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

200 Hz

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30 Hz

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240 Hz

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Solution

The correct option is C 200 Hz
The resonance length in a closed organ pipe is

$l = (2 n + 1) \frac{λ}{4}$

Hence, the first resonance length in a closed organ pipe is

$l_{1} = (2 (0) + 1) \frac{λ}{4}$

$l_{1} = \frac{λ}{4} = 50 c m$ .........................(given)

Now, the second resonance length in a closed organ pipe is

$l_{1} = (2 (1) + 1) \frac{λ}{4}$

$l_{1} = \frac{3 λ}{4} = (3) (50) = 150 c m$

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

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

The correct option is C 200 HzThe resonance length in a closed organ pipe isl=(2n+1)λ4Hence, the first resonance length in a closed organ pipe isl1=(2(0)+1)λ4l1=λ4=50cm .........................(given)Now, the second resonance length in a closed organ pipe isl1=(2(1)+1)λ4l1=3λ4=(3)(50)=150cm