Question

A stretched string resonates in fundamental node with tuning fork of frequency $512\text{Hz}$ when length of the string is $0.5\text{m}$. The length of the string required to vibrate in resonance with a tuning fork of frequency $256\text{Hz}$ in fundamental node would be

• A. $0.25\text{m}$
• B. $0.5\text{m}$
• C. $1\text{m}$
• D. $2\text{m}$

Answer 1

The correct option is C $1\text{m}$

For a stretched string, frequency of oscillation is
$f=\frac{1}{2l}\sqrt{\left[\frac{T}{\mu }\right]}$
$f\propto \frac{1}{l}$

When $f$ is halved, the length will be doubled.
Hence required length of string is $1\text{m}$