Halliday 2 Cap 18 - P18 063

Transcript

63. (a) Since ω = 2πf , Eq. 18-15 leads to ∆pm = vρ(2πf )sm =⇒ sm = 1.13 × 10−3 Pa 2π(1665 Hz)(343 m/s)(1.21 kg/m3 ) which yields sm = 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) We can plug into Eq. 18-27 or into its equivalent form, rewritten in terms of the pressure amplitude: 2
2 1.13 × 10−3 Pa 1 (∆pm ) 1 I= = = 1.5 nW/m2 . 2 ρv 2 (1.21 kg/m3 ) (343 m/s)