Two closed organ pipes A and B have the same length and A is wider than B. They resonate in the fundamental mode at frequencies n A and n B respectively, then

Question

Two closed organ pipes A and B have the same length and A is wider than B. They resonate in the fundamental mode at frequencies n A ​ and n B ​ respectively, then

TestHub · Accepted Answer

The fundamental frequency of a closed organ pipe is given by $n = \frac{v}{4 ( L + e )}$

where $v$ is the speed of sound, $L$ is the length of the pipe, and $e$ is the end correction.

The end correction $e$ is directly proportional to the radius $r$ of the pipe ( $e = 0.6 r$ ).

Since pipe A is wider than pipe B, its radius is larger, which implies $e_{A} > e_{B}$ .

Therefore, the effective length of pipe A is greater than that of pipe B, i.e., $(L + e_{A}) > (L + e_{B})$ .

As frequency is inversely proportional to the effective length, we get $n_{A} < n_{B}$ .

Answer: $n_{A} < n_{B}$

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