Question

An open organ pipe of length $l$ is sounded together with another organ pipe of length $l+x$ in their fundamental tones $(x<<l)$. The beat frequency heard will be $($speed of sound is $v):$

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

Answer 1

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

For an organ pipe with length $l$ the fundamental mode frequency is given by,

$\lambda =2l$

$f=\frac{v}{\lambda }=\frac{v}{2l}$

for length $l+x$

$f^{\prime }=\frac{v}{{\lambda }^{\prime }}=\frac{v}{2(l+x)}$

$f-f^{\prime }=\frac{v}{2l}-\frac{v}{2(l+x)}=\frac{v}{2}(\frac{1}{l}-\frac{1}{l+x})=\frac{vx}{2l(l+x)}\approx \frac{vx}{2l^2}$

Option "$C$" is correct.