An open organ pipe and closed organ pipe of same length produce 2 beats per second, when they are set into vibrations together, in fundamental mode. The length of open pipe is made half and that of closed pipe is doubled. The number of beats produced per second will be (neglect end correction)

Question

Aicharya · Accepted Answer

Let $L$ be the initial length of both pipes and $v$ the speed of sound. The fundamental frequencies are $f_o = \frac{v}{2L}$ for the open pipe and $f_c = \frac{v}{4L}$ for the closed pipe.

The initial beat frequency is the absolute difference: $f_{\text{beat1}} = |f_o - f_c| = \left|\frac{v}{2L} - \frac{v}{4L}\right| = \frac{v}{4L}$.

Given $f_{\text{beat1}} = 2$ Hz, this yields $\frac{v}{4L} = 2$, so $\frac{v}{L} = 8$ Hz.

After modification, the open pipe length becomes $\frac{L}{2}$ and the closed pipe length $2L$. The new frequencies are $f_o' = \frac{v}{2(L/2)} = \frac{v}{L}$ and $f_c' = \frac{v}{4(2L)} = \frac{v}{8L}$.

Substituting $\frac{v}{L} = 8$ gives $f_o' = 8$ Hz and $f_c' = 1$ Hz. The beat frequency becomes $|8 - 1| = 7$ Hz.

Final answer: $\boxed{7}$

An open organ pipe and closed organ pipe of same length produce 2 beats per second, when they are set into…

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