505_Physics ProblemsTechnical Physics

# 505_Physics ProblemsTechnical Physics - Chapter 17 P17.27(a...

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Chapter 17 507 P17.27 (a) 120 10 10 12 2 dB dB Wm = L N M M O Q P P log I I r r I == = 100 4 4 600 0691 2 . . . W 41 . 0 0 m 2 2 π ej We have assumed the speaker is an isotropic point source. (b) 01 0 10 12 dB dB ±Wm 2 = F H G I K J log I I r I = = × = 100 10 4 691 12 . . W . 0 0 Wm km 2 -12 2 We have assumed a uniform medium that absorbs no energy. P17.28 We begin with β 2 2 0 10 = F H G I K J log I I , and 1 1 0 10 = F H G I K J log I I , so ββ 21 2 1 10 −= F H G I K J log I I . Also, I r 2 2 2 4 = , and I r 1 1 2 4 = , giving I I r r 2 1 1 2 2 = F H G I K J . Then, 1 2 2 1 2 10 20 F H G I K J = F H G I K J log log r r r r . P17.29 Since intensity is inversely proportional to the square of the distance, II 40 4 1 100 = . and I P v 04 2 2 2 10 0 2 1 20 343 0121 . max . . . = ρ af a f 2 . The difference in sound intensity level is = F H G I K J =− = 10 10 2 00 20 0 4 log . . I I km 0.4 km
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