Question

A transverse wave is represented by the equation $y = {y_0}\sin \dfrac{{2\pi }}{\lambda }\left( {vt - x} \right)$. For what value of $\lambda$ is the maximum particle velocity equal to two times of the wave velocity?
A. $\lambda = 2\pi {y_0}$
B. $\lambda = \dfrac{{\pi {y_0}}}{3}$
C. $\lambda = \dfrac{{\pi {y_0}}}{2}$
D. $\lambda = \pi {y_0}$

Answer 1

Hint: The equation in the question, given that the equation of the displacement of the particle in the wave $\left( y \right)$. By differentiating this equation with respect to the time $t$, we get the value of the velocity of the particle $\left( {\dfrac{{dy}}{{dt}}} \right)$. By using that velocity, we can obtain the relation between the maximum velocity of the particle and the velocity of wave $v$. Using that relation, the value of $\lambda$ can be calculated.

Useful data:
In a transverse wave, the maximum velocity of the wave is two times the wave velocity.

Step by step solution:
Transverse waves:
The transverse waves are the waves which are in motion that have the oscillation of the wave is perpendicular or normal to the motion of the wave.
Vibration of string, sunlight are some examples of transverse waves.

Assume that,
The velocity of the wave is $v$
The velocity of particle in the wave is $\dfrac{{dy}}{{dt}}$
The maximum velocity of the particle in the wave is ${\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }}$

Given equation,
$y = {y_0}\sin \dfrac{{2\pi }}{\lambda }\left( {vt - x} \right)$
Differentiating the above equation with respect to the time $t$, we get
$\dfrac{{dy}}{{dt}} = {y_0}\cos \dfrac{{2\pi }}{\lambda }\left( {vt - x} \right) \times \dfrac{{2\pi }}{\lambda }v\;.......................................\left( 1 \right)$

To obtain the maximum value of $\dfrac{{dy}}{{dt}}$, the value of $\cos \dfrac{{2\pi }}{\lambda }\left( {vt - x} \right)$ must be $1$.
So, assume that, the value of $x$ is $vt$ and substitute the value of $x$ in equation (1),
${\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }} = {y_0}\cos \dfrac{{2\pi }}{\lambda }\left( {vt - vt} \right) \times \dfrac{{2\pi }}{\lambda }v\; \\\ {\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }} = {y_0}\cos \dfrac{{2\pi }}{\lambda }\left( 0 \right) \times \dfrac{{2\pi }}{\lambda }v\; \\\ {\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }} = {y_0}\cos 0 \times \dfrac{{2\pi }}{\lambda }v \\\$
Since, $\cos 0 = 1$
${\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }} = {y_0}\left( 1 \right) \times \dfrac{{2\pi }}{\lambda }v \\\ {\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }} = {y_0} \times \dfrac{{2\pi }}{\lambda }v\;..................................\left( 2 \right) \\\$
From useful data, ${\left( {\dfrac{{dy}}{{dt}}} \right)_{\max }} = 2v$, substitute in equation (2),
$2v = {y_0} \times \dfrac{{2\pi }}{\lambda }v \\\ \lambda = {y_0} \times \dfrac{{2\pi }}{{2v}}v \\\ \lambda = \pi {y_0} \\\$
Hence, the option (D) is correct.

Note: In the transverse wave, the maximum particle velocity should be twice the velocity of the wave. The given relation defines the displacement parameter of the particle. Hence by differentiating it with respect to time factor, the velocity of the particle will be obtained. Then, relating the maximum velocity of the particle to the wave velocity results in calculating the value of $\lambda$.