505_Physics ProblemsTechnical Physics

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online