Question

A transverse wave is represented as $y = A\sin (\omega \,t - kx)$. For what value of the wavelengths is the wave velocity equal to the maximum particle velocity?

Answer 1

Hint: Wave motion is defined as the motion of a disturbance through a medium from one place to another.
Mathematically, it is expressed as $y = A\sin (\omega \,t - kx)$ where y is the displacement of the wave, k is the wave number, A is the amplitude of the wave and $\omega $ is the angular frequency.
We can express angular frequency $\omega $ , in terms of its frequency as $\omega = 2\pi f$ .
There are two types of wave velocities.
1. Wave velocity
2. Particle velocity

Complete step by step solution:
 The mathematical expression for the wave velocity is ${v_w} = \dfrac{\omega }{k}$ .
The mathematical expression of the particle velocity is ${v_p} = \dfrac{{dy}}{{dt}}$ .
We know that the displacement of the wave is defined as $y = A\sin (\omega \,t - kx)$ .
Substituting in the equation for the particle velocity,
${v_p} = \dfrac{{dA\sin (\omega \,t - kx)}}{{dt}}$
$ \Rightarrow {v_p} = A\omega \cos (w\,t + kx)$ .
Now since $\cos (w\,t + kx)$ can range from -1 to 1, its maximum value is 1.
Hence the maximum value of the particle velocity would be when $\cos (w\,t + kx) = 1$ .
Thus, the maximum particle velocity is ${v_p} = A\omega $ .
Equating the maximum particle velocity with the wave velocity,
$\dfrac{\omega }{k} = A\omega $ .
$ \Rightarrow k = \dfrac{1}{A}$ .
We know that $k = \dfrac{{2\pi }}{\lambda }$
Substituting in the equation, we get
$\dfrac{1}{A} = \dfrac{{2\pi }}{\lambda }$
Rearranging the terms we get,
$\lambda = 2\pi A$
Hence option C is the correct answer.

Note:
$y = A\sin (\omega \,t - kx)$ represents the motion of wave along the positive direction of x axis though there is a negative sign with x. $y = A\sin (\omega \,t + kx)$ represents the motion of wave along the negative direction of x axis. We must keep in mind the direction of the motion of the wave while solving any numerical since it becomes imperative while solving a question to superpose the waves.