500_Physics ProblemsTechnical Physics

500_Physics ProblemsTechnical Physics - 502 P17.11 Sound...

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502 Sound Waves P17.11 (a) A = 200 . m µ λ π == = 2 15 7 0400 400 . .. m c m v k = ω 858 15 7 54 6 . . ms (b) s =− × = 157 00500 858 300 10 0433 3 .c o s . . . . af bg ej m (c) vA max = ωµ 858 172 1 s m m s P17.12 (a) P xt F H G I K J 127 340 .s i n Pa ms ππ (SI units) The pressure amplitude is: P max . = 127±Pa . (b) ωπ 23 4 0 f s, so f = 170 Hz (c) k 2 m, giving = . m (d) vf = 170 . m Hz 340 m s a fa f P17.13 k = 22 0100 62 8 1 . . m m a f = × 2 4 3 216 10 41 v m s . . Therefore, Px t × 0 200 62 8 2 16 10 4 i n . . Pa m s . P17.14 == = = × 2 2 4 3 4 f v m rad s a f . . s P v max max . . × ρω 0200 120 343 225 10 8 Pa kg m m s s m 3 k = 62 8 1 . . m m Therefore, ss k x t x t = × × max cos . cos . . e j 2 2 51 0 6 2 8 2 1 61 0 84 mm s . P17.15 Pv sv v s max max max
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