Question

A transverse wave is expressed as, $y=y_0\sin 2\pi ft$.
For what value of $\lambda$, maximum particle velocity equals to 4 times the wave velocity?

• A. $y_0\frac{\pi }{2}$
• B. $2y_0\pi$
• C. $y_0\pi$
• D. $y_0\frac{\pi }{4}$

Answer 1

The correct option is A $y_0\frac{\pi }{2}$

Wave velocity, $v=f\lambda$

Particle velocity is obtained by differentiating the given equation with respect to time.

$v_p=\frac{dy}{dt}=y_0\times \left(2\pi f\right)\times cos\left(2\pi ft\right)$

Particle velocity is maximum when $t=0$

Maximum particle velocity $=y_0\times \left(2\pi f\right)$

According to given condition in the question,

$\Rightarrow y_0\times \left(2\pi f\right)$$=4f\lambda$

This gives $\lambda =y_0\left(\frac{\pi }{2}\right)$

Hence, option A is correct.