Question

A transverse wave is described by the equation $y=y_0\sin 2\pi \left(ft-\frac{x}{\lambda }\right).$ 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. $\lambda =2\pi y_0$
• E. $\lambda =\frac{2\pi y_0}{3}$

Answer 1

Answer: (B)

$y=y_0\sin 2\pi \left(ft-\frac{x}{\lambda }\right)$ .. (i)
For particle velocity, $\frac{dy}{dt}=2\pi f{y}_0\cos 2\pi \left(ft-\frac{x}{\lambda }\right)$
Maximum particle velocity, ${\left(\frac{dy}{dt}\right)}_{\max }=2\pi f{y}_0$
Wave velocity $=f\lambda$
Accordingly,
$2\pi f{y}_0=4(f\lambda )$
Or $\lambda =\frac{\pi y_0}{2}$