Question

Two narrow organ pipes, one open (length $l_1$) and the other closed (length $l_2$) are sounded in the respective fundamental modes. The beat frequency heard is 5 Hz. If now the pipes are sounded in the first overtones, then also the beat frequency heard 5 Hz. Then

• A. $\frac{l_1}{l_2}=\frac{1}{2}$
• B. $\frac{l_1}{l_2}=\frac{1}{1}$
• C. $\frac{l_1}{l_2}=\frac{3}{2}$
• D. $\frac{l_1}{l_2}=\frac{2}{3}$

Answer 1

Answer: (B), (C)

. $\frac{l_1}{l_2}=\frac{1}{1}$.

The harmonic frequencies in an open pipe are given by,

${\nu }_n=\frac{nv}{2l_1}$

and that for a closed pipe are given by

${\nu }_n^{\prime }=\frac{(2n-1)v}{4l_2}$

For the beat frequency in fundamental mode:

$\frac{v}{2l_1}-\frac{v}{4l_2}=5$............(1)

For the beat frequency in first overtone:

$2\left(\frac{v}{2l_1}\right)-3\left(\frac{v}{4l_2}\right)=5$.......(2)
From Eqs. (1) and (2) we get,
$\frac{l_1}{l_2}=\frac{1}{1}=1:1$.