Question

A closed organ pipe and an open organ pipe of same length produce $2$ beats when they are set into vibrations simultaneously in their fundamental mode. The length of open organ pipe is now halved and of closed organ pipe is doubled, the number of beats produced will be

• A. $8$
• B. $7$
• C. $4$
• D. $2$

Answer 1

The correct option is B $7$

Given number of beats $f_o–f_c=2$
$f_0$ is fundamental frequency of open pipe

$f_c$ is fundamental frequency of closed pipe

or $\frac{v}{2l}–\frac{v}{4l}=2$

or $\frac{v}{4l}=2$

or $\frac{v}{l}=8$ . . . .(1)

When length of open organ pipe is halved and that of closed organ pipe is doubled.
New Fundamental frequecy is given by

$f_0^{\prime }=\frac{v}{l}$

$f_c^{\prime }=\frac{v}{8l}$

The beat frequency will be

$f_0^{\prime }–f_c^{\prime }=\frac{v}{l}–\frac{v}{8l}=\frac{7}{8}\frac{v}{l}$

Substituting $\frac{v}{l}=8$ from (1) beat frequency = $7$.