Question

A closed tube has a frequency $n$. If its length is doubled and radius is halved its frequency will become

• A. ${}^n{/}_3$
• B. ${}^n{/}_2$
• C. $n$
• D. $2n$

Answer 1

The correct option is B ${}^n{/}_2$

Let $v$ and $L$ be the velocity of sound and length of tube respectively.

Frequencies of different modes of vibration in a closed tube, $n_p=\frac{(2p-1)v}{4L}$

For a particular mode, frequency is given to be $n$.

Thus $n=\frac{(2p-1)v}{4L}$

Assume that there is no end correction, so frequency of vibration does not

depend on the radius of the tube.

For $L^{\prime }=2L$ $n_p=\frac{(2p-1)v}{4\times 2L}$

$⟹n_p=\frac{n}{2}$