Question

A tranverse wave is described by the equation $Y=Y_{0}sin2\pi (ft-x/\lambda )$. The maximum particle velocity is equal to four times the wave velocity and $Y_0=1/\pi$, then value of $\lambda$ is

Answer 1

$v=\frac{dY}{dt}=Y_{0}cos\left[2\pi (ft-\frac{x}{\lambda })\right]\times 2\pi f$
or $v=2\pi f{Y}_{0}cos\left[2\pi (ft-\frac{x}{\lambda })\right]$

The particle velocity is maximum, when $cos\left[2\pi (ft-\frac{x}{\lambda })\right]=1$
$\therefore v_{max}=2\pi f{Y}_0$ $\cdots \left(1\right)$

We know that, $Y=asin(\omega t-kx)$
The wave velocity V is given by
$V=\frac{\omega }{k}=\frac{2\pi f}{2\pi /\lambda }=f\lambda$ $\cdots \left(2\right)$
Given that $V_{max}=4V$
$\therefore 2\pi f{Y}_0=4f\lambda$
or $\lambda =\frac{\pi Y_o}{2}$
As, $Y_0=1/\pi ⟹\lambda =0.5$