CHAPTER 16: Sound. Responses to Questions

Transcription

1 CHAPTER 6: Sound Responses to Questions 3. Sound waes generated in the irst cup cause the bottom o the cup to ibrate. These ibrations excite ibrations in the stretched string which are transmitted down the string to the second cup, where they cause the bottom o the second cup to ibrate, generating sound waes which are heard by the second child. 4. The waelength will change. The requency cannot change at the boundary since the media on both sides o the boundary are oscillating together. I the requency were to somehow change, there would be a pile-up o wae crests on one side o the boundary. 5. I the speed o sound in depended signiicantly on requency, then the sounds that we hear would be separated in time according to requency. For example, i a chord were played by an orchestra, we would hear the high notes at one time, the middle notes at another, and the lower notes at still another. This eect is not heard or a large range o distances, indicating that the speed o sound in does not depend signiicantly on requency. 6. Helium is much less dense than, so the speed o sound in the helium is higher than in. The waelength o the sound produced does not change, because it is determined by the length o the ocal cords and other properties o the resonating caity. The requency thereore increases, increasing the pitch o the oice. 0. A tube will hae certain resonance requencies associated with it, depending on the length o the tube and the temperature o the in the tube. Sounds at requencies ar rom the resonance requencies will not undergo resonance and will not persist. By choosing a length or the tube that isn t resonant or speciic requencies you can reduce the amplitude o those requencies. 3. Standing waes are generated by a wae and its relection. The two waes hae a constant phase relationship with each other. The intererence depends only on where you are along the string, on your position in space. Beats are generated by two waes whose requencies are close but not equal. The two waes hae a arying phase relationship, and the intererence aries with time rather than position.

2 6. (a) The closer the two component requencies are to each other, the longer the waelength o the beat. I the two requencies are ery close together, then the waes ery nearly oerlap, and the distance between a point where the waes interere constructiely and a point where they interere destructiely will be ery large. 8. No. The Doppler shit is caused by relatie motion between the and obserer. I the wind is blowing, both the waelength and the elocity o the sound will change, but the requency o the sound will not. Solutions to Problems 3. (a) 7 m.7 0 m 0 Hz 0 khz 4 0 Hz.00 Hz So the range is rom.7 cm to 7 m. 5 (b).30 m Hz 7. The total time T is the time or the stone to all t down plus the time or the sound to come back to the top o the cli t up : T t t. Use constant acceleration relationships or an object up down dropped rom rest that alls a distance h in order to ind t down, with down as the positie direction. Use the constant speed o sound to ind t up or the sound to trael a distance h. down: h y y t at h gt up: h t t 0 0 down down down up up h down up h gt g T t g T h T h T 0 g This is a quadratic equation or the height. This can be soled with the quadratic ormula, but be sure to keep seeral signiicant digits in the calculations.

3 h 3.0s h 3.0s m s 6 h h h 6068 m m m, 4m The larger root is impossible since it takes more than 3.0 sec or the rock to all that distance, so the correct result is h 4m. 8. The two sound waes trael the same distance. The sound will trael aster in the concrete, and thus take a shorter time. concrete d t t concrete concrete concrete t 0.75s t 0.75s d t concrete concrete 0.75s concrete The speed o sound in concrete is obtained rom Table 6- as 3000 m/s. d 3000m s 0.75s 90 m 3000m s 3. The pressure wae is P Pa sin 0.38 m x 350 s t. (a) k 0.38 m 5.3m (b) 350s 675Hz 350s (c) 3553ms 3600m s k 0.38 m (d) Use Eq. 6-5 to ind the displacement amplitude. P A A M P Pa 3 M m 300 kg m 3553m s 675 Hz

4 I 0 db 0 log W m.0 W m I 0 4. I I 0 0 I db 0log I 0 I W m W m 0 0 I The pain leel is times more intense than the whisper. 8. Compare the two power output ratings using the deinition o decibels. P 50W P 00W 50 0log 0log.8dB 00 This would barely be perceptible. 34. For a ibrating string, the requency o the undamental mode is gien by F T. L L m L T 4 F =4L m m 440 Hz 3.50 kg 87 N T L F m L 45. The tension and mass density o the string do not change, so the wae speed is constant. The requency ratio or two adjacent notes is to be /. The requency is gien by. l

5 uningered st ret uningered uningered ; x l l nd / / nth / nth uningered nth uningered ret ret ret ret n l st st ret ret l / uningered 65.0 cm l 6.35cm st / / l uningered ret l l l l n / l l l / x / 3/ x 4 / 5 / x 6 / x 65.0 cm 3.6 cm ; 65.0 cm 7.cm x x 65.0 cm 0.3cm ; 65.0 cm 3.4 cm cm 6.3cm ; 65.0 cm 9.0 cm The beat period is.0 seconds, so the beat requency is the reciprocal o that, 0.50 Hz. Thus the other string is o in requency by second string is too high or too low Hz. The beating does not tell the tuner whether the 56. (a) Since the sounds are initially 80 out o phase, another 80 o phase must be added by a path A x d x B length dierence. Thus the dierence o the distances rom the speakers to the point o constructie intererence must be hal o a waelength. See the diagram. d x x d x d 0.583m min 94 Hz This minimum distance occurs when the obserer is right at one o the speakers. I the speakers are separated by more than m, the location o constructie intererence will be moed away rom the speakers, along the line between the speakers. (b) Since the sounds are already 80 out o phase, as long as the listener is equidistant rom the speakers, there will be completely destructie intererence. So een i the speakers hae a tiny separation, the point midway between them will be a point o completely destructie intererence. The minimum separation between the speakers is (a) Obserer moing towards stationary.

6 30.0m s obs 350 Hz 470 Hz (b) Obserer moing away rom stationary. 30.0m s obs 350 Hz 30 Hz 63. (a) For the 8 m/s relatie elocity: 300 Hz 47 Hz 430 Hz moing 8m s 8m s obserer moing 300 Hz 4Hz 40 Hz The requency shits are slightly dierent, with obserer moing moing. The two requencies are close, but they are not identical. As a means o comparison, calculate the spread in requencies diided by the original requency. moing obserer moing 47 Hz 4Hz % 300 Hz (b) For the 60 m/s relatie elocity: 300 Hz 43Hz 430 Hz moing 60 m s 60 m s obserer moing 300 Hz 337 Hz 3370 Hz The dierence in the requency shits is much larger this time, still with obserer. moing moing moing obserer moing 43Hz 337 Hz % 300 Hz (c) For the 30 m/s relatie elocity:

7 300 Hz 34,300 Hz moing 30 m s 30m s obserer moing 300 Hz 4446 Hz 4450 Hz The dierence in the requency shits is quite large, still with obserer. moing moing moing obserer moing 34,300 Hz 4446 Hz % 300 Hz (d) The Doppler ormulas are asymmetric, with a larger shit or the moing than or the moing obserer, when the two are getting closer to each other. In the ollowing deriation, assume, = and use the binomial expansion. obserer moing moing 65. (a) The obserer is stationary, and the is moing. First the is approaching, then the is receding. m s 0.0 km h 33.33m s 3.6 km h 80 Hz 40 Hz moing 33.33m s towards 80 Hz 70 Hz moing 33.33m s away (b) Both the obserer and the are moing, and so use Eq. 6-. m s 90.0km h 5m s 3.6km h

8 obs approaching obs receding 5m s 80 Hz 50 Hz 33.33m s 5m s 80 Hz 080 Hz 33.33m s (c) Both the obserer and the are moing, and so again use Eq. 6-. m s 80.0 km h. m s 3.6 km h. m s 33.33m s obs police car approaching 80 Hz 330 Hz. m s 33.33m s obs police car receding 80 Hz 40 Hz 66. The wall can be treated as a stationary obserer or calculating the requency it receies. The is lying toward the wall. wall Then the wall can be treated as a stationary emitting the requency moing obserer, lying toward the wall., and the as a wall wall 7.0 m s 7.0 m s Hz Hz