Question

A string of length $l$ is fixed at both ends and is vibrating in second harmonic. The tension in string is $T$ and the linear mass density of string is $\mu$. The ratio of magnitude of maximum velocity of particle and the magnitude of maximum acceleration is

• A. $\frac{\lambda }{2\pi }\sqrt{\frac{\mu }{T}}$
• B. $\frac{\lambda }{\pi }\sqrt{\frac{\mu }{T}}$
• C. $\frac{\lambda }{2\pi }\sqrt{\frac{2\mu }{T}}$
• D. $\frac{\lambda }{2\pi }\sqrt{\frac{\mu }{2T}}$

Answer 1

The correct option is A $\frac{\lambda }{2\pi }\sqrt{\frac{\mu }{T}}$

In second harmonic $\lambda =2L/n;n=2$

so, $\lambda =\frac{2L}{2}=L$

The velocity is $v=\sqrt{T/\mu }$

and frequency is $\frac{v}{\lambda }=\frac{\sqrt{T/\mu }}{L}$

so angular frequency is $\omega =2\pi \nu =2\pi \frac{\sqrt{T/\mu }}{L}-\left(1\right)$

Equation of standing wave is given as:

$y=\left(2a\sin kx\right)\cos \omega t$

so velocity is $v\left(t\right)=\frac{dy}{dt}=-\left(2a\omega \sin kx\right)\sin \omega t$

its maximum value is $\left(2a\omega \sin kx\right)$

and acceleration is $a\left(t\right)=\frac{dv}{dt}=\left(2a{\omega }^2\sin kx\right)\cos \omega t$

its maximum value is $\left(2a{\omega }^2\sin kx\right)$

The ratio of magnitude of maximum velocity to maximum acceleration is

$\frac{1}{\omega }=\frac{L}{2\pi }\sqrt{\mu /T}$ (using $\left(1\right)$)