Question

A closed organ pipe of length $L$ $\frac{L}{4}$ is in resonance with a tuning fork. If a hole is made in the pipe at a distance L/4 from closed-end, it will be in resonance again, when :

• A. Tuning fork is replaced by another of high frequency.
• B. Tuning fork is replaced by another of lower frequency.
• C. It will be resonance with same tuning fork.
• D. Now the pipe will never resonate with any tuning fork.

Answer 1

Answer: (D)

Standing wave cannot be formed. It becomes combination of two pipes one closed and other open pipe. They will not satisfy the condition of standing wave simultaneously.
$\frac{n_{1}V}{2(\frac{3L}{4})}=\frac{n_{2}V}{4(\frac{L}{4})}$
$\frac{n_{1}2V}{3L}=\frac{n_{2}V}{L}=\frac{V}{4L}$
$n_1=\frac{3}{8}{n}_2=\frac{1}{4}$ Which are not integer so not possible