Halliday 2 Cap 18 - P18 074

Transcript

74. (a) The wavelength of the sound wave is λ= v 343 m/s = = 0.343 m . f 1000 Hz (b) From ∆pm = v 2 ρksm = 2πvρf sm we find sm = ∆pm 10.0 Pa = = 3.83 × 10−6 m . 3 2πvρf (2π)(343 m/s)(1.21 kg/m )(1000 Hz) (c) The velocity of the particle is the derivative of the sinusoidal wave function with respect to time. Its maximum value is vm = 2πf sm = (3.60 × 10−6 m)(2π)(1000 Hz) = 2.41 × 10−2 m/s. (d) From Eq. 18-38, we obtain L= 0.343 m λ = = 0.172 m . 2 2