The fundamental frequency of a closed pipe of length L is equal to the second overtone of a pipe open at both the ends of length (XL). The value of X is (Neglect end correction)

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The fundamental frequency of a closed pipe of length $L$ is given by:

$n_{c} = \frac{v}{4 L}$

For a pipe open at both ends of length $X L$ , the fundamental frequency is:

$n_{o} = \frac{v}{2 X L}$

The second overtone of an open pipe corresponds to the third harmonic, which is:

$n_{o, 2} = 3 n_{o} = \frac{3 v}{2 X L}$

According to the given condition, the fundamental frequency of the closed pipe is equal to the second overtone of the open pipe:

$\frac{v}{4 L} = \frac{3 v}{2 X L}$

Cancelling $v$ and $L$ from both sides, we get:

$\frac{1}{4} = \frac{3}{2 X}$

$2 X = 12$

$X = 6$

Answer: $6$

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