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Question: Consider a plane standing sound wave of frequency 10<sup>3</sup> Hz in air at 300 K. Suppose the amp...

Consider a plane standing sound wave of frequency 103 Hz in air at 300 K. Suppose the amplitude of pressure variation associated with this wave is 1 dyne/cm2. The equilibrium pressure is 106 dyne/cm2. The amplitude of displacement of air molecules associated with this wave is :

(Given speed of sound : 340 m/s

Molar mass of air : 29 × 10–3 kg/mol)

A

4 × 10–6 m

B

40 × 10–6 m

C

400 × 10–6 m

D

40000 × 10–6 m

Answer

400 × 10–6 m

Explanation

Solution

y = A sin (wt – kx)

DP = –B ∂y∂x\frac{\partial y}{\partial x} = + BAk cos (wt – kx)

DP = DPm cos (wt – kx)

DPm = BAk = BAων\frac{BA\omega}{\nu}

A = ΔPmvBω\frac{\Delta P_{m}v}{B\omega} = ΔPmvρv2ω\frac{\Delta P_{m}v}{\rho v^{2}\omega}

A = ΔPmρv2πf\frac{\Delta P_{m}}{\rho v2\pi f}

f = 103 Hz

DPm = 1 dyne/m2

v = 340 m/s

M = 29 × 10–3 kg/mole

T = 300 K

where ρ=PMRT\begin{matrix} \rho = \frac{PM}{RT} \end{matrix}