04 midterm ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm Question 1 part 1 of 1 10 points The diagrams below show dierent standing waves on a 49.5 cm string. Which of these waves has a 33 cm wavelength? 8. 1. correct
1
7.
2.
9.
3.
4.
5.
6.
Explanation: A standing wave on a string has nodes where the string does not vibrate; these nodes are spaced half-wavelength from each other: x0 = 0, x1 = , x2 = 2 , 2 2 x3 = 3 , x4 = 4 , . . . . 2 2 A xed end of the string cannot vibrate, so it must be a node of the standing wave. For the string xed at both ends, both ends must be nodes of the wave, hence L = n , n = 1, 2, 3, . . . , (1) 2 where n is the number of vibrating segments of the string. More precisely, the wave on the string has n antinodes and n + 1 nodes: One node at each end of the string, plus n 1 nodes in the middle. The wave in question has wavelength = 33 cm on a string of length L = 49.5 cm, hence according to eq. (1), n= 49.5 cm L = = 3, /2 (33 cm)/2
which means the wave has 3 antinodes and 4 nodes: One at each end, and 2 in the middle.
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm In other words, it looks like 6. 4 = 5 dB 7. 4 = 20 dB . 8. 4 = 40 dB 9. 4 = 60 dB Question 2 part 1 of 1 10 points Sound in air can best be described as which of the following type of wave? 1. Torsional 2. Transverse 3. Polarized 4. Electromagnetic 5. Longitudinal correct Explanation: A sound wave in the air is propagated by the oscillation of air molecules. It is best described as a longitudinal wave. Question 3 part 1 of 1 10 points The sound of a man shouting at the top of his lungs from a rather large distance away from your ear has loudness of only 20 decibels. What would be the decibel level of four men shouting at the top of their (equally powerful) lungs from the same distance away from you ear? Assume that there is no interference from superposed waves. 1. 4 = 160 dB 2. 4 = 26 dB correct 3. 4 = 80 dB 4. 4 = 6 dB 5. 4 = 14 dB 10. 4 = 10 dB
2
Explanation: The decibel level is a logarithmic measure of sound intensity I : = 10 log10 I I0 .
Since we have assumed that there is no interference from superposed waves, four men shouting together produce a sound intensity four times greater than that of just one man (I4 = 4 I1), so 4 = 10 log10 4 I1 I0 I1 = 10 log10 + 10 log10 (4) I0 = 1 + 10 log 10(4) = 20 dB + 10 log10 (4) = 26.0206 dB .
Question 4 part 1 of 1 10 points A(n) 0.323 kg mass is attached to a spring and undergoes simple harmonic motion with a period of 0.0526 s . The total energy of the system is 4.4 J . Find the amplitude of the motion. 1. 0.0347154 m 2. 0.0363089 m 3. 0.0375827 m 4. 0.0387585 m 5. 0.040258 m 6. 0.0418031 m 7. 0.0436964 m correct 8. 0.0464428 m
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm 9. 0.0486466 m 10. 0.0505099 m Explanation: Basic Concepts: From T = 2 we have k= 4 2 m T2 4 2 (0.323 kg) = (0.0526 s)2 = 4608.83 N/m . m , k 2. 4.29536 s 3. 4.43381 s 4. 4.57687 s 5. 4.72413 s 6. 4.8917 s 7. 5.04575 s 8. 5.21071 s 9. 5.3781 s 10. 5.54771 s correct
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The total energy E is, E= 1 k A2 , 2
Explanation: The hoop oscillates as a physical pendulum. Its center of mass is in its geometric center, at distance R = 3.82 m from the pivot. So when it swings by a small angle away from the equilibrium position (center directly below the pivot), there is a torque due to gravity force, = mg R sin (1) mgr for small . The hoop has moment of inertia I0 = mR2 around its center of mass. The momend of inertia relative to the pivot follows via the parallel axis theorem I = I0 + mR2 = 2mR2 . (2)
where A is the amplitude of motion. Thus A= = 2E k 2 (4.4 J) (4608.83 N/m)
= 0.0436964 m . Consequently, the equation of motion for the swinging hoop Question 5 part 1 of 1 10 points A thin hoop of radius 3.82 m and mass 8 kg is suspended from a pivot on the hoop itself as shown in the gure. I becomes 2mR2 d2 = mgR , dt2 (4) d2 = dt2 (3)
which simplies to
g d2 = 2 dt 2R
(5)
This is a harmonic oscillator equation with The acceleration of gravity is 9.8 m/s2 . Find the period T of small oscillations of this hoop around = 0. 1. 4.16199 s 2 = g . 2R (6)
Translating this angular frequency into the
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm oscillation period, we have T= 2 2R g 2(3.82 m) 9.8 m/s2 (7)
4
obtained by dierentiating x twice, where t = 2 s: a = 2 A sin( t + ) = (3 rad/s)2 (6 m) sin[(3 rad/s) (2 s) + (1.0472 rad)] = 37.3585 m/s2 . The phase (given in radians) incorporates the initial condition where the body started (t = 0), meaning it started at x0 = A sin = (6 m) sin(1.0472 rad) = 5.19615 m and it is now at x = A sin( t + ) = (6 m) sin[(3 rad/s) (2 s) + (1.0472 rad)] = 4.15094 m . (These two last facts are not needed to solve the problem but clarify the physical picture.) Question 7 part 1 of 1 10 points A light spring with a spring constant of 16.6 N/m rests vertically on the bottom of a large beaker of water, as shown in (a). The acceleration of gravity is 9.81 m/s2 . A 0.00324 kg block of wood with a density of 662.1 kg/m3 is connected to the spring, and the mass-spring system is allowed to come to static equilibrium, as shown in (b). The magnitude of the force pulling the spring back to its unstretched position equals k x.
= 2 = 2
= 5.54771 s. Note that this period does not depend on the hoops mass m. Question 6 part 1 of 1 10 points A body oscillates with simple harmonic motion along the x-axis. Its displacement varies with time according to the equation, x(t) = A sin( t + ). If A = 6 m, = 3 rad/s, and = 1.0472 rad, what is the magnitude of the acceleration of the body at t = 2 s? Note: The argument of the sine function is given here in radians rather than degrees. 1. 44.7636 m/s2 2. 43.1841 m/s2 3. 41.535 m/s2 4. 39.8623 m/s2 5. 38.558 m/s2 6. 37.3585 m/s2 correct 7. 36.2372 m/s2 8. 34.8795 m/s2 9. 33.6229 m/s2 10. 32.5609 m/s2 Explanation: x = A sin( t + ) v= a= dx = A cos( t + ) dt
m x k k (a) (b)
dv = 2 A sin( t + ) dt The basic concepts above are enough to solve the problem. Just use the formula for a
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm How much does the spring stretch? 1. 0.000801841 m 2. 0.000854621 m 3. 0.000882307 m 4. 0.000914499 m 5. 0.000944873 m 6. 0.000977171 m correct 7. 0.00100756 m 8. 0.00103963 m 9. 0.00107226 m 10. 0.0011072 m Explanation: Basic Concepts: Fnet = FB Fg,bl Fspring = 0 FB = mf luid g Fg = m g Fspring = k x m = V Given: k = 16.6 N/m mbl = 0.00324 kg bl = 662.1 kg/m3 water = 1000 kg/m3 g = 9.81 m/s2 Solution: water V g mbl g k x = 0 mbl g mbl g bl x = k mbl g water 1 = bl k 3 1000 kg/m = 1 662.1 kg/m3 (0.00324 kg) (9.81 m/s2 ) 16.6 N/m water = 0.000977171 m . Question 8 part 1 of 1 10 points
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An ambulance is traveling north at 46.6 m/s, approaching a car that is also traveling north at 33 m/s. The ambulance driver hears his siren at a frequency of 652 cycles/s. The velocity of sound is 343 m/s. 46.6 m/s 33 m/s Ambulance Car
What is the wavelength at the car drivers position for the sound from the ambulances siren? 1. 0.377361 m 2. 0.389163 m 3. 0.401657 m 4. 0.414306 m 5. 0.427259 m 6. 0.440573 m 7. 0.454601 m correct 8. 0.468844 m 9. 0.483391 m 10. 0.499473 m Explanation:
Let : vcar vamb vsound f
= 33 m/s , = 46.6 m/s , = 343 m/s , and = 652 cycles/s .
By the Doppler Eect, the wavelength of the sound created by a source with rest frequency f and speed vsource is = vsound vsource . f
The wave speed relative to a moving observer is v = vsound vobserver and the observed frequency is v f= .
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm Note: The wavelength is specied in the reference frame of the medium of propagation. Sound waves always travel at a given speed with respect to their medium of propagation. Note: Both the ambulance and car drivers, as well as any observers at rest (on the side of the road, for example), will measure the same wavelength for the sound from the siren, because length measurements will not depend on the velocity of the measurer. However, as sources/observers move through the medium at dierent velocities, they see the sound waves move past them at dierent velocities. As a result, the number of wavefronts passing them in a given time interval (i.e., the frequency of the sound) must change. The relationship between observed frequency and observed wavelength is always given by f = vrel , where vrel is the relative speed of the sound wave and the observer/source. Therefore, the wavelength of the sound emitted in front of the ambulance is =
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3.1 kg What is the wave speed when the suspended mass is 2.2 kg? 1. 22.0617 m/s 2. 22.7454 m/s correct 3. 23.4694 m/s 4. 24.2061 m/s 5. 24.9592 m/s 6. 25.7917 m/s 7. 26.6382 m/s 8. 27.4766 m/s 9. 28.4368 m/s 10. 29.5322 m/s Explanation: Let : m1 = 3.1 kg , m2 = 2.2 kg , and v1 = 27 m/s . The linear density is v1 = = F , so (1)
vsound vamb f 343 m/s 46.6 m/s = 652 cycles/s = 0.454601 m .
The negative sign arises because the ambulance driver is traveling in the same direction as these sound waves and therefore perceives them as being slower than sound waves emitted when the ambulance is at rest. This results in a smaller wavelength; intuitively, the wavefronts are compressed together by the motion of the siren. Question 9 part 1 of 1 10 points Tension is maintained in a string as in the gure. The observed wave speed is 27 m/s when the suspended mass is 3.1 kg . The acceleration of gravity is 9.8 m/s2 .
F 2 v1 m1 g =2 v1 (3.1 kg) (9.8 m/s2 ) = (27 m/s)2 = 0.0416735 kg/m .
The velocity is v2 = v1 = F , m2 g so (2)
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm = (2.2 kg) (9.8 m/s2 ) 0.0416735 kg/m
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On a standing wave, the distance of the nearest two nodes is d= , 2
= 22.7454 m/s . Alternate Solution: Plugging from Eq. 1 into Eq. 2, we have v1 = = = v1 m2 g
2 m2 g v1 m1 g
(2)
where is the wavelength. In this problem, x = 0 is a node, which can be seen by adding y1 and y2 . So the smallest positive value of x corresponding to a node is 1 2 1 = (0.3125 m) 2 = 0.15625 m .
m2 m1 (2.2 kg) (3.1 kg)
(3)
x=
= (27 m/s)
= 22.7454 m/s . Question 10 part 1 of 1 10 points A standing wave is a superposition of two harmonic waves described by y1 = A sin(k x + t) y2 = A sin(k x t) , and
Question 11 part 1 of 1 10 points Note: Patm = 101300 Pa/atm. The viscosity of the uid is negligible and the uid is incompressible.. A liquid of density 1304 kg/m3 ows with speed 1.15 m/s into a pipe of diameter 0.14 m. The diameter of the pipe decreases 0.05 to m at its exit end. The exit end of the pipe is 4.01 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.4 atm. The acceleration of gravity is 9.8 m/s2 . P1
where A = 6.92637 cm, k = 20.1062 m1 and = 20.1062 s1 . Determine the smallest positive value of x corresponding to a node . 1. 0.151515 m 2. 0.15625 m correct 3. 0.16129 m 4. 0.166667 m 5. 0.172414 m 6. 0.178571 m 7. 0.185185 m 8. 0.192308 m 9. 0.2 m 10. 0.208333 m Explanation: 2 = k =
1.15 m/s 0.14 m
P2 1.4 atm
4.01 m
v2 0.05 m Applying Bernoullis principle, what is the pressure P1 at the entrance end of the pipe? 1. 142713 Pa correct 2. 147275 Pa
2 20.1062 m1 = 0.3125 m .
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm 3. 152518 4. 157291 5. 165147 6. 170718 7. 177206 8. 183235 9. 190468 10. 197290 Pa Pa Pa Pa Pa Pa Pa Pa
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to the right length so as to resonate at its fundamental frequency when placed in this sound wave? 1 1. = 4 correct 1 8 3 3. = 2 2. = 4. = 2 5. = 3 4
Explanation: Applying Bernoullis principle to the uid ow at the entrance and exit of the pipe gives 12 12 v1 = P2 + g y2 + v2 2 2 1 P1 = P2 + g (y2 y1 ) + (v22 v1 2 ) . 2 We also have y2 y1 = h, since the entrance height y1 is greater than the exit height y2 . Therefore 1 P1 = P2 g h + [v22 v1 2 ] 2 = (141820 Pa) (1304 kg/m3 ) (9.8 m/s2 ) (4.01 m) 1 + (1304 kg/m3 ) 2 [(9.016 m/s)2 (1.15 m/s)2 ] P1 + g y 1 + = 142713 Pa . Question 12 part 1 of 1 10 points This picture shows the displacements S of the air molecules in a traveling sound wave as a function of distance, x. Sound Wave +A
6. = 1 2 3 8. = 8 7. = Explanation: The tube is closed at the left end requiring an node in the wave form. The tube is open at the right end requiring an antinode in the wave form. The displaced between the node on the left and the antinode on the right is shown in the gure below. This correspondance can be 1 represented by the distance between = 2 3 and = as shown. 4 Sound Wave +A
S S A A /2 3 /2 0 2 Which of the following tubes, closed at the left end and open at the right end, is closest /2 3 /2 0 2 So the length of a tube to produce the 1 fundamental resonance should be = . 4
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm First of four versions. Question 13 part 1 of 1 10 points A harmonic wave y = A sin[k x t ] , where A = 1 meter, k has units of m1 , has units of s1 , and has units of radians, is plotted in the diagram below. At the time t = 0 +1 A (meters) 10. y = A sin 2 15 m xt 2 3
9
Explanation: From the diagram of the wave function the wave-length = 3 m (6 horizontal scale divisions of 0.5 m each, see diagram below). Notice: Since one wave-length is 2 ra2 dians, each horizontal division is = 6 3 radians. The given wave function (sine function with t = 0) y = A sin k x = A sin = A sin 2 x 2 x 3m
1 x (meters) 2 4 6
(dark curve in diagram below) is shifted 1 divisions to the right (negative phase shift) of a no-phase-shift sine function y = sin 2 3m x
Which wave function corresponds best to the diagram? 1. y = A sin rect 2. y = A sin 3. y = A sin 4. y = A sin 5. y = A sin 6. y = A sin 7. y = A sin 8. y = A sin 9. y = A sin 2 9m 2 3m 2 3m 2 9m 2 15 m 2 9m 2 15 m 2 9m 1 3 2 xt 3 4 xt 3 4 xt 3 5 xt 3 5 xt 3 4 xt 3 2 xt 3 xt 2 3m 1 x t 3 cor-
(gray curve in diagram below), therefore =1 1 = radians . 3 3
Checking the wave function y at x = 0, we have agreement with the diagram below y = sin 0 1 3 = 0.866025 m .
At the time t = 0 +1 A (meters) 1 x (meters) =3 3
6
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm Therefore, the wave function is y = A sin 2 3m xt 1 3 . earthquake center being d= vP vS t vP vS (8160 m/s) (5380 m/s) (53 s) = (8160 m/s) (5380 m/s) = 836.958 km . away from the seismic station. Question 15 part 1 of 1 10 points
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keywords: Question 14 part 1 of 1 10 points Earthquakes produce two kinds of seismic waves: The longitudinal primary waves (called P waves) and the transverse secondary waves (called S waves). Both S waves and P waves travel through the Earths crust and mantle, but do so at different speeds; the P waves are always faster than the S waves, but their exact speeds depend on depth and location. For the purpose of this exercise, we assume the P waves speed to be vP = 8160 m/s while the S waves travel at a slower speed of vS = 5380 m/s. Suppose a seismic station detects a P wave and then t = 53 s later detects an S wave. How far away is the earthquake center? 1. 739.317 km 2. 762.433 km 3. 786.835 km 4. 811.416 km 5. 836.958 km correct 6. 864.327 km 7. 891.591 km 8. 920.575 km 9. 950.82 km 10. 980.431 km Explanation: Suppose the earthquake happens at time t = 0 at some distance d. The P wave and the S wave are both emitted at the same time t = 0, but they arrive at dierent times, respectively tP = d/vP and tS = d/vS . The S wave is slower, so it arrives later than the P wave, the time dierence being t = d d d(vP vS ) = . vS vP vP vS
The gure below shows a complex wave pattern on a string moving towards a rigid hook at the wall on the right. After some time, the wave is reected from the wall. v
Select the wave pattern for the reected wave. v 1. correct
v 2.
v 3.
4. v 5.
Consequently, given this time dierence and the two waves speeds vP and vS , we nd the
Explanation:
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm Consider the wave pattern image reected about the rigid hook on the wall. v
11
3.
v After the time it takes for the wave to be reected from the wall, this image is the wave pattern traveling to the left along the string. Note: Reection about a point (hook) is the same as reection about the y -axis (wall) followed by reection about the x-axis (string). The leading part of the wave must remain in front and the wave is ipped over. v
4.
5.
Explanation: Notice that the two pulses have the same width and amplitude. Choosing the point P (the same as point Q when the two pulses coincide) as the origin, the two pulses can be described as: P: Q: y1 = y2 = A , d x d A , d x < 0 A , 0<x<d
v This is the fourth wave pattern of four possible wave patterns presented by this question. Question 16 part 1 of 1 10 points The gure shows two wave pulses that are approaching each other.
Using the principle of superposition, the resultant pulse is y = y1 + y2 = 2 A , d x < 0 0 , 0<x<d .
P
Q
P Q
Which of the following best shows the shape of the resultant pulse when the centers of the pulses, points P and Q, coincide?
1.
P +Q Question 17 part 1 of 1 10 points
2. correct
A skyrocket explodes 92 m above the ground. Three observers are spaced 106 m apart, with observer A directly under the point of the explosion.
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm IA r2 h2 + d2 = B= 2 IB h2 rA (92 m)2 + (106 m)2 = (92 m)2 = 2.3275 . A 106 m B 106 m C Question 18 part 1 of 1 10 points
12
92 m
Find the ratio of the sound intensity heard by observer A to that heard by observer B. The gure is not drawn to scale. For practical purposes, you may treat each observer as a point on the ground. 1. 1.92398 2. 1.98467 3. 2.04819 4. 2.1142 5. 2.18313 6. 2.25087 7. 2.3275 correct 8. 2.40108 9. 2.47611 10. 2.55549 Explanation:
A U-tube of constant cross-sectional area, open to the atmosphere, is partially lled with a heavy liquid with density 11.3 g/cm3 . A light liquid with density 1.79 g/cm3 is then poured into both arms.
h light liquid 1.79 g/cm3 1.16 cm heavy liquid 11.3 g/cm3 If the equilibrium conguration of the tube is as shown in the gure, with a dierence in the height of the heavy liquid of 1.16 cm, determine the value of the dierence in height of the light liquid h . 1. 5.59949 cm 2. 5.77519 cm 3. 5.97606 cm 4. 6.16291 cm correct 5. 6.36771 cm 6. 6.56816 cm 7. 6.78 cm 8. 7 cm 9. 7.2407 cm 10. 7.56732 cm Explanation: Let : = 1.79 g/cm3 , h = 11.3 g/cm3 , hh = 1.16 cm ,
Let : h = 92 m and d = 106 m . Source rA = 92 m rC rB B
C 106 m 106 m The intensity at a distance r from the source is P I= , 4 r2 and distances from the source to points A and B are rA = h rB = and h2 + d2 , so
A
and
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm Also, let h be the height of the light liquid column added to the right side of the U-tube. Consider the pressure at the elevation of the light-heavy liquid interface in the left column and at the same elevation in the right column. By Pascals Principle, the absolute pressure at that elevation is the same in both columns. We have, Pright = Patm + h g hh + g h and Plef t = Patm + g (hh + h + h ) . From Pascals Principle, Plef t = Pright (hh + h ) = h (hh ) , or h h = 1 hh 11.3 g/cm3 = 1 1.79 g/cm3 = 6.16291 cm . Question 19 part 1 of 1 10 points The small piston of a hydraulic lift has a cross-sectional area of 7.4 cm2 and the large piston has an area of 50 cm2 , as in the gure below. F area 7.4 cm2 50 cm2 6. 11100 N correct 7. 11446.2 N 8. 11802.6 N 9. 12342.9 N 10. 12817.9 N Explanation: = 7.4 cm2 , = 50 cm2 , = 75 kN , and =F.
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Let :
A1 A2 W F
According to Pascals law, the pressure exerted on A1 must be equal to the one exerted F on A2 . The pressure P1 = must be equal A1 W to the pressure P2 = due to the load. A2 hh W F = , so A1 A2 A1 F= W A2 (7.4 cm2 ) (75000 N) = (50 cm2 ) = 11100 N .
Question 20 part 1 of 1 10 points A block of unknown mass is attached to a spring of spring constant 6.14 N/m and undergoes simple harmonic motion with an amplitude of 15.2 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 35.6 cm/s. Calculate the mass of the block. 1. 0.738686 kg 2. 0.762884 kg 3. 0.787803 kg 4. 0.813529 kg 5. 0.839492 kg correct 6. 0.86648 kg 7. 0.893815 kg 8. 0.922009 kg
What force F must be applied to the small piston to raise a load of 75 kN? 1. 9442.62 N 2. 9752 N 3. 10064.5 N 4. 10384.6 N 5. 10721.9 N
midterm 04 ALIBHAI, ZAHID Due: May 2 2007, 11:00 pm 9. 0.951758 kg 10. 0.983792 kg Explanation: Basic Concepts Energy conservation: If K is kinetic energy and U is potential energy, Ki + Ui = Kf + Uf Kinetic energy of particle with mass m and speed v : 1 K = m v2 2 Mass m on spring with constant k : = k m
14
and potential energy of a spring at displacement x: 1 U = k x2 2 Period 2 T= Solution: Call the maximum displacement (amplitude) A. The halfway displacement is A/2. Energy conservation requires 1 1 1 0 + k A2 = m v 2 + k 2 2 2 or k A2 = m v 2 + so m= 3 k A2 4 v2 (3) (6.14 N/m) (0.152 m)2 = (4) (0.356 m/s)2 = 0.839492 kg . 1 k A2 4 A 2
2