Question

The equation of a longitudinal standing wave due to superposition of the progressive waves produced by two sources of sound is $s=-20\sin \left(10\pi x\right)\times \sin \left(100\pi t\right)$ where $s$ is the displacement from mean position measured in $mm,x$ is in meters and $t$ is in seconds. The specific gravity of the medium is ${10}^{-3}$. Density of water $={10}^{3}kg/m^3$. Find:
(a) Wavelength, frequency and velocity of the progressive waves.
(b) Bulk modulus of the medium and the pressure amplitude.
(c)Minimum distance between pressure antinode and a displacement antinode
(d)intensity at the displacement node.

Answer 1

Given,

$S=-20\sin \left(10\pi x\right)\sin \left(100\pi t\right)=20\left[\frac{\cos (10\pi x+100\pi t)-\cos (10\pi x-100\pi t)}{2}\right]=10\cos (10\pi x+10\pi t)-10\cos (10\pi x-100\pi t)=10\sin (10\pi x+100\pi t+\pi /2)-10\sin (10\pi x-100xt+\pi /2)=10\sin (10\pi x+100\pi t+\pi /2)+10\sin (10\pi x-100\pi t+3\pi /2)$

$\Rightarrow A=10,\omega =100\pi ,k=10\pi f=\frac{\omega }{2\pi }=50Hz,T=\frac{1}{f}=0.02sec$

Now, $\omega =kv\Rightarrow v=\frac{\omega }{k}\Rightarrow v=\frac{100\pi }{10\pi }=10m/s\lambda =\frac{v}{f}=\frac{10}{50}=0.2m\left(a\right)\lambda =0.2m,f=50Hz,v=10m/s\left(b\right)k=10\pi ,$

Pressure amplitude $P_m=\left(ve\omega \right)sm=10\times 1\times 100x\times 10={10}^4\pi$

$\left(c\right)$ Minimum dist$=\frac{\lambda }{4}=\frac{0.2}{4}=0.05m$

$\left(d\right)I=\frac{1}{2}e\times v\times (A\omega {)}^2=0$

$\because$ At displacement node $y=0$