# A transverse wave is given by the equation y=y_0 sin 2pi(ft-x/lambda) The maximum particle velocity will be 4 times the wave velocity if, A. lambda =(pi y_0)/4 B.lambda =(pi y_0)/2 C.lambda =pi y_0 D.lambda =2 pi y_0 ?

Mar 16, 2018

$B$

#### Explanation:

Comparing the given equation with $y = a \sin \left(\omega t - k x\right)$ we get,

amplitude of particle motion is $a = {y}_{o}$ , $\omega = 2 \pi f$ ,$\nu = f$ and wavelength is $\lambda$

Now,maximum particle velocity i.e maximum velocity of S.H.M is $v ' = a \omega = {y}_{o} 2 \pi f$

And,wave velocity $v = \nu \lambda = f \lambda$

Given condition is $v ' = 4 v$

so,${y}_{o} 2 \pi f = 4 f \lambda$

or,$\lambda = \frac{\pi {y}_{o}}{2}$