Physics 41 Superposition Chapter 21 Knight HW # 4

Transcription

1 Physics 4 uperposition Chapter Knight HW # 4 Knight nd Ed Exercises and Probles:,, 9,, 4, 6, 8,,3, 7, 8, 9, 4, 49, 68, 74, 77 ) Left: The graph at t.0 s differs fro the graph at t 0.0 s in that the left wave has oved to the right by.0 and the right wave has oved to the left by.0. This is because the distance covered by the wave pulse in.0 s is.0. The snapshot graphs at t.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right oving waves. The overlapping parts of the two waves are shown by the dotted lines. ) Right: The snapshot graph at t.0 s differs fro the graph t 0.0 s in that the left wave has oved to the right by.0 and the right wave has oved to the left by.0. This is because the distance covered by each wave in.0 s is.0. The snapshot graphs at t.0 s, 3.0 s, and 4.0 s are a superposition of the left and the right oving waves. The overlapping parts of the two waves are shown by the dotted lines.

2 .9. Model: A string fixed at both ends supports standing waves. olve: (a) We have f a 4 Hz f where f is the fundaental frequency that corresponds to. The next successive frequency is f b 36 Hz ( ) f. Thus, f f b 36 Hz 4 Hz.5.5 f Hz f f 4 Hz a L The wave speed is v f f.0 Hz 4 /s (b) The frequency of the third haronic is 36 Hz. For 3, the wavelength is L Model: A string fixed at both ends fors standing waves. olve: (a) The wavelength of the third haronic is calculated as follows: L L (b) The speed of waves on the string is v3f3(0.807 )(80 Hz) 45.3 /s. The speed is also given by v T /, so the tension is T v v L kg 45.3 /s 69.7 N olve: (a) For the open-open tube, the two open ends exhibit antinodes of a standing wave. The possible wavelengths for this case are L,, 3, The three longest wavelengths are (b) In the case of an open-closed tube, 4L, 3, 5, The three longest wavelengths are

3 .6. olve: For the open-open tube, the fundaental frequency of the standing wave is f 500 Hz when the tube is filled with vheliu 970 /s heliu gas at 0C. Using L, f heliu L iilarly, when the tube is filled with air, f v 33 /s f 33 /s 33 /s f 500 Hz 5 Hz 970 /s 970 /s air air air air L f heliu Assess: Note that the length of the tube is one-half the wavelength whether the tube is filled with heliu or air..8. Model: Reflections at the string boundaries cause a standing wave on a stretched string. olve: Because the vibrating section of the string is.9 long, the two ends of this vibrating wire are fixed, and the string is vibrating in the fundaental haronic. The wavelength is L L The wave speed along the string is v f (7.5 Hz)(3.80 ) 04.5 /s. The tension in the wire can be found as follows: T ass kg v T v v 04.5 /s 80 N length.00.. Model: The interference of two waves depends on the difference between the phases ( ) of the two waves. olve: (a) Because the speakers are in phase, 0 0 rad. Let d represent the path-length difference. Using 0 for the sallest d and the condition for destructive interference, we get x 0 rad 0,,, 3 x d v 343 /s 0 rad 0 rad rad d 0.5 f 686 Hz (b) When the speakers are out of phase, 0. Using for the sallest d and the condition for constructive interference, we get d v 343 /s d 0.5 f 686 Hz x 0 0,,, 3,

4 .3. Model: Reflection is axiized if the two reflected waves interfere constructively. olve: The fil thickness that causes constructive interference at wavelength is given by Equation.3: 9 nd C C d n.39 6 n where we have used to calculate the thinnest fil. Assess: The fil thickness is uch less than the wavelength of visible light. The above forula is applicable because n air n fil n glass..7. Model: The two speakers are identical, and so they are eitting circular waves in phase. The overlap of these waves causes interference. Visualize: olve: Fro the geoetry of the figure, r r o, r r r The phase difference between the sources is 0 0 rad and the wavelength of the sound waves is v 340 /s f 800 Hz Thus, the phase difference of the waves at the point 4.0 in front of one source is r rad 5 rad.5( rad) This is a half-integer ultiple of π rad, so the interference is perfect destructive.

5 .8. Model: The two radio antennas are eitting out-of-phase, circular waves. The overlap of these waves causes interference. Visualize: olve: Fro the geoetry of the figure, r 800 and r o, r r r 00 and 0 rad. The wavelength of the waves is 8 c 3.00 /s 00 6 f 3.00 Hz Thus, the phase difference of the waves at the point (300, 800 ) is r 00 0 rad 5 rad.5( rad) 00 This is a half-integer ultiple of π rad, so the interference is perfect destructive..9. olve: The beat frequency is f f f 3 Hz f 00 Hz f 03 Hz beat f is larger than f because the increased tension increases the wave speed and hence the frequency.

6 .4. Model: The stretched string with both ends fixed fors standing waves. Visualize: olve: The astronauts have created a stretched string whose vibrating length is L.0. The weight of the hanging ass creates a tension T Mg in the string, where M.0 kg. As a consequence, the wave speed on the string is v T where ( kg)/(.5 ) kg/ is the linear density. The astronauts then observe standing waves at frequencies of 64 Hz and 80 Hz. The first is not the fundaental frequency of the string because 80 Hz 64 Hz. But we can easily show that both are ultiples of 6 Hz: 64 Hz 4 f and 80 Hz 5 f. Both frequencies are also ultiples of 8 Hz. But 8 Hz cannot be the fundaental frequency because, if it were, there would be a standing wave resonance at 9(8 Hz) 7 Hz. o the fundaental frequency is f 6 Hz. The fundaental wavelength is L 4.0. Thus, the wave speed on the string is vf 64.0 /s. Now we can find g on Planet X: Mg kg/ v 64 /s 8. /s g v M.0 kg Mg.49. Model: A tube fors standing waves. olve: (a) The fundaental frequency cannot be 390 Hz because 50 Hz and 650 Hz are not integer ultiples of it. But we note that the difference between 390 Hz and 50 Hz is 30 Hz as is the difference between 50 Hz and 650 Hz. We see that 390 Hz 3 30 Hz 3f, 50 Hz 4f, and 650 Hz 5f. o we are seeing the third, fourth, and fifth haronics of a tube whose fundaental frequency is 30 Hz. According to Equation.7, this is an open-open tube because f f with,, 3, 4, For an open-closed tube has only odd values. (b) Knowing f, we can now find the length of the tube: v 343 /s L.3 f 30 Hz (c) 50 Hz is the fourth haronic. 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 fundaental frequency is v 80 /s f 06 Hz L.3 Thus the frequencies of the n 3, 4, and 5 odes are f 3 3f 38 Hz, f 4 4f 44 Hz, and f 5 5f 530 Hz.

7 .68. Model: The changing sound intensity is due to the interference of two overlapped sound waves. olve: Miniu intensity iplies destructive interference. Destructive interference occurs where the path length difference for the r. We have assued 0 0 rad for two speakers playing exactly the sae tone. The wavelength of the two waves is sound is v f sound 343 /s 686 Hz Consider a point that is a distance x in front of the top speaker. Let r be the distance fro the top speaker to the point and r the distance fro the botto speaker to the point. We have Destructive interference occurs at distances x such that r x r x 3 r x 9 x To solve for x, isolate the square root on one side of the equation and then square: Evaluating x for different values of : 9 x x x x 9 x 0 3 x () Because you start at x.5 and walk away fro the speakers, you will only hear inia for values x.5. Thus, to correct significant figures, inia will occur at distances of 3.0, 5.6, and 8.

8 .74. Model: The superposition of two slightly different frequencies gives rise to beats. olve: The third haronic of note A and the second haronic of note E are (b) The beat frequency between the first haronics is The beat frequency between the second haronics is f 3 f Hz 30 Hz f f 659 Hz 38 Hz 3A A E E f3a fe 30 Hz 38 Hz Hz f E f A 659 Hz 440 Hz 9 Hz f E f A 38 Hz 880 Hz 438 Hz The beat frequency between f 3A and f E is Hz. It therefore eerges that the tuner looks for a beat frequency of Hz. (c) If the beat frequency is 4 Hz, then the second haronic frequency of the E string is f 30 Hz 4 Hz 36 Hz f 36 Hz 658 Hz E E Note that the second haronic frequency of the E string could also be f E 30 Hz 4 Hz 34 Hz f 66 Hz This higher frequency can be ruled out because the tuner started with low tension in the E string and we know that T vstring f f T E.77. Model: The frequency of the loudspeaker s sound in the back of the pick-up truck is Doppler shifted. As the truck oves away fro you, its frequency is decreased. olve: Because you hear 8 beats per second as the truck drives away fro you, the frequency of the sound fro the speaker in the pick-up truck is f 400 Hz 8 Hz 39 Hz. This frequency is f0 f v v v 400 Hz 343 /s 39 Hz v 7.0 /s That is, the velocity of the source v and hence the pick-up truck is 7.0 /s.