Unformatted text preview: PHYS 132, WEEK #3
21.2. Model: The principle of superposition comes into play whenever the waves overlap. Visualize: The snapshot graph at t = 1 s differs from the graph t = 0 s in that the left wave has moved to the right by 1 m and the right wave has moved to the left by 1 m. This is because the distance covered by each wave in 1 s is 1 m. The snapshot graphs at t = 2 s, 3 s, and 4 s are a superposition of the left and the right moving waves. The overlapping parts of the two waves are shown by the dotted lines. 21.8. Model: Reflections at the string boundaries cause a standing wave on the string.
Solve: Figure Ex21.8 indicates one and a half full wavelengths on the string. Hence l = 2 ( 60 cm) = 40 cm. Thus 3
v = lf = (0.40 m )(100 Hz ) = 40 m / s 21.10. Model: A string fixed at both ends supports standing waves.
Solve: (a) A standing wave can exist on the string only if its wavelength is 2L lm = m = 1, 2, 3, K m
2 (2.40 m ) 1 2 (2.40 m ) 2 2 (2.40 m ) 3 The three longest wavelengths for standing waves will therefore correspond to m = 1, 2, and 3. Thus,
l1 =
= 4.80 m l2 = = 2.40 m l3 = = 1.60 m (b) Because the wave speed on the string is unchanged from one m value to the other, fl (50 Hz )(2.40 m ) f 2 l 2 = f3 l 3 fi f 3 = 2 2 = = 75 Hz l3 1.60 m 21.13. Model: A string fixed at both ends forms standing waves.
Solve: (a) Three antinodes means the string is vibrating as the m = 3 standing wave. The frequency is f3 = 3f1, so the fundamental frequency is f1 = 1 (420 Hz) = 140 Hz. The fifth harmonic will have the frequency f5 = 5f1 = 700 Hz. 3 (b) The wavelength of the fundamental mode is l1 = 2L = 1.20 m. The wave speed on the string is v = l1f1 = (1.20 m)(140 Hz) = 168 m/s. Alternatively, the wavelength of the n = 3 mode is l3 = 1 (2L) = 0.40 m, from which v = l3f3 = (0.40 m)(420 Hz) = 3 168 m/s. The wave speed on the string is given by
v= TS m fi TS = m v = ( 0.0020 kg / m )(168 m / s ) = 56.4 N 2 2 Assess: You must remember to use the linear density in SI units of kg/m. Also, the speed is the same for all modes, but you must use a matching l and f to calculate the speed. 21.14. Model: The laser light forms a standing wave inside the cavity.
Solve: The wavelength of the laser beam is
lm =
2L m fi l100 ,000 = 2( 0.5300 m ) 100, 000 = 10 .6 mm The frequency is
f100 ,000 = c l100 ,000 = 3.0 10 8 m / s 10.6 10 6 m = 2.83 10 13 Hz 21.15. Model: We have an openopen tube that forms standing sound waves.
Visualize: Please refer to Figure Ex21.15. Solve: The gas molecules at the ends of the tube exhibit maximum displacement, making antinodes at the ends. There is another antinode in the middle of the tube. Thus, this is the m = 2 mode and the wavelength of the standing wave is equal to the length of the tube, that is, l = 0.80 m. Since the frequency f = 500 Hz, the speed of sound in this case is v = fl = (500 Hz)(0.80 m) = 400 m/s. Assess: The experiment yields a reasonable value for the speed of sound. 21.18. Solve: For the openopen tube, the fundamental frequency of the standing wave is f1 = 1500 Hz when the tube
is filled with helium gas at 0C. Using l m = 2L m ,
f1 helium = v helium = 970 m / s 2L l1 Similarly, when the tube is filled with air,
f1 air = vair l1 = 331 m / s 2L fi f1 air f1 helium = 331 m / s 331 m / s ^ fi f1 air = ~ (1500 Hz ) = 512 Hz 970 m / s 970 m / s Assess: Note that the length of the tube is onehalf the wavelength whether the tube is filled with helium or air. 21.22. Model: Interference occurs according to the difference between the phases ( Df ) of the two waves.
Solve: (a) A separation of 20 cm between the speakers leads to maximum intensity on the xaxis, but a separation of 60 cm leads to zero intensity. That is, the waves are in phase when (Dx)1 = 20 cm but out of phase when (Dx)2 = 60 cm. Thus, ( Dx ) 2  (Dx )1 = l
2 fi l = 2 (60 cm  20 cm ) = 80 cm (b) If the distance between the speakers continues to increase, the intensity will again be a maximum when the separation between the speakers that produced a maximum has increased by one wavelength. That is, when the separation between the speakers is 20 cm + 80 cm = 100 cm. 21.24. Model: Reflection is maximized if the two reflected waves interfere constructively.
Solve: The film thickness that causes constructive interference at wavelength l is given by Equation 21.32:
lC =
2nd m fid= l Cm
2n = ( 600 10 9 m (1 ) ) (2 )(1.39 ) = 216 nm where we have used m = 1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair < nfilm < nglass. 21.25. Model: Reflection is maximized if the two reflected waves interfere constructively.
Solve: The film thickness that causes constructive interference at wavelength l is given by Equation 21.32:
lC =
2nd m fid= l Cm
2n = ( 500 10 9 m (1 ) ) (2 )(1.25) = 200 nm where we have used m = 1 to calculate the thinnest film. Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicable because nair < noil < nwater. 21.28. Model: The two speakers are identical, and so they are emitting circular waves in phase. The overlap of these
waves causes interference. Visualize: Solve: From the geometry of the figure,
r2 = r1 + (2.0 m ) =
2 2 (4.0 m ) + (2.0 m ) = 4.472 m 2 2 So, Dr = r2  r1 = 4.472 m  4.0 m = 0.472 m . The phase difference between the sources is Df 0 = 0 rad and the wavelength of the sound waves is v 340 m / s l = = = 0.1889 m f 1800 Hz Thus, the phase difference of the waves at the point 4.0 m in front of one source is
Df = 2p Dr l + D f0 = 2 p ( 0.472 m ) 0.1889 m + 0 rad = 5p rad = 2.5(2 p rad) This is a halfinteger multiple of 2 rad, so the interference is perfect destructive. 21.31. Solve: The flute player's initial frequency is either 523 Hz + 4 Hz = 527 Hz or 523 Hz  4 Hz = 519 Hz. Since
she matches the tuning fork's frequency by lengthening her flute, she is increasing the wavelength of the standing wave in the flute. A wavelength increase means a decrease of frequency because v = fl. Thus, her initial frequency was 527 Hz. 21.35. Model: The wavelength of the standing wave on a string vibrating at its secondharmonic frequency is equal to
the string's length. Visualize: Solve: The length of the string L = 2.0 m, so l = L = 2.0 m. This means the wave number is 2p 2p k = = = p rad m l 2.0 m According to Equation 21.5, the displacement of a medium when two sinusoidal waves superpose to give a standing wave is D( x ,t ) = A (x ) cos wt , where A (x ) = 2 asin kx = A max sin kx . The amplitude function gives the amplitude of oscillation from point to point in the medium. For x = 10 cm,
A (x = 10 cm ) = (2.0 cm ) sin (p rad m)( 0.10 m ) = 0.62 cm [ Similarly, A (x = 20 cm ) = 1.18 cm , A (x = 30 cm ) = 1.62 cm , A (x = 40 cm ) = 1.90 cm , and A (x = 50 cm ) = 2.00 cm . Assess: Consistent with the above figure, the amplitude of oscillation is a maximum at x = 0.50 m. 21.41. Model: The wave on a stretched string with both ends fixed is a standing wave. For vibration at its
fundamental frequency, l = 2L. Solve: The wavelength of the wave reaching your ear is 39.1 cm = 0.391 m. So the frequency of the sound wave is f = vair l
1 = 344 m / s 0.391 m = 879.8 Hz This is also the frequency emitted by the wave on the string. Thus,
879.8 Hz = vstring l = TS l
1 2 m = 1 150 N 0.0006 kg / m l fi l = 0.568 m fiL= l = 0.284 m = 28.4 cm 21.43. Model: The stretched string with both ends fixed forms standing waves. Visualize: Solve: The astronauts have created a stretched string whose vibrating length is L = 2.0 m. The weight of the hanging mass creates a tension TS = Mg in the string, where M = 1.0 kg. As a consequence, the wave speed on the string is
v= TS m = Mg m where m = (0.0050 kg)/(2.5 m) = 0.0020 kg/m is the linear density. The astronauts then observe standing waves at frequencies of 64 Hz and 80 Hz. The first is not the fundamental frequency of the string because 80 Hz 2 64 Hz. But we can easily show that both are multiples of 16 Hz: 64 Hz = 4 f1 and 80 Hz = 5 f1 . Both frequencies are also multiples of 8 Hz. But 8 Hz cannot be the fundamental frequency because, if it were, there would be a standing wave resonance at 9(8 Hz) = 72 Hz. So the fundamental frequency is f1 = 16 Hz. The fundamental wavelength is l1 = 2L = 4.0 m. Thus, the wave speed on the string is v = l1 f1 = 64.0 m / s . Now we can find g on Planet X:
v= Mg m fig= m
M v = 2 0.0020 kg / m 1.0 kg (64 m / s ) = 8.19 m / s 2 2 21.48. Model: A tube forms standing waves.
Solve: (a) The fundamental frequency cannot be 390 Hz because 520 Hz and 650 Hz are not integer multiples of it. But we note that the difference between 390 Hz and 520 Hz is 130 Hz as is the difference between 520 Hz and 650 Hz. We see that 390 Hz = 3 130 Hz = 3f1, 520 Hz = 4f1, and 650 Hz = 5f1. So we are seeing the third, fourth, and fifth harmonics of a tube whose fundamental frequency is 130 Hz. According to Equation 21.17, this is an openopen tube because fm = mf1 with m = 1, 2, 3, 4, ... For an openclosed tube m has only odd values. (b) Knowing f1, we can now find the length of the tube: v 343 m / s L= = = 1.32 m 2 f 1 2 (130 Hz ) (c) 520 Hz is the fourth harmonic. This is a sound wave, not a wave on a string, so the wave will have four nodes and will have antinodes at the ends, as shown. (d) With carbon dioxide, the new fundamental frequency is v 280 m / s f1 = = = 106 Hz 2 L 2 (1.32 m ) Thus the frequencies of the n = 3, 4, and 5 modes are f3 = 3f1 = 318 Hz, f4 = 4f1 = 424 Hz, and f5 = 5f1 = 530 Hz. 21.51. Model: The openclosed tube forms standing waves. Visualize: Solve: When the air column length L is the proper length for a 580 Hz standing wave, a standing wave resonance will be created and the sound will be loud. From Equation 21.18, the standing wave frequencies of an openclosed tube are fm = m(v/4L), where v is the speed of sound in air and m is an odd integer: m = 1, 3, 5, ... The frequency is fixed at 580 Hz, but as the length L changes, 580 Hz standing waves will occur for different values of m. The length that causes the mth standing wave mode to be at 580 Hz is m (343 m / s ) L= (4 )( 580 Hz ) We can place the values of L, and corresponding values of h = 1 m  L, in a table: h=1mL m L 1 3 5 7 0.148 m 0.444 m 0.739 m 1.035 m 0.852 m = 85.2 cm 0.556 m = 55.6 cm 0.261 m = 26.1 cm h can't be negative So water heights of 26.1 cm, 55.6 cm, and 85.2 cm will cause a standing wave resonance at 580 Hz. The figure shows the m = 3 standing wave at h = 55.6 cm. 21.58. Model: Constructive or destructive interference occurs according to the phases of the two waves. Visualize: Solve: (a) To go from destructive to constructive interference requires moving the speaker Dx = 1 l , equivalent to a 2 phase change of p rad. Since Dx = 40 cm, we find l = 80 cm. (b) Destructive interference at Dx = 10 cm requires
2p Dx 3p 10 cm ^ + Df 0 = 2p rad ~ rad + D f 0 = p rad fi D f 0 = 80 cm l 4 (c) When side by side, with Dx = 0, the phase difference is Df = Df 0 = 3p/4 rad. The amplitude of the superposition of the two waves is 3p Df ^ a = 2a cos = 2a cos = 0.765a 2 8 21.66. Model: A light wave that reflects from a boundary at which the index of refraction increases has a phase shift
of rad. Solve: (a) Because nfilm > nair, the wave reflected from the outer surface of the film (called 1) is inverted due to the phase shift of rad. The second reflected wave does not go through any phase shift of rad because the index of refraction decreases at the boundary where this wave is reflected, which is on the inside of the soap film. We can write for the phases f 1 = kx1 + f 10 + p rad f 2 = kx2 + f 20 + 0 rad
fi D f = f 2  f 1 = k( x 2  x1 ) + (f 20  f 10 )  p rad = k Dx + Df 0  p rad = k Dx  p rad
Df 0 = 0 rad because the sources are identical. For constructive interference, 2p ^ Df = 2m p rad fi k Dx  p rad = 2m p rad fi ~ (2d ) = (2m + 1)p rad l film fi l film = lC
n = 2d m+
1 2 fi lC = 2nd m+
1 2 = 2.66 d m+
1 2 m = 0, 1, 2, 3, ... (b) For m = 0 the wavelength for constructive interference is
lC = (2.66 )(390 nm ) ( 1) 2 = 2075 nm For m = 1 and 2, l C = 692 nm ( ~ red ) and l C = 415 nm (~ violet ) . Red and violet together give a purplish color. 21.68. Model: The changing sound intensity is due to the interference of two overlapped sound waves.
Visualize: Please refer to Figure P21.68. Solve: Minimum intensity implies destructive interference. Destructive interference occurs where the path length difference for the two waves is Dr = (m + 1 )l . We have assumed Df 0 = 0 rad for two speakers playing "exactly the 2 same" tone. The wavelength of the sound is l = vsound f = (343 m / s ) 686 Hz = 0.500 m . Consider a point that is a distance x in front of the top speaker. Let r1 be the distance from the top speaker to the point and r2 the distance from the bottom speaker to the point. We have
r1 = x r2 = x + (3 m )
2 2 Destructive interference occurs at distances x such that
Dr = x +9m
2 2  x = (m + 1 2 )l
= 9 m  (m + 2 (m +
1 2 1 2 To solve for x, isolate the square root on one side of the equation and then square:
x + 9 m = x + (m +
2 [ 1 2 )l ] 2 = x + 2( m + 2 1 2 )lx + ( m + 1 ) 2 l2 fi x 2
x (m) 17.88 5.62 2.98 1.79 )2 l2 )l Evaluating x for different values of m: m 0 1 2 3 Because you start at x = 2.5 m and walk away from the speakers, you will only hear minima for values x > 2.5 m. Thus, minima will occur at distances of 2.98 m, 5.62 m, and 17.88 m. 21.71. Model: The two radio transmitters are sources of outofphase, circular waves. The overlap of these waves
causes interference. Visualize: Please refer to Figure 21.29b. Solve: The phase difference of the waves at point P is given by Dr Df = 2p + D f0 l
Dr = (3000 m ) + (85 m )  ( 3000 m ) + (35 m ) = 0.99976 m 2 2 2 2 The intensity at P is a maximum. Using m = 1 for the first maximum, and Df 0 = p rad since the transmitters are out of phase, the condition for constructive interference is Df = 2m p = 2 p . Thus,
2p rad = 2p Dr l + p rad fi l = 2Dr = 2( 0.99976 m ) fi f = c l = 3 108 m / s 2 ( 0.99976 m ) = 150 MHz 21.75. Model: The superposition of two slightly different frequencies creates beats.
Solve: (a) The wavelength of the sound initially created by the flutist is 342 m / s l = = 0.77727 m 440 Hz When the speed of sound inside her flute has increased due to the warming up of the air, the new frequency of the A note is 346 m / s f= = 445 Hz 0.77727 m Thus the flutist will hear beats at the following frequency:
f  f = 445 Hz  440 Hz = 5 beats /s Note that the wavelength of the A note is determined by the length of the flute rather than the temperature of air or the increased sound speed. (b) The initial length of the flute is L = 1 l = 1 (0.77727 m ) = 0.3886 m . The new length to eliminate beats needs to be 2 2
L = l
2 = 1 v ^ 1 346 m / s ^ = 0.3932 m ~= 2 f 2 440 Hz Thus, she will have to extend the "tuning joint" of her flute by
0.3932 m  0.3886 m = 0.0046 m = 4.6 mm 21.84. Solve: (a) The wavelengths of the standing wave modes are
lm =
fi l1 = 2 (10.0 m ) 1 = 20.0 m 2L m m = 1, 2, 3, ....
= 10.0 m l2 = 2 (10.0 m ) 2 l3 = 2(10.0 m ) 3 = 6.67 m The depth of the pool is 5.0 m. Clearly the standing waves with l2 and l3 are "deep water waves" because the 20 m depth is larger than onequarter of the wavelength. The wave with l1 barely qualifies to be a deep water standing wave. (b) The wave speed for the first standing wave mode is
v1 = gl 1 2p = (9.8 m / s )(20.0 m )
2 2p = 5.59 m / s Likewise, v 2 = 3.95 m / s and v3 = 3.22 m / s . (c) We have
v= gl m 2p = f m lm fi fm =
g 2pl m = mg 4 pL Please note that m is the mode and not the mass. (d) The period of oscillation for the first standing wave mode is calculated as follows (1)(9.8 m / s 2 ) f1 = = 0.279 Hz fi T1 = 3.58 s 4p (10.0 m )
Likewise, T2 = 2.53 s and T3 = 2.07 s . ...
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This note was uploaded on 05/29/2008 for the course PHYS 131133 taught by Professor All during the Spring '08 term at Cal Poly.
 Spring '08
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