Question

A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire?

Answer 1

Let the length and tension in the wire are $L$ and $T$ respectively.

$n^{th}$ harmonic frequency is given by:

${\nu }_n^{th}=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$

Fundamental frequency $\left({\nu }_o\right)$ when Length is $L$

${\nu }_o=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

This is the fundamental frequency of tuning fork.

Wire of twice the length $\left(2L\right)$ vibrates in its second harmonic to resonate with tuning fork

${\nu }_n^{th}=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$

${\nu }_2^{nd}=\frac{2}{4L}\sqrt{\frac{T}{\mu }}$

Wire of twice the length vibrates in its second harmonic. Thus, if the tuning fork resonates at $L$, it will resonate at $2L$ under same condition in different harmonics.

Final Answer: Under same condition in different harmonics.