Question

A transverse wave is described by the equation $Y=Y_0\sin 2\pi (ft-\frac{x}{\lambda })$. The maximum particle velocity is equal to four times the wave velocity if:

• A. $\lambda =\frac{\pi Y_0}{4}$
• B. $\lambda =\frac{\pi Y_0}{2}$
• C. $\lambda =\pi Y_0$
• D. $\lambda =2\pi Y_0$

Answer 1

The correct option is B $\lambda =\frac{\pi Y_0}{2}$

$\text{Solution}:$

$y=y_{0}sin2\pi [ft-\frac{x}{\lambda }]$

$\therefore \frac{dy}{dt}=\left[y_{0}cos2\pi \right(ft-\frac{x}{\lambda }]\times 2\pi f$

$⟹[\frac{dy}{dt}{]}_{max}=y_0\times 2\pi f$

Given that the maximum particle velocity is equal to four times the wave velocity $(c=f\lambda )$

$\therefore y_0\times 2\pi f=4f\times \lambda$

$\lambda =\frac{\pi y_0}{2}$

$\text{Hence B is the correct option}$