Question

A stretched string of length $2m$ is found to vibrate in resonance with a tuning fork of frequency $420Hz.$ The next higher frequency for which resonance occurs is $490Hz$. The velocity of the transverse wave along this string is

• A. $140m/s$
• B. $360m/s$
• C. $340m/s$
• D. $280m/s$

Answer 1

The correct option is D $280m/s$

$f_1=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$
$f_2=\frac{(n-1)}{2l}\sqrt{\frac{T}{\mu }}$
$\therefore f_1-f_2=70=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$v=\sqrt{\frac{T}{\mu }}=70\times 2l=70\times 4=280m{s}^{-1}$