Question

A plane wave propagates along positive x-direction in a homogeneous medium of density $\rho =200kg/m^3.$ Due to propagation of the wave medium particles oscillate. Space density of their oscillation energy is $E=0.16{\pi }^{2}J/m^3$ and maximum shear strain produced in the medium is ${\varphi }_0=8\pi \times {10}^{-5}.$ If at an instant, phase difference between two particles located at points (1m , 1m , 1m) and (2 m , 2m , 2m) is $\Delta \theta ={144}^{\circ },$ assuming at $t=0$ phase of particles at $x=0$ to be zero,
Equation of wave is

• A. $y={10}^{-4}sin\pi (2000t-0.8x)$
• B. $y={10}^{-4}sin\pi (400t-0.8x)$
• C. $y={10}^{-4}sin\pi (100t-8x)$
• D. $y={10}^{-4}sin\pi (100t-2x)$

Answer 1

Answer: (B)

$y={10}^{-4}sin\pi (400t-0.8x)$

Since, the wave is a plane travelling wave, intensity at every point will be the same.
Since, initial phase of particle at $x=0$ is zero and the wave is travelling along positive x-direction equation of the wave will be of the form
$\delta =asin\omega (t-\frac{x}{v})$ (i)
Let intensity of the wave be I, then space density of oscillation energy of medium particles will be equal to
$E=\frac{I}{v}$
But, $I=2{\pi }^2{n}^2{a}^2\rho v$
Therefore $E=2{\pi }^2{n}^2{a}^2\rho =0.16{\pi }^{2}J/m^3$
$a^2{n}^2=4\times {10}^{-4}$
or, $an=0.02$
Shear strain of the medium is
$\varphi =\frac{d}{dx}\delta$
Differentiating Eq. (i),
$\varphi =-\frac{a\omega }{v}cos\omega (t-\frac{x}{v})$
Modulus of shear strain $f$ will be maximum when
$cos\omega (t-\frac{x}{v})=\pm 1$
$\therefore$ Maximum shear strain $8\pi \times {10}^{-5}$
${\varphi }_0=\frac{a\omega }{v}$
but it is equal to
$\frac{a\omega }{v}=8\pi \times {10}^{-5}$
where
$\omega =2\pi n$
$an=4v\times {10}^{-5}$ (iii)
Solving Eqs. (ii) and (iii), $v=500m/s$
Since, the wave is travelling along positive x-direction, therefore, phase difference between particles at points (1m , 1m, 1m) and (2m , 2m, 2m) is due to difference between their x-coordinates only.
The phase difference is given by
$\Delta \theta =2\pi \frac{\Delta x}{\lambda }$
$\Delta x=(x_2-x_1)=(2-1)m=1m$
$\lambda =\frac{2\pi \Delta x}{\Delta \theta }=2.5m$
But $v=n\lambda ,$ therefore,
$n=\frac{v}{\lambda }=200Hz$
Substituting $n=200Hz$ in Eq. (ii),
$a=1\times {10}^{-4}m$
Angular frequency, $\omega =2\pi n=400\pi rad/s.$ Substituting all these values in Eq. (i),
$\delta ={10}^{-4}sin\pi (400t-0.8x)m$
Since, due to propagation of the wave, shear strain is produced in the medium, the wave is a plane transverse wave.