P18_063.PDF

63. (a) Since ω =2 πf , Eq. 18-15 leads to

∆ p m = vρ (2 πf ) s m =

⇒

s m =

10 − 3 Pa

2 π (1665Hz)(343m / s)(1 . 21kg / m 3 )

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which yields s m =0 . 26 nm. The nano preﬁx represents 10 − 9 . We use the speed of sound and air

density values given at the beginning of the exercises and problems section in the textbook.

(b) WecanplugintoEq.18-27orintoitsequivalentform, rewrittenintermsofthepressureamplitude:

I = 1

(∆ p m ) 2

ρv

= 1

10 − 3 Pa 2

(1 . 21kg / m 3 )(343m / s) =1 . 5nW / m 2 .

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