Question

If equation od a transverse wave is $y=x_0\cos 2\pi (nt-\frac{x}{\lambda })$. Maximum velocity of particle is twice of wave velocity, if $\lambda$ is:-

• A. $\frac{\pi }{2x_0}$
• B. $2\pi x_0$
• C. $\pi x$
• D. $\pi x_0$

Answer 1

The correct option is D $\pi x_0$

Given,

$y=x_{o}cos2\pi (nt-\frac{x}{\lambda })$

and $v_y=2\times vwhere,v_y=particlevelocity\&v=wavevelocity$

Comparing the given equation with a standard harmonic equation of the form

$y=r\cos (kx-\omega t)$

We get,

$\omega =2\pi n\Rightarrow 2\pi f=2\pi n\Rightarrow f=n$

Maximum particle velocity $v_y=\frac{\partial y}{\partial t}$

$v_y=-2x_o\pi n\cos 2\pi (nt-\frac{x}{\lambda })$

Max particle velocity is $v_{ymax}=2x_o\pi n$

So,

$2\times v=v_y$

$2\times \lambda f=2x_o\pi n(v=\lambda fandn=f)$

$\lambda =\pi x_o$

So D is the correct answer