Question

The length of two open organ pipes are $l$ and $(l+\Delta l)$ respectively . Neglecting end correction, the frequency of beats between them will be approximately
(Here $v$ is the speed of sound)

• A. $\frac{v}{2l}$
• B. $\frac{v}{4l}$
• C. $\frac{v\Delta l}{2l^2}$
• D. $\frac{v\Delta l}{l}$

Answer 1

The correct option is C $\frac{v\Delta l}{2l^2}$

We can consider wavelength of organ pipes as ${\lambda }_1=2l,{\lambda }_2=2l+2\Delta l$
hence frequency of organ pipes will be $n_1=\frac{v}{2l},n_2=\frac{v}{2l+2\Delta l}$
No. of beats heard is
$=n_1-n_2=\frac{v}{2}(\frac{1}{l}-\frac{1}{l+\Delta l})=\frac{v\Delta l}{2l^2}$