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Unformatted text preview: Review: Yesterday
! Hooke's Force Law: F = !kx
d2x ! Equation of motion: m 2 = !kx dt Physics for Scientists & Engineers 1
Spring Semester 2006 Lecture 46 ! Solution: or: ! Amplitude: Phase: d2x k + x=0 dt 2 m x(t) = Asin(! 0t) + B cos(! 0t) x(t) = C sin(! 0t + " 0 )
C = A2 + B2 ! 0 = arctan(B / A)
k m
2 ! Angular Frequency: ! 0 =
April 14, 2006 Physics for Scientists&Engineers 1 1 April 14, 2006 Physics for Scientists&Engineers 1 Initial Conditions
! How does one determine constants A and B, or C and !0? ! One uses initial conditions to determine these constants Initial position Example: Initial Conditions
Question 1: ! A spring with spring constant 56.0 N/m has a mass of 1.00 kg attached to its end. The mass is pulled +5.5 cm from its equilibrium position and pushed so that it receives an initial velocity of 0.32 m/s. What is the equation of motion for this oscillation? x0 = x(t = 0) Initial velocity v0 = v(t = 0) = (dx / dt) t = 0
! Two equations for two unknowns !
April 14, 2006 Physics for Scientists&Engineers 1 3 April 14, 2006 Physics for Scientists&Engineers 1 4 Example: Initial Conditions (2)
Answer 1: ! Equation of motion
x(t) = Asin(! 0t) + B cos(! 0t) with ! 0 = k/m Example: Initial Conditions (3)
Question 2: ! What is the amplitude of this oscillation? Answer 2: ! Although we have the constants A and B, they are not the amplitude. ! Have to use: ! We can already calculate the angular frequency ! Find equation for velocity by taking the time derivate
x(t) = Asin(! 0t) + B cos(! 0t) " v(t) = ! 0 A cos(! 0t) # ! 0 Bsin(! 0t) ! 0 = k / m = (56.0 N/m)/(1.00 kg) = 7.48 s1 C = A 2 + B 2 = 0.0432 + 0.055 2 m = 0.070 m
! Please note that amplitude (7 cm) is bigger than initial elongation from equilibrium (5.5 cm) ! Initial push added to amplitude, even though it was toward the equilibrium position
5 April 14, 2006 Physics for Scientists&Engineers 1 6 ! At time 0, because sin(0)=0 and cos(0)=1, we have
x0 = x(t = 0) = B " ' B = x0 = 0.055 m #%( v0 = v(t = 0) = ! 0 A $ ) A = v0 / ! 0 = &0.043 m
April 14, 2006 Physics for Scientists&Engineers 1 Biography: Robert Hooke
! Robert Hooke (16351703) ! Contemporary of Newton ! Curator of experiments for Royal Society ! Anticipated universal gravitation ! Finest mechanic of his time ! Invented spiral spring in watches ! Most important publication: Micrographia (1665) ! Coined the term "cell" as smallest parts of living organisms Position, Velocity, Acceleration
! Position: ! Velocity: ! Acceleration: x(t) = C sin(! 0t + " 0 ) v(t) = ! 0C cos(! 0t + " 0 ) 2 a(t) = !" 0 C sin(" 0t + # 0 )
Note: 1. Phase shifts of 90 between x & v, v & a 2. Acceleration always in opposite direction to displacement 3. Ratio of amplitudes (v/x, a/v) is !0 April 14, 2006 Physics for Scientists&Engineers 1 7 April 14, 2006 Physics for Scientists&Engineers 1 8 Period and Frequency
! Sinusoidal functions are periodic ... adding 2" to the argument does not change the value of the function Period and Frequency for Spring
! Now we use our result for the mass on a spring, ! This time interval over which function repeats itself is called period, T 2" sin(! t) = sin(2" + ! t) = sin(! ( + t)) ! !0 =
! Period: k m T=
! Frequency, f: 2! " T=
! Frequency: 2! = "0
f = 2! m = 2! k k/m
k m 1 f = T !0 1 = 2" 2" (So far this is just the same as for circular motion) ! This relationship was used for timekeeping in antique clocks (since R. Hooke, and before quartz)
9 April 14, 2006 Physics for Scientists&Engineers 1 10 April 14, 2006 Physics for Scientists&Engineers 1 Tunnel Through the Moon
! Suppose we would drill a tunnel straight through the center of the moon, from one end to the other ! The fact that the moon is composed of solid rock makes this scenario slightly more realistic than drilling a tunnel through the center of Earth ! Question: ! If we go to one end of this tunnel and release a steel ball of mass 5.0 kg from rest, what can we say about the motion of this ball? ! Answer: ! In our chapter on gravity, we have found that the magnitude of the gravitational force inside a spherical mass distribution of constant density and radius R is Tunnel Through the Moon (2)
! The gravitational force inside a homogeneous spherical mass distribution follows Hooke's Law
F(x) = !kx with a "spring constant" given by k = mg / R ! Here g = gM is the gravitational acceleration experienced at the surface of the moon ! We start by calculating the acceleration of gravity on the surface of the moon MM = 7.35#1022 kg, RM = 1735 km = 1.735#106 m Fg (r < R) = mgr / R
! This force points toward the center of the moon, in the opposite direction from the displacement
April 14, 2006 Physics for Scientists&Engineers 1 11 6.67 !10 "11 m 3kg 1s "2 7.35 !10 22 kg GM M gM = = = 1.63 m/s 2 2 2 RM 1.735 !10 6 m ( )( ) ( ) April 14, 2006 Physics for Scientists&Engineers 1 12 Tunnel Through the Moon (3)
! We have found that the solution to the equation of motion with a spring force is the oscillatory motion x(t) = Asin(! 0t) + B cos(! 0t) ! Releasing it from rest from the surface of the moon implies that A=0 and B=x0. Thus our equation of motion is in this case: x(t) = RM cos(! 0t) ! The angular frequency of the oscillation is: k gM 1.63 m/s 2 !0 = = = = 9.69 "10 #4 Hz m RM 1.735 "10 6 m ! Please note the mass of the steel ball turned out to be irrelevant. The period of this oscillation is 2! T= = 6485 s "0
April 14, 2006 Physics for Scientists&Engineers 1 13 Tunnel Through the Moon (4)
! Our steel ball will come to the surface at the other side of the moon 3242 s after release, and then oscillate back ! To transit the entire moon in a little less than an hour would make this mode of transportation extremely efficient ! The velocity of the steel ball during this oscillation would be Especially because no power supply would be needed v(t) = ! The maximum velocity would be reached as the ball crosses the center of the moon and would have the numerical value of dx = !" 0 RM sin(" 0t) dt vmax = ! 0 RM = (9.69 "10 #4 s1 )(1.735 "10 6 m) = 1680 m/s = 3760 mph
! If the tunnel were big enough, the same motion could be obtained for a vehicle holding people, providing very efficient transportation to the other side of the moon without the need for propulsion ! During the entire journey the people inside the vehicle would feel absolute weightlessness!
April 14, 2006 Physics for Scientists&Engineers 1 14 Relationship with Circular Motion (1) Relationship with Circular Motion (2)
! Cartesian projections of circular motion perform simple harmonic oscillations during circular motion with constant angular velocity ! In circular motion with constant angular velocity: ! (t) = " t + ! 0
x(t) = r cos(! t + " 0 ) y(t) = r sin(! t + " 0 )
April 14, 2006 Physics for Scientists&Engineers 1 15 April 14, 2006 Physics for Scientists&Engineers 1 16 Quiz 46.1
! The driver of a car decides to smoke a cigar. ! Which window should be opened so that the passenger, sitting beside the driver, does not have to breathe the secondhand smoke? Pendulum Motion
! There is a second important system that oscillates with which we are all familiar: the pendulum ! Ideal pendulum: a thin, massless string with a massive object attached to its end ! Let's look at the equation of motion first. ! The differential equation for a ball on a rope of length l at an angle ! relative to the vertical is A) The driverside window B) The passengerside window d 2! g + !=0 dt 2 l
(for small angles ! ) April 14, 2006 Physics for Scientists&Engineers 1 17 April 14, 2006 Physics for Scientists&Engineers 1 18 Pendulum Motion (3)
! We recognize that this equation is similar to our equation of motion for a mass on a spring
d2x k + x=0 dt 2 m
d 2! g + !=0 dt 2 l with position replaced with angle x !" and the angular frequency replaced by ! This differential equation has the solution
k /m ! g/l ! (t) = Asin(" 0t) + B cos(" 0t) with " 0 = g / l April 14, 2006 Physics for Scientists&Engineers 1 19 ...
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This note was uploaded on 03/19/2008 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.
 Spring '08
 Wolf
 Physics, Force

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