Question

Plane harmonic wave of frequency $500\, Hz$ are produced in air with displacement amplitude of $10$ micron. Given that density of air is $1.29\, kgm ^{-3}$ and speed of sound in air is $340\, ms ^{-1}$ :

• A. The pressure amplitude is $13.76\, Nm ^{-2}$
• B. The energy density is $6.45 \times 10^{-4}\, Jm ^{-3}$
• C. The energy flux is $0.22\, Jm ^{-2} s ^{-1}$
• D. Only (A) and (C) are correct

Answer 1

Answer: (A), (B), (C)

Given that
$n=500\, Hz$
$A=10 \times 10^{-6}\, m$
$\rho=1.29\, kg / m ^{3}$
$v=340\, m / s$
We use $v=\sqrt{\frac{B}{\rho}}$
$\Rightarrow B=\rho v^{2}=1.29 \times(340)^{2}=1.49 \times 10^{5} N / m ^{2}$
Wavelength $\lambda=\frac{v}{\lambda}=\frac{340}{500}=0.68\, m$
Pressure amplitude
$\Delta P=\frac{2 \pi}{\lambda} A B$
$=\frac{2 \pi}{0.68} \times 10 \times 10^{-6} \times 1.49 \times 10^{5}=13.76\, m$
Energy density $\frac{P}{S v} =2 \pi^{2} n^{2} A^{2} \rho$
$=2 \times 10 \times(500)^{2} \times\left(10 \times 10^{-6}\right)^{2} \times 1.29$
$=6.45 \times 10^{-4} J / m ^{3}$
Energy flux $u_{\phi}=\frac{P}{S}=2 \pi^{2} n^{2} A^{2} \rho v$
$=0.22\, J / m ^{2}- s$