Question

A point source of power $50\, \pi$ watts is producing sound waves of frequency $1875\, Hz$. The velocity of sound is $330\, m / s$, atmospheric pressure is $1.0 \times 10^{5} Nm ^{-2}$, density of air is $\frac{400}{99 \pi} \,kgm ^{-3}$. Then the displacement amplitude at $r=\sqrt{330} \,m$ from the point source is

• A. $0.5\, \mu m$
• B. $0.2\, \mu m$
• C. $1\, \mu m$
• D. $2\, \mu m$

Answer 1

Answer: (C)

$P_{0}=B K S_{0} ; k=\frac{2 \pi}{\lambda} ;$
$\lambda=\frac{v}{f} ; v=\sqrt{\frac{B}{\rho}}$
Using above, we get
$S_{0}=\frac{P_{0}}{2 \rho v \pi f}$
$=\frac{5}{2 \times 1 \times 330 \times 3.14 \times 1875}$
$=1 \mu$ meter