Question

The first overtone frequency of a closed organ pipe $P_1$ is equal to the fundamental frequency of an open organ pipe $P_2$. If the length of the pipe $P_1$ is $60\text{cm}$, what will be the length of $P_2$?

• A. $20\text{cm}$
• B. $40\text{cm}$
• C. $60\text{cm}$
• D. $80\text{cm}$

Answer 1

The correct option is B $40\text{cm}$

Modes of vibration of air column in a closed organ pipe are given by

$f_n=\frac{(2n-1)v}{4l}$

First overtone of a closed organ pipe is obtained for $n=2$

$f_2=\frac{3v}{4l_1}$

where $l_1$ is the length of the closed organ pipe.

Modes of vibration of air column in an open organ pipe is given by

$f_n=\frac{nv}{2l}$

Fundamental frequency of an open organ pipe is obtained for $n=1$

$f_1=\frac{v}{2l_2}$

where $l_2$ is the length of the open organ pipe.

From the data given in the question, $f_1=f_2$
From this, $\frac{3v}{4l_1}=\frac{v}{2l_2}\Rightarrow l_2=\frac{2}{3}{l}_1$

Given, $l_1=60\text{cm}$
$\therefore l_2=\frac{2}{3}\times 60=40\text{cm}$

Thus, option (b) is the correct answer.