An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher by 100 Hz than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is

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Detailed Solution

f₁ $λ$ ₁ = v; f₁ $(\frac{4 L}{3}) = v$

$f_{1} = \frac{3 v}{4 L}$ ; f₂(2L)= v $\Rightarrow$ f₂ = $\frac{v}{2 L}$

Given that f₁ – f₂ = 100

$\Rightarrow \frac{3 v}{4 L} - \frac{v}{2 L} = 100$ $\Rightarrow \frac{v}{2 L} = 200 H z$

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