Question

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

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

Answer 1

The correct option is A $1\mu m$

Given, power of point source, $P_o=50\pi$ Watts
Frequency of sound wave, $f=1875Hz$
Velocity of sound, $v=330m/s$
Density of air, $\rho =\frac{400}{99\pi }kg/m^3$
As we know that the relation between pressure amplitude and displacement amplitude is given by $P_o=Bk{S}_o$
$\therefore S_0=\frac{P_0}{Bk}$ ... (i)
Where $B=$ Bulk modulus, $k=2\pi /\lambda$ and $\lambda =v/f$
Therefore $f=2\pi f/v$
From Newton's formula the speed of sound is
$v=\sqrt{\frac{B}{\rho }}$
$v^2\rho =B$
So, from equation (i)
$S_0=\frac{P_0}{\frac{v^2\rho 2\pi f}{v}}$
Now the intensity is given by
$T=\frac{P_0^2}{2\rho v}$
$\frac{P}{4\pi r^2}=\frac{P_0^2}{2\rho v}$
$P_0=\sqrt{\frac{P\rho v}{2\pi r^2}}$
Substituting the values, we get
$P_0=5$
$S_0=\frac{P_0}{\rho 2\pi fv}$
$S_0=\frac{5}{2\times 3.14\times 1875\times 330}$
$S_0=1\mu m$