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Suggested Answers of Exercise 15 (Acoustics) ---------------------------------------------- 1. (a) Power emitted as sound = 50 x 0.05 = 2.5 W (d) (i) Interchange the wire Hence sound intensity at 3 m connections to one of the loudspeakers. (ii) The positions of maximum and minimum intensity are interchanged. P 2.5 ---------------------------------------------- 0.0442 W m -2 2r 2 2 (3) 2 4. (a) The formation of heaps of powder is due to the formation of longitudinal stationary waves set up along the tube. The (b) Intensity level middle of each heap is a node position 0.0442 10 log 106 dB where the air does vibrate whereas the air on 10 -2 both sides vibrates, causing the heap to (c) The minimum intensity that the remain at the node position. sound can be barely heard is 10-12 W m-2. (b) Longitudinal stationary waves are 2.5 formed in the tube only when the length of 10 12 2r 2 the glass tube reaches some particular values. 2.5 (c) v = f = 2 f L where L is the r 6.31 105 m 2 10 12 distance between successive heaps (d) There is attenuation of sound. A = 2 x 500 x 0.33 = 330 m s-1 more realistic distance is a few kilometers. (d) Their velocities and wavelengths ---------------------------------------------- are different. 2. When they are sounded together, the ---------------------------------------------- intensity received is doubled, i.e. I’ = 2I. 5. (a) Required intensity level 2I I 10 log 10 log 2 10 log 3.01 70 73 dB Io Io 256 Hz ---------------------------------------------- 3. (a) This is to ensure that the phase difference between the two loudspeakers stays constant with time. The frequencies of distance two separate signal generators are not exactly the same. (b) (b) (i) Sound of maximum intensity is v vr heard. The intensity decreases along AB. vv fr fs s (ii) Maximum and minimum intensity of sound is heard alternately. 330 270 256 330 v v s 17.1 m s -1 Distance between successive maxima s D vD 340 20 3.33 m (c) f r 256 330 a fa 680 3 330 (17.1) 243 Hz (c) There is a minimum intensity of ------------------------------------------------ sound at Z because a wave from L1 arrives 6. (a) Microwaves there exactly out of phase with a wave from f 2u 1.03 103 2u L2. Destructive interference occurs. When L2 is disconnected, no such cancellation (b) f c 1 1010 3 108 occurs, so only one set of wave arrives, so u 15.4 m s -1 the intensity increases locally. At other points where there was originally a maximum intensity signal, there will be a reduction in intensity when L2 is disconnected. Overall energy is conserved. = 10 log I / Io = 60 – 10 log 4 = 54 dB (c) Beat frequency f = (2u/c) f ---------------------------------------------------- % error of beat frequency 3. 50 = 10 log I / Io where = % error of u I = P / 4π(2)2 The lower limit of the car speed Now P 2P, 2 m 6 m, so = 15.4 x 0.97 = 14.9 m s-1 I 2P / 4π (6)2 = (2/9) I Local speed limit = 50 x 1000/3600 So required intensity level = 13.9 m s-1 = 10 log (2/9) I / Io So the car speed still exceeds the = 10 log 2/9 + 50 = 43.5 dB speed limit. ---------------------------------------------------- ---------------------------------------------- 4. Stationary waves are formed because 7. (a) The reflected sound can be regarded the sound intensity varies. The formation of as emitted from the image of the bat which stationary wave is due to superposition of is moving towards the wall at the same two waves, of same frequency, (one incident speed 10.0 m s-1. and one reflected) traveling in opposite directions. So statement 1 is incorrect. v vr 340 0 Positions of soft sound corresponds to fr fs vv 60000 s 340 10 nodes. 6.18 kHz Wavelength = v/f = 340 / 1000 = 0.34 m So distance between successive nodes is (b) The bat itself is a moving receiver. 0.34 / 2 = 0.17 m. So statement 2 is incorrect. v vr 340 (10) fr fs vv 60000 Increasing frequency will lower the s 340 10 wavelength, so positions of soft sound 63.6 kHz (nodes) become closer. ---------------------------------------------------- ---------------------------------------------------- Multiple Choice 5. Statement 2 is correct because ultrasonic has high frequency, so the 1-5 CBBBA wavelength is short, as a result, diffraction 6-10 B D D C C effect is less significant. (Longer 11-13 B C cancelled or A wavelength has more significant 14-15 B C diffraction). ---------------------------------------------------- ---------------------------------------------------- 1. 6. In a stationary longitudinal wave, v vr displacement antinode = pressure node vv fr fs ---------------------------------------------------- s 7. 60 = 10 log (I60 / Io) => 106 = I60 / Io 336 0 40 = 10 log (I40 / Io) => 104 = I40 / Io 300 fs => I60 / I40 = 106 / 104 = 100 336 v s ---------------------------------------------------- 336 0 8. 150 fs 336 (v s ) On solving, we have vs = 112 m/s ---------------------------------------------------- 2. Let I be the intensity due to one loudspeaker. Since intensity is proportional to (amplitude)2, so the resulting intensity at a point of constructive interference will be 4I (not 2I). 4I 60 10 log =10 log 4 + 10 log I / Io Suppose observer O stands at O. Io Without loss of generality, suppose the boy Required intensity level is moving in anticlockwise direction and is at A at t = 0. Then at A, to observer O, he is neither approaching nor receding, so the 15. R, S and T are positions of nodes. frequency remains unchanged at fo. = 3 cm = 0.03 m At B, the boy is moving away from O, and f = v / = 3 x 108 / 0.03 = 1 x 1010 Hz the drop in frequency must be the largest. At ---------------------------------------------------- C, he is again neither approaching nor receding, so the frequency is unchanged. At D, the boy is approaching towards O, and the rise in frequency would be the largest. The time for travelling from B to D is obviously longer that that from D to B, so the answer is D. ---------------------------------------------------- 9. Wavelength is unchanged for stationary source. = v/f = 300 / 600 = 0.5 m ---------------------------------------------------- 10. For a closed tube, fo = v / 4l, f1 = 3fo. ---------------------------------------------------- 11. Δf / f = 2u / c 2uf 2u f c 2 100 103 / 3600 f 1900Hz 0.03 ---------------------------------------------------- 12. v vr vv fr fs s 300 0 462 500 300 ( v ) s vs = 24.7 m/s ---------------------------------------------------- 13. The first resonance position corresponds to the lowest frequency with simplest mode of vibration. = 4 l => fo = v / = v / 4l For a closed tube, only odd harmonics are present. f1 = v / = v / (4l / 3) = 3v / 4l = 3 fo So if the length is unchanged, the next higher frequency corresponds to 3fo. Now if l is increased to 3l (by lowering the water level), then f will be kept unchanged. So the required length should be 5 cm x 3 = 15 cm ---------------------------------------------------- 14. 70 = 10 log I / Io Required intensity level = 10 log 2I / Io = 10 log 2 + 10 log I / Io = 3.01 + 70 = 73.01 dB ----------------------------------------------------

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User's Manual, Legal terms, Local Area, Brisbane Water, Catchment Study, Water Catchment, ILOG Solver, Transportation Industry Solutions, Exercise 3, comment out

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posted: | 3/28/2011 |

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