but the water isnt carried with it. When we observe a water wave, we see a rearrangement of the waters surface. Without the water, there wouldnt be a wave. Similarly, a wave traveling on a string wouldnt exist without the string. Sound waves travel through air as a result of pressure variations from point to point. Therefore, we can consider a wave to be the motion of a disturbance. In a later chapter we will discuss electromagnetic waves, which dont require a medium. The mechanical waves discussed in this chapter require (1) some source of disturbance, (2) a medium that can be disturbed, and (3) some physical connection or mechanism through which adjacent portions of the medium can inuence each other. All waves carry energy and momentum. The amount of energy transmitted through a medium and the mechanism responsible for the transport of energy differ from case to case. The energy carried by ocean waves during a storm, for example, is much greater than the energy carried by a sound wave generated by a single human voice.
Applying Physics 13.2 Burying Bond
At one point in On Her Majestys Secret Service, a James Bond lm from the 1960s, Bond was escaping on skis. He had a good lead and was a hard-to-hit moving target. There was no point in wasting bullets shooting at him, so why did the bad guys open re? Explanation These misguided gentlemen had a good understanding of the physics of waves. An impulsive sound, like a gunshot, can cause an acoustical disturbance that propagates through the air. If it impacts a ledge of snow that is ready to break free, an avalanche can result. Such a disaster occurred in 1916 during World War I when Austrian soldiers in the Alps were smothered by an avalanche caused by cannon re. So the bad guys, who have never been able to hit Bond with a bullet, decided to use the sound of gunre to start an avalanche.
Types of Waves
One of the simplest ways to demonstrate wave motion is to ip one end of a long rope that is under tension and has its opposite end xed, as in Figure 13.21. The bump (called a pulse) travels to the right with a denite speed. A disturbance of this type is called a traveling wave. The gure shows the shape of the rope at three closely spaced times. such As a wave pulse travels along the rope, each segment of the rope that is disturbed moves in a direction perpendicular to the wave motion. Figure 13.22 illustrates this point for a particular tiny segment P. The rope never moves in the direction of the wave. A traveling wave in which the particles of the disturbed medium move in a direction perpendicular to the wave velocity is called a transverse wave. Figure 13.23a illustrates the formation of transverse waves on a long spring. In another class of waves, called longitudinal waves, the elements of the medium undergo displacements parallel to the direction of wave motion. Sound
Figure 13.21 A wave pulse traveling along a stretched rope. The shape of the pulse is approximately unchanged as it travels.
13.7
Waves
443
waves in air are longitudinal. Their disturbance corresponds to a series of highand low-pressure regions that may travel through air or through any material medium with a certain speed. A longitudinal pulse can easily be produced in a stretched spring, as in Figure 13.23b. The free end is pumped back and forth along the length of the spring. This action produces compressed and stretched regions of the coil that travel along the spring, parallel to the wave motion. Waves need not be purely transverse or purely longitudinal: ocean waves exhibit a superposition of both types. When an ocean wave encounters a cork, the cork executes a circular motion, going up and down while going forward and back. Another type of wave, called a soliton, consists of a solitary wave front that propagates in isolation. Ordinary water waves generally spread out and dissipate, but solitons tend to maintain their form. The study of solitons began in 1849, when the Scottish engineer John Scott Russell noticed a solitary wave leaving the turbulence in front of a barge and propagating forward all on its own. The wave maintained its shape and traveled down a canal at about 10 mi/h. Russell chased the wave two miles on horseback before losing it. Only in the 1960s did scientists take solitons seriously; they are now widely used to model physical phenomena, from elementary particles to the Giant Red Spot of Jupiter.
P
P
P
P
Figure 13.22 A pulse traveling on a stretched rope is a transverse wave. Any element P on the rope moves (blue arrows) in a direction perpendicular to the direction of propagation of the wave motion (red arrows).
(a) Transverse wave
y vt Compressed Compressed v x Stretched (b) Longitudinal wave Figure 13.23 (a) A transverse wave is set up in a spring by moving one end of the spring perpendicular to its length. (b) A longitudinal pulse along a stretched spring. The displacement of the coils is in the direction of the wave motion. For the starting motion described in the text, the compressed region is followed by a stretched region. t=0 t Stretched
Picture of a Wave
Active Figure 13.24 shows the curved shape of a vibrating string. This pattern is a sinusoidal curve, the same as in simple harmonic motion. The brown curve can be thought of as a snapshot of a traveling wave taken at some instant of time, say, t 0; the blue curve is a snapshot of the same traveling wave at a later time. This picture can also be used to represent a wave on water. In such a case, a high point would correspond to the crest of the wave and a low point to the trough of the wave. The same waveform can be used to describe a longitudinal wave, even though no up-and-down motion is taking place. Consider a longitudinal wave traveling on a spring. Figure 13.25a is a snapshot of this wave at some instant, and Figure 13.25b shows the sinusoidal curve that represents the wave. Points where the coils of the spring are compressed correspond to the crests of the waveform, and stretched regions correspond to troughs. The type of wave represented by the curve in Figure 13.25b is often called a density wave or pressure wave, because the crests, where the spring coils are compressed, are regions of high density, and the troughs, where the coils are stretched, are regions of low density. Sound waves are longitudinal waves, propagating as a series of high- and low-density regions.
ACTIVE FIGURE 13.24 A one-dimensional sinusoidal wave traveling to the right with a speed v. The brown curve is a snapshot of the wave at t 0, and the blue curve is another snapshot at some later time t.
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.24 to watch the wave move and to take snapshots of it at various times.
(a)
Density
(b)
Figure 13.25 (a) A longitudinal wave on a spring. (b) The crests of the waveform correspond to compressed regions of the spring, and the troughs correspond to stretched regions of the spring.
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13.8 FREQUENCY, AMPLITUDE, AND WAVELENGTH
Active Figure 13.26 illustrates a method of producing a continuous wave or a steady stream of pulses on a very long string. One end of the string is connected to a blade that is set vibrating. As the blade oscillates vertically with simple harmonic motion, a traveling wave moving to the right is set up in the string. Active Figure 13.26 consists of views of the wave at intervals of one-quarter of a period. Note that each small segment of the string, such as P, oscillates vertically in the y-direction with simple harmonic motion. This must be the case, because each segment follows the simple harmonic motion of the blade. Every segment of the string can therefore be treated as a simple harmonic oscillator vibrating with the same frequency as the blade that drives the string. The frequencies of the waves studied in this course will range from rather low values for waves on strings and waves on water, to values for sound waves between 20 Hz and 20 000 Hz (recall that 1 Hz 1 s 1), to much higher frequencies for electromagnetic waves. These waves have different physical sources, but can be described with the same concepts. The horizontal dashed line in Active Figure 13.26 represents the position of the string when no wave is present. The maximum distance the string moves above or below this equilibrium value is called the amplitude A of the wave. For the waves we work with, the amplitudes at the crest and the trough will be identical. Active Figure 13.26b illustrates another characteristic of a wave. The horizontal arrows show the distance between two successive points that behave identically. This distance is called the wavelength (the Greek letter lambda). We can use these denitions to derive an expression for the speed of a wave. We start with the dening equation for the wave speed v: x v t The wave speed is the speed at which a particular part of the wave say, a crest moves through the medium. A wave advances a distance of one wavelength in a time interval equal to one period of the vibration. Taking x and t T, we see that v
T Because the frequency is the reciprocal of the period, we have
Wave speed
v
f
[13.17]
This important general equation applies to many different types of waves, such as sound waves and electromagnetic waves.
ACTIVE FIGURE 13.26 One method for producing traveling waves on a continuous string. The left end of the string is connected to a blade that is set vibrating. Every part of the string, such as point P, oscillates vertically with simple harmonic motion. (a) Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.26 to adjust the frequency of the blade.
l y A P Vibrating blade (b) P
P
P (c) (d)
13.8
Frequency, Amplitude, and Wavelength
445
EXAMPLE 13.8 A Traveling Wave
Goal Obtain information about a wave directly from its graph. Problem A wave traveling in the positive x-direction is pictured in Figure 13.27a. Find the amplitude, wavelength, speed, and period of the wave if it has a frequency of 8.00 Hz. In Figure 13.27a, x 40.0 cm and y 15.0 cm.
y(cm) x y x(cm) y(cm) x y x(cm)
(a) Figure 13.27
(b)
Strategy The amplitude and wavelength can be read directly from the gure: The maximum vertical displacement is the amplitude, and the distance from one crest to the next is the wavelength. Multiplying the wavelength by the frequency gives the speed, while the period is just the reciprocal of the frequency. Solution The maximum wave displacement is the amplitude A: The distance from crest to crest is the wavelength: Multiply the wavelength by the frequency to get the speed of the wave. v f
(a) (Example 13.8) (b) (Exercise 13.8)
A
y x
15.0 cm 40.0 cm
0.150 m 0.400 m 3.20 m/s
(8.00 Hz)(0.400 m)
Take the reciprocal of the frequency to get the period:
T
1 f
1 s 8.00
0.125 s
Exercise 13.8 A wave traveling in the positive x-direction is pictured in Figure 13.27b. Find the amplitude, wavelength, speed, and period of the wave if it has a frequency of 15.0 Hz. In the gure, x 72.0 cm and y 25.0 cm. Answers A 0.25 m, 0.720 m, v 10.8 m/s, T 0.066 7 s
EXAMPLE 13.9 Sound and Light
Goal Perform elementary calculations using speed, wavelength, and frequency. Problem A wave has a wavelength of 3.00 m. Calculate the frequency of the wave if it is (a) a sound wave and (b) a light wave. Take the speed of sound as 343 m/s and the speed of light as 3.00 108 m/s. Solution (a) Find the frequency of a sound wave with
3.00 m. f v 343 m/s 3.00 m 114 Hz (1)
Solve Equation 3.17 for the frequency and substitute: (b) Find the frequency of a light wave with 3.00 m.
Substitute into Equation (1), using the speed of light for c:
f
c
3.00
108 m/s 3.00 m
1.00
108 Hz
Remark The same equation can be used to nd the frequency in each case, despite the great difference between the physical phenomena. Notice how much larger frequencies of light waves are than frequencies of sound waves.
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Exercise 13.9 (a) Find the wavelength of an electromagnetic wave with frequency 9.00 GHz 9.00 109 Hz (G giga 109), which is in the microwave range. (b) Find the speed of a sound wave in an unknown uid medium if a frequency of 567 Hz has a wavelength of 2.50 m. Answers (a) 0.0333 m (b) 1.42 103 m/s
13.9 THE SPEED OF WAVES ON STRINGS
In this section we focus our attention on the speed of a transverse wave on a stretched string. For a vibrating string, there are two speeds to consider. One is the speed of the physical string that vibrates up and down, transverse to the string, in the y-direction. The other is the wave speed, which is the rate at which the disturbance propagates along the length of the string in the x-direction. We wish to nd an expression for the wave speed. If a horizontal string under tension is pulled vertically and released, it starts at its maximum displacement, y A, and takes a certain amount of time to go to y A and back to A again. This amount of time is the period of the wave, and is the same as the time needed for the wave to advance horizontally by one wavelength. Dividing the wavelength by the period of one transverse oscillation gives the wave speed. For a xed wavelength, a string under greater tension F has a greater wave speed because the period of vibration is shorter, and the wave advances one wavelength during one period. It also makes sense that a string with greater mass per unit length, , vibrates more slowly, leading to a longer period and a slower wave speed. The wave speed is given by
v
F
[13.18]
A P P L I C AT I O N Bass Guitar Strings
where F is the tension in the string and is the mass of the string per unit length, called the linear density. From Equation 13.18, its clear that a larger tension F results in a larger wave speed, while a larger linear density gives a slower wave speed, as expected. According to Equation 13.18, the propagation speed of a mechanical wave, such as a wave on a string, depends only on the properties of the string through which the disturbance travels. It doesnt depend on the amplitude of the vibration. This turns out to be generally true of waves in various media. A proof of Equation 13.18 requires calculus, but dimensional analysis can easily verify that the expression is dimensionally correct. The dimensions of F are ML/T 2, and the dimensions of are M/L. The dimensions of F/ are therefore L2/T 2, so those of F/ are L/T, giving the dimensions of speed. No other combination of F and is dimensionally correct, so in the case where the tension and mass density are the only relevant physical factors, we have veried Equation 13.18 up to an overall constant. According to Equation 13.18, we can increase the speed of a wave on a stretched string by increasing the tension in the string. Increasing the mass per unit length, on the other hand, decreases the wave speed. These physical facts lie behind the metallic windings on the bass strings of pianos and guitars. The windings increase the mass per unit length, , decreasing the wave speed and hence the frequency, resulting in a lower tone. Tuning a string to a desired frequency is a simple matter of changing the tension in the string.
13.10
Interference of Waves
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INTERACTIVE EXAMPLE 13.10 A Pulse Traveling on a String
Goal Calculate the speed of a wave on a string.
5.00 m
Problem A uniform string has a mass M of 0.030 0 kg and a length L of 6.00 m. Tension is maintained in the string by suspending a block of mass m 2.00 kg from one end (Fig. 13.28). (a) Find the speed of a transverse wave pulse on this string. (b) Find the time it takes the pulse to travel from the wall to the pulley. Strategy The tension F can be obtained from Newtons second law for equilibrium applied to the block, and the mass per unit length of the string is M/L. With these quantities, the speed of the transverse pulse can be found by substitution into Equation 13.18. Part (b) requires the formula d vt. Solution (a) Find the speed of the wave pulse. Apply the second law to the block: the tension F is equal and opposite to the force of gravity. Substitute expressions for F and into Equation 13.18: v F F mg 0:F mg
1.00 m
2.00 kg Figure 13.28 (Interactive Example 13.10) The tension F in the string is maintained by the suspended block. The wave speed is given by the expression v F/ .
F 62.6 m/s d v
mg M/L
(2.00 kg)(9.80 m/s2) (0.030 0 kg)/(6.00 m)
19.6 N 0.005 00 kg/m
(b) Find the time it takes the pulse to travel from the wall to the pulley. Solve the distance formula for time: t 5.00 m 62.6 m/s 0.0799 s
Exercise 13.10 To what tension must a string with mass 0.010 0 kg and length 2.50 m be tightened so that waves will travel on it at a speed of 125 m/s? Answer 62.5 N Investigate the transmission of such pulses by logging into PhysicsNow at www.cp7e.com and going to Interactive Example 13.10.
13.10 INTERFERENCE OF WAVES
Many interesting wave phenomena in nature require two or more waves passing through the same region of space at the same time. Two traveling waves can meet and pass through each other without being destroyed or even altered. For instance, when two pebbles are thrown into a pond, the expanding circular waves dont destroy each other. In fact, the ripples pass through each other. Likewise, when sound waves from two sources move through air, they pass through each other. In the region of overlap, the resultant wave is found by adding the displacements of the individual waves. For such analyses, the superposition principle applies: When two or more traveling waves encounter each other while moving through a medium, the resultant wave is found by adding together the displacements of the individual waves point by point.
Superposition principle
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(a)
(b)
(c) Figure 13.29 Constructive interference. If two waves having the same frequency and amplitude are in phase, as in (a) and (b), the resultant wave when they combine (c) has the same frequency as the individual waves, but twice their amplitude.
Experiments show that the superposition principle is valid only when the individual waves have small amplitudes of displacement an assumption we make in all our examples. Figures 13.29a and 13.29b show two waves of the same amplitude and frequency. If at some instant of time these two waves were traveling through the same region of space, the resultant wave at that instant would have a shape like that shown in Figure 13.29c. For example, suppose the waves are water waves of amplitude 1 m. At the instant they overlap so that crest meets crest and trough meets trough, the resultant wave has an amplitude of 2 m. Waves coming together like this are said to be in phase and to exhibit constructive interference. Figures 13.30a and 13.30b show two similar waves. In this case, however, the crest of one coincides with the trough of the other, so one wave is inverted relative to the other. The resultant wave, shown in Figure 13.30c, is seen to be a state of complete cancellation. If these were water waves coming together, one of the waves would exert an upward force on an individual drop of water at the same instant the other wave was exerting a downward force. The result would be no motion of the water at all. In such a situation, the two waves are said to be 180 out of phase and to exhibit destructive interference. Figure 13.31 illustrates the interference of water waves produced by drops of water falling into a pond. Active Figure 13.32 shows constructive interference in two pulses moving toward each other along a stretched string; Active Figure 13.33 shows destructive interference in two pulses. Notice in both gures that when the two pulses separate, their shapes are unchanged, as if they had never met!
(a)
13.11 REFLECTION OF WAVES
In our discussion so far, we have assumed that waves could travel indenitely without striking anything. Often, such conditions are not realized in practice. Whenever a traveling wave reaches a boundary, part or all of the wave is reected. For example, consider a pulse traveling on a string that is xed at one end (Active Fig. 13.34). When the pulse reaches the wall, it is reected. Note that the reected pulse is inverted. This can be explained as follows: When the pulse meets the wall, the string exerts an upward force on the wall. According to Newtons third law, the wall must exert an equal and opposite (downward) reaction force on the string. This downward force causes the pulse to invert on reection. Now suppose the pulse arrives at the strings end, and the end is attached to a ring of negligible mass that is free to slide along the post without friction (Active Fig. 13.35). Again the pulse is reected, but this time it is not inverted. On reaching
(b)
(c) Figure 13.30 Destructive interference. When two waves with the same frequency and amplitude are 180 out of phase, as in (a) and (b), the result when they combine (c) is complete cancellation.
(a)
(d)
(b)
Martin Dohrn/SPL/Photo Researchers, Inc.
(e)
(c) ACTIVE FIGURE 13.32 Two wave pulses traveling on a stretched string in opposite directions pass through each other. When the pulses overlap, as in (b), (c), and (d), the net displacement of the string equals the sum of the displacements produced by each pulse.
Figure 13.31 Interference patterns produced by outward-spreading waves from many drops of liquid falling into a body of water.
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.32 to choose the amplitude and orientation of each of the pulses and study the interference between them as they pass each other.
Summary
449
Incident pulse (a) (d) (a)
(b)
(e)
(b)
(c) (c) ACTIVE FIGURE 13.33 Two wave pulses traveling in opposite directions with displacements that are inverted relative to each other. When the two overlap, as in (c), their displacements subtract from each other. (d)
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.33 to choose the amplitude and orientation of each of the pulses and study the interference between them as they pass each other.
(e)
the post, the pulse exerts a force on the ring, causing it to accelerate upward. The ring is then returned to its original position by the downward component of the tension in the string. An alternate method of showing that a pulse is reected without inversion when it strikes a free end of a string is to send the pulse down a string hanging vertically. When the pulse hits the free end, its reected without inversion, just as is the pulse in Active Figure 13.35. Finally, when a pulse reaches a boundary, its partly reected and partly transmitted past the boundary into the new medium. This effect is easy to observe in the case of two ropes of different density joined at some boundary.
Incident pulse (a) (c) Reflected pulse (b) (d)
Reflected pulse
ACTIVE FIGURE 13.34 The reection of a traveling wave at the xed end of a stretched string. Note that the reected pulse is inverted, but its shape remains the same.
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.34 to adjust the linear mass density of the string and the transverse direction of the initial pulse. ACTIVE FIGURE 13.35 The reection of a traveling wave at the free end of a stretched string. In this case, the reected pulse is not inverted.
Log into PhysicsNow at www.cp7e.com, and go to Active Figure 13.35 to adjust the linear mass density of the string and the transverse direction of the initial pulse.
SUMMARY
Take a practice test by logging into PhysicsNow at www.cp7e.com and clicking on the Pre-Test link for this chapter. When an object moves with simple harmonic motion, its acceleration as a function of position is a k x m [13.2]
13.1
Hookes Law
Simple harmonic motion occurs when the net force on an object along the direction of motion is proportional to the objects displacement and in the opposite direction: Fs kx [13.1]
13.2 Elastic Potential Energy
The energy stored in a stretched or compressed spring or in some other elastic material is called elastic potential energy: PEs
1 2 2 kx
[13.3]
This is called Hookes law. The time required for one complete vibration is called the period of the motion. The reciprocal of the period is the frequency of the motion, which is the number of oscillations per second.
The velocity of an object as a function of position, when the object is moving with simple harmonic motion, is v
k (A2 m
x 2)
[13.6]
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13.4 Position, Velocity, and Acceleration as a Function of Time
The period of an object of mass m moving with simple harmonic motion while attached to a spring of spring constant k is T 2
In a longitudinal wave the elements of the medium move parallel to the direction of the wave velocity. An example is a sound wave.
m k
13.8 Frequency, Amplitude, and Wavelength
The relationship of the speed, wavelength, and frequency of a wave is v f [13.17]
[13.8]
where T is independent of the amplitude A. The frequency of an object spring system is f The angular frequency of the system in rad/s is 2f
1/T.
k m
This relationship holds for a wide variety of different waves. [13.11]
13.9 The Speed of Waves on Strings
The speed of a wave traveling on a stretched string of mass per unit length and under tension F is v
When an object is moving with simple harmonic motion, the position, velocity, and acceleration of the object as a function of time are given by x v a A cos(2 ft) A sin(2 ft) A
2
[13.14a] [13.14b] [13.14c]
F
[13.18]
cos(2 ft)
13.10 Interference of Waves
The superposition principle states that if two or more traveling waves are moving through a medium, the resultant wave is found by adding the individual waves together point by point. When waves meet crest to crest and trough to trough, they undergo constructive interference. When crest meets trough, the waves undergo destructive interference.
13.5
Motion of a Pendulum
A simple pendulum of length L moves with simple harmonic motion for small angular displacements from the vertical, with a period of T 2
L g
[13.15]
13.7
Waves
13.11 Reection of Waves
When a wave pulse reects from a rigid boundary, the pulse is inverted. When the boundary is free, the reected pulse is not inverted.
In a transverse wave the elements of the medium move in a direction perpendicular to the direction of the wave. An example is a wave on a stretched string.
CONCEPTUAL QUESTIONS
1. If one end of a heavy rope is attached to one end of a light rope, the speed of a wave will change as the wave goes from the heavy rope to the light one. Will the speed increase or decrease? What happens to the frequency? To the wavelength? 2. If a spring is cut in half, what happens to its spring constant? 3. An object spring system undergoes simple harmonic motion with an amplitude A. Does the total energy change if the mass is doubled but the amplitude isnt changed? Are the kinetic and potential energies at a given point in its motion affected by the change in mass? Explain. 4. The speed of sound in air as given in this chapter (343 m/s) is an enormous speed compared to the speed of common objects. Yet the speed of sound is of the same order of magnitude as the rms speed of air molecules at 1 atmosphere and 20 C as given by the kinetic theory of gases. Is this just a remarkable coincidence? Explain. 5. An object is hung on a spring, and the frequency of oscillation of the system, f, is measured. The object, a second identical object, and the spring are carried to space in the Space Shuttle. The two objects are attached to the ends of the spring, and the system is taken out into space on a space walk. The spring is extended, and the system is released to oscillate while oating in space. The coils of the spring dont bump into one another. What is the frequency of oscillation for this system, in terms of f ? 6. If an object spring system is hung vertically and set into oscillation, why does the motion eventually stop? 7. Is a bouncing ball an example of simple harmonic motion? Is the daily movement of a student from home to school and back simple harmonic motion? 8. If a pendulum clock keeps perfect time at the base of a mountain, will it also keep perfect time when it is moved to the top of the mountain? Explain. 9. A pendulum bob is made from a sphere lled with water. What would happen to the frequency of vibration of this pendulum if the sphere had a hole in it that allowed the water to leak out slowly? 10. If a grandfather clock were running slow, how could we adjust the length of the pendulum to correct the time? 11. A grandfather clock depends on the period of a pendulum to keep correct time. Suppose such a clock is calibrated
Problems
451
correctly and then the temperature of the room in which it resides increases. Does the clock run slow, fast, or correctly? [Hint: A material expands when its temperature increases.] 12. If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of the pulse if you stretch the hose more tightly? What happens to the speed if you ll the hose with water? 13. In a long line of people waiting to buy tickets at a movie theater, when the rst person leaves, a pulse of motion occurs as people step forward to ll in the gap. The gap moves through the line of people. What determines the speed of this pulse? Is it transverse or longitudinal? How about the wave at a baseball game, where people in the stands stand up and shout as the wave arrives at their location (Fig. Q13.13) and the pulse moves around the stadium what determines the speed of this pulse? Is it transverse or longitudinal?
14. As part of a physics open house, a department sets up a bungee jump from the top of the physics building. Assume that one end of the elastic band will be rmly attached to the top of the building and the other to the waist of a courageous participant. The participant will step off the edge of the building, to be slowed down and brought back up by the elastic band before hitting the ground. Estimate the length and spring constant of the elastic you would recommend. (Question 14 is courtesy of Edward F. Redish. For more questions of this type, see http://www.physics.umd.edu/perg/.) 15. In mechanics, massless strings are often assumed. Why is this not a good assumption when discussing waves on strings? 16. What happens to the wavelength of a wave on a string when the frequency is doubled? Assume that the tension in the string remains the same. 17. Explain why the kinetic and potential energies of an object spring system can never be negative. 18. What happens to the speed of a wave on a string when the frequency is doubled? Assume that the tension in the string remains the same. 19. By what factor would you have to multiply the tension in a stretched spring in order to double the wave speed? 20. The left end of a spring is attached to a wall, and its right end is attached to a cart lying on a frictionless horizontal surface. An experimenter pulls the cart away from the wall and holds it there. (a) What forces are acting on the spring? What is the total force on the spring? (b) What forces are acting on the cart? What is the total force on the cart? (c) If the cart is released, describe its motion.
Gregg Adams/Stone/Getty Images
Figure Q13.13
PROBLEMS
1, 2, 3 = straightforward, intermediate, challenging = full solution available in Student Solutions Manual/Study Guide = coached problem with hints available at www.cp7e.com = biomedical application Section 13.1 Hookes Law 1. A 0.40-kg object is attached to a spring with force constant 160 N/m so that the object is allowed to move on a horizontal frictionless surface. The object is released from rest when the spring is compressed 0.15 m. Find (a) the force on the object and (b) its acceleration at that instant. 2. A load of 50 N attached to a spring hanging vertically stretches the spring 5.0 cm. The spring is now placed horizontally on a table and stretched 11 cm. (a) What force is required to stretch the spring by that amount? (b) Plot a graph of force (on the y-axis) versus spring displacement from the equilibrium position along the x-axis. 3. A ball dropped from a height of 4.00 m makes a perfectly elastic collision with the ground. Assuming that no mechanical energy is lost due to air resistance, (a) show that the motion is periodic and (b) determine the period of the motion. (c) Is the motion simple harmonic? Explain. 4. A small ball is set in horizontal motion by rolling it with a speed of 3.00 m/s across a room 12.0 m long between two walls. Assume that the collisions made with each wall are perfectly elastic and that the motion is perpendicular to the two walls. (a) Show that the motion is periodic and determine its period. (b) Is the motion simple harmonic? Explain. 5. A spring is hung from a ceiling, and an object attached to its lower end stretches the spring by a distance of 5.00 cm from its unstretched position when the system is in equilibrium. If the spring constant is 47.5 N/m, determine the mass of the object. 6. An archer must exert a force of 375 N on the bowstring shown in Figure P13.6a (page 452) such that the string makes an angle of 35.0 with the vertical. (a) Determine the tension in the bowstring. (b) If the applied force is replaced by a stretched spring as in Figure P13.6b, and the spring is stretched 30.0 cm from its unstretched length, what is the spring constant?
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k
(b) Answer part (a) if the coefcient of kinetic friction between block and tabletop is 0.200.
20.0 N
Figure P13.12 (a) Figure P13.6 (b)
Section 13.2 Elastic Potential Energy 7. A slingshot consists of a light leather cup containing a stone. The cup is pulled back against two parallel rubber bands. It takes a force of 15 N to stretch either one of these bands 1.0 cm. (a) What is the potential energy stored in the two bands together when a 50-g stone is placed in the cup and pulled back 0.20 m from the equilibrium position? (b) With what speed does the stone leave the slingshot? 8. An archer pulls her bowstring back 0.400 m by exerting a force that increases uniformly from zero to 230 N. (a) What is the equivalent spring constant of the bow? (b) How much work is done in pulling the bow? 9. A childs toy consists of a piece of plastic attached to a spring (Fig. P13.9). The spring is compressed against the oor a distance of 2.00 cm, and the toy is released. If the toy has a mass of 100 g and rises to a maximum height of 60.0 cm, estimate the force constant of the spring.
13. A 10.0-g bullet is red into, and embeds itself in, a 2.00-kg block attached to a spring with a force constant of 19.6 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block and the block slides on a frictionless surface? [Note: You must use conservation of momentum in this problem. Why?] 14. A 1.5-kg block is attached to a spring with a spring constant of 2 000 N/m. The spring is then stretched a distance of 0.30 cm and the block is released from rest. (a) Calculate the speed of the block as it passes through the equilibrium position if no friction is present. (b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 2.0 N retards its motion. (c) What would be the strength of the frictional force if the block reached the equilibrium position the rst time with zero velocity? Section 13.3 Comparing Simple Harmonic Motion with Uniform Circular Motion Section 13.4 Position, Velocity, and Acceleration as a Function of Time 15. A 0.40-kg object connected to a light spring with a force constant of 19.6 N/m oscillates on a frictionless horizontal surface. If the spring is compressed 4.0 cm and released from rest, determine (a) the maximum speed of the object, (b) the speed of the object when the spring is compressed 1.5 cm, and (c) the speed of the object when the spring is stretched 1.5 cm. (d) For what value of x does the speed equal one-half the maximum speed? 16. An object spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and the object has a mass of 0.50 kg, determine (a) the mechanical energy of the system, (b) the maximum speed of the object, and (c) the maximum acceleration of the object. 17. At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 20.0 cm on a spring with a force constant of 16.0 N/m. It is observed that the maximum speed of the bunch of bananas is 40.0 cm/s. What is the weight of the bananas in newtons? 18. A 50.0-g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 25.0 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless? 19. While riding behind a car traveling at 3.00 m/s, you notice that one of the cars tires has a small hemispherical bump on its rim, as in Figure P13.19. (a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion. (b) If the radius of the cars tires is 0.30 m, what is the bumps period of oscillation?
Figure P13.9
10. An automobile having a mass of 1 000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with constant 5.00 106 N/m and is compressed 3.16 cm as the car is brought to rest. What was the speed of the car before impact, assuming that no energy is lost in the collision with the wall? 11. A simple harmonic oscillator has a total energy E. (a) Determine the kinetic and potential energies when the displacement is one-half the amplitude. (b) For what value of the displacement does the kinetic energy equal the potential energy? 12. A 1.50-kg block at rest on a tabletop is attached to a horizontal spring having constant 19.6 N/m, as in Figure P13.12. The spring is initially unstretched. A constant 20.0-N horizontal force is applied to the object, causing the spring to stretch. (a) Determine the speed of the block after it has moved 0.300 m from equilibrium if the surface between the block and tabletop is frictionless.
Problems
453
Bump
Figure P13.19
20. An object moves uniformly around a circular path of radius 20.0 cm, making one complete revolution every 2.00 s. What are (a) the translational speed of the object, (b) the frequency of motion in hertz, and (c) the angular speed of the object? 21. Consider the simplied single-piston engine in Figure P13.21. If the wheel rotates at a constant angular speed , explain why the piston rod oscillates in simple harmonic motion.
v Piston A x= A x (t )
5.00 N/m. The object is displaced 3.00 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released? (b) How many times does the object oscillate in 3.50 s? 28. A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is pulled horizontally so that it stretches the spring 5.00 cm and is then released from rest at t 0. (a) What is the force constant of the spring? (b) What are the angular frequency , the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (e) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t 0.500 s. Given that x A cos ( t) is a sinusoidal 29. function of time, show that v (velocity) and a (acceleration) are also sinusoidal functions of time. [Hint: Use Equations 13.6 and 13.2.] Section 13.5 Motion of a Pendulum 30. A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the oor and that its period is 15.5 s. (a) How tall is the tower? (b) If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s2, what is the period there? 31. A simple 2.00-m-long pendulum oscillates at a location where g 9.80 m/s2. How many complete oscillations does it make in 5.00 min? 32. An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0 C. (a) When placed in a room at a temperature of 5.0 C, will it gain time or lose time? (b) How much time will it gain or lose every hour? [Hint: See Chapter 10.] 33. A pendulum clock that works perfectly on Earth is taken to the Moon. (a) Does it run fast or slow there? (b) If the clock is started at 12:00 midnight, what will it read after one Earth day (24.0 h)? Assume that the free-fall acceleration on the Moon is 1.63 m/s2. 34. A simple pendulum is 5.00 m long. (a) What is the period of simple harmonic motion for this pendulum if it is located in an elevator accelerating upward at 5.00 m/s2? (b) What is its period if the elevator is accelerating downward at 5.00 m/s2? (c) What is the period of simple harmonic motion for the pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s2? 35. The free-fall acceleration on Mars is 3.7 m/s2. (a) What length of pendulum has a period of 1 s on Earth? What length of pendulum would have a 1-s period on Mars? (b) An object is suspended from a spring with force constant 10 N/m. Find the mass suspended from this spring that would result in a period of 1 s on Earth and on Mars. Section 13.6 Damped Oscillations Section 13.7 Waves Section 13.8 Frequency, Amplitude, and Wavelength 36. A cork on the surface of a pond bobs up and down two times per second on ripples having a wavelength of
Figure P13.21
22. The frequency of vibration of an object spring system is 5.00 Hz when a 4.00-g mass is attached to the spring. What is the force constant of the spring? 23. A spring stretches 3.9 cm when a 10-g object is hung from it. The object is replaced with a block of mass 25 g that oscillates in simple harmonic motion. Calculate the period of motion. 24. When four people with a combined mass of 320 kg sit down in a car, they nd that the car drops 0.80 cm lower on its springs. Then they get out of the car and bounce it up and down. What is the frequency of the cars vibration if its mass (when it is empty) is 2.0 103 kg? 25. A cart of mass 250 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.5 N/m is attached between the cart and the left end of the track. When in equilibrium, the cart is located 12 cm from the left end of the track. If the cart is displaced 4.5 cm from its equilibrium position, nd (a) the period at which it oscillates, (b) its maximum speed, and (c) its speed when it is 14 cm from the left end of the track. 26. The motion of an object is described by the equation x (0.30 m) cos t 3
Find (a) the position of the object at t 0 and t 0.60 s, (b) the amplitude of the motion, (c) the frequency of the motion, and (d) the period of the motion. 27. A 2.00-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is
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8.50 cm. If the cork is 10.0 m from shore, how long does it take a ripple passing the cork to reach the shore? 37. A wave traveling in the positive x-direction has a frequency of 25.0 Hz, as in Figure P13.37. Find the (a) amplitude, (b) wavelength, (c) period, and (d) speed of the wave.
18.0 cm
10.0 cm Figure P13.37
A simple pendulum consists of a ball of mass 5.00 kg hanging from a uniform string of mass 0.0600 kg and length L. If the period of oscillation of the pendulum is 2.00 s, determine the speed of a transverse wave in the string when the pendulum hangs vertically. 48. A string is 50.0 cm long and has a mass of 3.00 g. A wave travels at 5.00 m/s along this string. A second string has the same length, but half the mass of the rst. If the two strings are under the same tension, what is the speed of a wave along the second string? 49. Tension is maintained in a string as in Figure P13.49. The observed wave speed is 24 m/s when the suspended mass is 3.0 kg. (a) What is the mass per unit length of the string? (b) What is the wave speed when the suspended mass is 2.0 kg? 47.
38. A bat can detect small objects, such as an insect, whose size is approximately equal to one wavelength of the sound the bat makes. If bats emit a chirp at a frequency of 60.0 kHz, and if the speed of sound in air is 340 m/s, what is the smallest insect a bat can detect? 39. If the frequency of oscillation of the wave emitted by an FM radio station is 88.0 MHz, determine (a) the waves period of vibration and (b) its wavelength. (Radio waves travel at the speed of light, 3.00 108 m/s.) 40. The distance between two successive maxima of a transverse wave is 1.20 m. Eight crests, or maxima, pass a given point along the direction of travel every 12.0 s. Calculate the wave speed. 41. A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 40.0 vibrations in 30.0 s. Also, a given maximum travels 425 cm along the rope in 10.0 s. What is the wavelength? 42. Ocean waves are traveling to the east at 4.0 m/s with a distance of 20 m between crests. With what frequency do the waves hit the front of a boat (a) when the boat is at anchor and (b) when the boat is moving westward at 1.0 m/s? Section 13.9 The Speed of Waves on Strings 43. A phone cord is 4.00 m long and has a mass of 0.200 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.800 s. What is the tension in the cord? 44. A circus performer stretches a tightrope between two towers. He strikes one end of the rope and sends a wave along it toward the other tower. He notes that it takes the wave 0.800 s to reach the opposite tower, 20.0 m away. If a 1-m length of the rope has a mass of 0.350 kg, nd the tension in the tightrope. 45. Transverse waves with a speed of 50.0 m/s are to be produced on a stretched string. A 5.00-m length of string with a total mass of 0.060 0 kg is used. (a) What is the required tension in the string? (b) Calculate the wave speed in the string if the tension is 8.00 N. 46. An astronaut on the Moon wishes to measure the local value of g by timing pulses traveling down a wire that has a large object suspended from it. Assume a wire of mass 4.00 g is 1.60 m long and has a 3.00-kg object suspended from it. A pulse requires 36.1 ms to traverse the length of the wire. Calculate g Moon from these data. (You may neglect the mass of the wire when calculating the tension in it.)
3.0 kg Figure P13.49
50. The elastic limit of a piece of steel wire is 2.70 109 Pa. What is the maximum speed at which transverse wave pulses can propagate along the wire without exceeding its elastic limit? (The density of steel is 7.86 103 kg/m3.) 51. Transverse waves travel at 20.0 m/s on a string that is under a tension of 6.00 N. What tension is required for a wave speed of 30.0 m/s in the string? Section 13.10 Interference of Waves Section 13.11 Reection of Waves 52. A series of pulses of amplitude 0.15 m is sent down a string that is attached to a post at one end. The pulses are reected at the post and travel back along the string without loss of amplitude. What is the amplitude at a point on the string where two pulses are crossing (a) if the string is rigidly attached to the post? (b) if the end at which reection occurs is free to slide up and down? 53. A wave of amplitude 0.30 m interferes with a second wave of amplitude 0.20 m traveling in the same direction. What are (a) the largest and (b) the smallest resultant amplitudes that can occur, and under what conditions will these maxima and minima arise? ADDITIONAL PROBLEMS 54. The position of a 0.30-kg object attached to a spring is described by x (0.25 m) cos(0.4 t)
Find (a) the amplitude of the motion, (b) the spring constant, (c) the position of the object at t 0.30 s, and (d) the objects speed at t 0.30 s. 55. A large block P executes horizontal simple harmonic motion as it slides across a frictionless surface with a
Problems
455
frequency f 1.50 Hz. Block B rests on it, as shown in Figure P13.55, and the coefcient of static friction between the two is s 0.600. What maximum amplitude of oscillation can the system have if block B is not to slip?
400 m/s
ms B P 5.00 cm v
Figure P13.55
56. A 500-g block is released from rest and slides down a frictionless track that begins 2.00 m above the horizontal, as shown in Figure P13.56. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a spring constant of 20.0 N/m. Find the maximum distance the spring is compressed.
500 g
Figure P13.58
2.00 m
k
59. A 25-kg block is connected to a 30-kg block by a light string that passes over a frictionless pulley. The 30-kg block is connected to a light spring of force constant 200 N/m, as in Figure P13.59. The spring is unstretched when the system is as shown in the gure, and the incline is smooth. The 25-kg block is pulled 20 cm down the incline (so that the 30-kg block is 40 cm above the oor) and is released from rest. Find the speed of each block when the 30-kg block is 20 cm above the oor (that is, when the spring is unstretched).
Figure P13.56
57. A 3.00-kg object is fastened to a light spring, with the intervening cord passing over a pulley (Fig. P13.57). The pulley is frictionless, and its inertia may be neglected. The object is released from rest when the spring is unstretched. If the object drops 10.0 cm before stopping, nd (a) the spring constant of the spring and (b) the speed of the object when it is 5.00 cm below its starting point.
25 kg 30 kg 40 20 cm
Figure P13.59
3.00 kg k
Figure P13.57
58. A 5.00-g bullet moving with an initial speed of 400 m/s is red into and passes through a 1.00-kg block, as in Figure P13.58. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 900 N/m. If the block moves 5.00 cm to the right after impact, nd (a) the speed at which the bullet emerges from the block and (b) the mechanical energy lost in the collision.
60. A spring in a toy gun has a spring constant of 9.80 N/m and can be compressed 20.0 cm beyond the equilibrium position. A 1.00-g pellet resting against the spring is propelled forward when the spring is released. (a) Find the muzzle speed of the pellet. (b) If the pellet is red horizontally from a height of 1.00 m above the oor, what is its range? 61. A 2.00-kg block hangs without vibrating at the end of a spring (k 500 N/m) that is attached to the ceiling of an elevator car. The car is rising with an upward acceleration of g/3 when the acceleration suddenly ceases (at t 0). (a) What is the angular frequency of oscillation of the block after the acceleration ceases? (b) By what amount is the spring stretched during the time that the elevator car is accelerating? This distance will be the amplitude of the ensuing oscillation of the block. 62. An object of mass m is connected to two rubber bands of length L, each under tension F, as in Figure P13.62. The object is displaced vertically by a small distance y. Assuming the tension does not change, show that (a) the restoring force is (2F/L)y and (b) the system exhibits simple harmonic motion with an angular frequency 2F/mL .
456
Chapter 13
Vibrations and Waves
y L
Figure P13.62
L
63. A light balloon lled with helium of density 0.180 kg/m3 is tied to a light string of length L 3.00 m. The string is tied to the ground, forming an inverted simple pendulum (Fig. P13.63a). If the balloon is displaced slightly from equilibrium, as in Figure P13.63b, show that the motion is simple harmonic, and determine the period of the motion. Take the density of air to be 1.29 kg/m3. [Hint: Use an analogy with the simple pendulum discussed in the text, and see Chapter 9.]
He Air Air He
L g g
u
L
(a) Figure P13.63
(b)
64. A light string of mass 10.0 g and length L 3.00 m has its ends tied to two walls that are separated by the distance D 2.00 m. Two objects, each of mass M 2.00 kg, are suspended from the string as in Figure P13.64. If a wave pulse is sent from point A, how long does it take to travel to point B?
D L 4 A L 2 B L 4
Assume Earth has a uniform density . Write down Newtons law of gravitation for an object at a distance r from the center of the Earth, and show that the force on it is of the form of Hookes law, F kr, with an effective force constant of k (4) Gm , where G is the gravitational constant. 3 66. A 60.0-kg reghter slides down a pole while a constant frictional force of 300 N retards his motion. A horizontal 20.0-kg platform is supported by a spring at the bottom of the pole to cushion the fall. The reghter starts from rest 5.00 m above the platform, and the spring constant is 2500 N/m. Find (a) the reghters speed just before he collides with the platform and (b) the maximum distance the spring is compressed. Assume that the frictional force acts during the entire motion. [Hint: The collision between the reghter and the platform is perfectly inelastic.] An object of mass m1 9.0 kg is in 67. equilibrium while connected to a light spring of constant k 100 N/m that is fastened to a wall, as in Figure P13.67a. A second object, of mass m 2 7.0 kg, is slowly pushed up against m1, compressing the spring by the amount A 0.20 m, as shown in Figure P13.67b. The system is then released, causing both objects to start moving to the right on the frictionless surface. (a) When m1 reaches the equilibrium point, m 2 loses contact with m1 (Fig. P13.67c) and moves to the right with velocity :. Dev termine the magnitude of :. (b) How far apart are the v objects when the spring is fully stretched for the rst time (Fig. P13.67d)? [Hint: First determine the period of oscillation and the amplitude of the m1spring system after m 2 loses contact with m1.]
m1
k (a)
(b)
k
m1 m 2
A
v
M Figure P13.64
M (c) k
m1 m 2
65. Assume that a hole is drilled through the center of the Earth. It can be shown that an object of mass m at a distance r from the center of the Earth is pulled toward the center only by the material in the shaded portion of Figure P13.65.
Earth
(d)
k
m1
v m2
D m r Figure P13.67
Figure P13.65
68. An 8.00-kg block travels on a rough horizontal surface and collides with a spring. The speed of the block just before the collision is 4.00 m/s. As it rebounds to the left with the spring uncompressed, the block travels at 3.00 m/s. If the coefcient of kinetic friction between the block and the surface is 0.400, determine (a) the loss in
Problems
457
mechanical energy due to friction while the block is in contact with the spring and (b) the maximum distance the spring is compressed. 69. Two points, A and B, on Earth are at the same longitude and 60.0 apart in latitude. An earthquake at point A sends two waves toward B. A transverse wave travels along the surface of Earth at 4.50 km/s, and a longitudinal wave travels through Earth at 7.80 km/s. (a) Which wave arrives at B rst? (b) What is the time difference between the arrivals of the two waves at B? Take the radius of Earth to be 6.37 106 m. 70. Figure P13.70 shows a crude model of an insect wing. The mass m represents the entire mass of the wing, which pivots about the fulcrum F. The spring represents the surrounding connective tissue. Motion of the wing corresponds to vibration of the spring. Suppose the mass of the wing is 0.30 g and the effective spring constant of the tissue is 4.7 10 4 N/m. If the mass m moves up and down a distance of 2.0 mm from its position of equilibrium, what is the maximum speed of the outer tip of the wing?
3.00 mm m 1.50 cm
ACTIVITIES A.1. Construct a simple pendulum by tying a metal bolt to one end of a string and taping the other end to the top of a doorframe. Adjust the length of the pendulum to be about 1.0 m, and measure it precisely. Obtain the period by timing 25 complete oscillations, making sure the string always makes small angles with the vertical. Repeat the measurements for precisely measured pendulum lengths ranging from 0.4 m to 1.6 m in increments of 0.2 m. Plot the square of the period versus the length of the pendulum and measure the slope of the line best tting your data points. Does your slope agree with that predicted by T 2 (4 2/g)L? What value of g do you obtain from your data? While you have your pendulum in position, use a procedure similar to the preceding to verify that the period is also independent of the amplitude for small angles and that the period is also independent of the mass. A.2. Attach one end of a rope (or a spring such as a Slinky ) to a wall, stretch it taut, and use the system to study the following aspects of wave motion: (a) Send a pulse down the rope by striking it sharply from the side. An observer watching from the side can measure the time elapsed during 3 5 trips of the pulse from one end to the other. Dividing the total distance traveled by the elapsed time yields the wave speed. To get reliable results, take a number of readings and average them. (b) Use the same setup as in part (a) to test whether the initial amplitude of the pulse changes the wave speed. (c) Have two people hold opposite ends of the rope (or spring). Then, at the same instant, have both people hit the rope sharply from the side. Observe what happens when the two pulses meet. The superposition lasts only for the short time that the pulses overlap, and you must look carefully to see the effect. (d) Using the same setup, devise a way to check whether pulses traveling toward one another pass through when they collide or reect off each other. (e) Tie one end of a rope to a doorknob and send a pulse down it. Observe what happens when the pulse reects from the door. Does it return on the same side of the rope, or does it invert?
F
Figure P13.70
71. A 1.6-kg block on a horizontal surface is attached to a spring with a force constant of 1.0 103 N/m, as in Active Figure 13.1. The spring is compressed a distance of 2.0 cm, and the block is released from rest. (a) Calculate the speed of the block as it passes through the equilibrium position, x 0, if the surface is frictionless. (b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 4.0 N retards its motion. (c) How far does the block travel before coming to rest in part (b)?

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Today: Vertebrates, Part II (Chapter 15) Next Time: History of Geologic Thought (Review Ch. 11-15)Chapter 15: Vertebrates 2 Next lecture: Chapters 11-15 TetrapodsAmphibians Amniotes Reptiles (Diapsids) Dinosaurs Swimming reptiles Flying reptiles Birds S

University of Texas - CMS - 306M

Click to edit Master subtitle styleOr ganizing a Successful Pr esentation Chapter 119/16/09Why organize a presentation?9/16/09Why organize a presentation? Necessary to communicate ideaseffectively. Keeps audience's interest andattention. Creates

University of Texas - CMS - 306M

Foundations of CommunicationClick to edit Master subtitleOne Chapter style9/16/09Defining CommunicationCommunication: the process whereby oneperson stimulates meaning in the mind of another through verbal and/or nonverbal means (pg. 4-5).9/16/09The