Question

A plane wave is described by the equation y = 3 cos(x/4-10t-π/2). The maximum velocity of the particles of the medium due to this wave is
A. $30$
B. $\dfrac{{3\pi }}{2}$
C. $\dfrac{3}{4}$
D. $40$

Answer 1

Hint: The maximum velocity of a wave is obtained at its maximum displacement i.e., amplitude. The product of amplitude and angular velocity is the maximum velocity of the particles of the medium.

Complete step by step answer:A simple plane wave is represented by the equation y = a sin(ωt-kx) where y is the displacement of the particle at any point t, ω is the angular velocity, k is known as propagation constant or angular wave number and its value is equal to 2π/λ where λ is the wavelength , x is a distance and a is the amplitude of vibration of the particles of a wave. The value of k gives us the change of phase per unit path difference. The path difference is the difference in the path traversed by the two waves and phase difference is the difference in the phase angle of the two waves.
In the equation, $y = 3\cos \left( {\dfrac{x}{4} - 10t - \dfrac{\pi }{2}} \right) = 3\cos \left\{ { - \left( {\dfrac{\pi }{2} + 10t - \dfrac{x}{4}} \right)} \right\} = 3\cos \left( {\dfrac{\pi }{2} + 10t - \dfrac{x}{4}} \right)\left[ {\cos \left( { - x} \right) = \cos x} \right]$
So, now the equation is $y = 3\cos \left( {10t - \dfrac{x}{4} + \dfrac{\pi }{2}} \right)$ where $a = 3,\omega = 10,k = \dfrac{1}{4}$ and $\dfrac{\pi }{2}$is the initial phase difference of the wave.
The wave attains its maximum velocity at its amplitude and let the maximum velocity be v.
$v = a\omega \left[ {\because v = r\omega } \right]$
$v = 3 \times 10 = 30$
Hence, the maximum velocity of the particles of the medium due to this wave is 30.
Therefore, option A is correct.

Note:The maximum velocity of the vibrating particles of the wave is obtained at the point of its maximum displacement from its still position to the top of a crest (upward displacement) or to the bottom of a trough (downward displacement).