Question

In a transverse wave of amplitude A, the maximum particle velocity is four times wave velocity, then the wave length of the wave is :

• A. ($\pi$ A)/4
• B. ($\pi$ A)/2
• C. $\pi$ A
• D. 2$\pi$ A

Answer 1

The correct option is B ($\pi$ A)/2

$k=\frac{\omega }{V}=\frac{2\pi }{\lambda }$
$y=Asin\omega t$ ; $Vp=A\omega cos\omega t$
$V{p}_{max}=A\omega$ $=\frac{A\times 2\pi }{T}$
$Vp=4V$ $\to \frac{A\times 2\pi }{T}=4\times \frac{\lambda }{T}$
$\lambda =\frac{A\pi }{2}$