# 16 - CHAPTER 16 Sound 5 If we let L1 represent the...

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CHAPTER 16 – Sound 5. If we let L 1 represent the thickness of the top layer, the total transit time is t = ( L 1 / v 1 ) + [( L L 1 )/ v 2 )]; 4.5 s = [ L 1 /(331 m/s)] + [(1500 m – L 1 )/(343 m/s)], which gives L 1 = 1200 m . Thus the bottom layer is 1500 m – 1200 m = 300 m . 7. We find the displacement amplitude from Δ P M = 2 π ρ vD M f . ( a ) For the frequency of 100 Hz, we have 3 . 0 × 10 –3 Pa = 2 π (1.29 kg/m 3 )(331 m/s) D M (100 Hz), which gives D M = 1.1 × 10 –8 m . ( b ) For the frequency of 10 kHz, we have 3 . 0 × 10 –3 Pa = 2 π (1.29 kg/m 3 )(331 m/s) D M (10 × 10 3 Hz), which gives D M = 1.1 × 10 –10 m . 11. ( a ) We find the intensity level from β = 10 log 10 ( I / I 0 ) = 10 log 10 [(8.5 × 10 –8 W/m 2 )/(10 –12 W/m 2 )] = 49 dB . ( b ) We find the intensity from = 10 log 10 ( I / I 0 ); 25 dB = 10 log( I /10 –12 W/m 2 ), which gives I = 3.2 × 10 –10 W/m 2 . 21. The intensity with one firecracker is one-half that of two firecrackers. Thus the change in intensity level is 2 1 = 10 log 10 ( I 2 / I 1 ); 2 – 90 dB = 10 log 10 ( ½ ), which gives 2 = 87 dB . 25. ( a ) We find the intensity of the sound from = 10 log 10 ( I / I 0 ); 130 dB = 10 log( I /10 –12 W/m 2 ), which gives I = 10 W/m 2 . The power output of the speaker is P = IA = I 4 π r 2 = (10 W/m 2 )4 π (3.4 m) 2 = 1.5 × 10 3 W . ( b ) We find the intensity of the sound from = 10 log 10 ( I / I 0 ); 90 dB = 10 log( I /10 –12 W/m 2 ), which gives I = 1.0 × 10 –3 W/m 2 . We find the distance from

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16 - CHAPTER 16 Sound 5 If we let L1 represent the...

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