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Standing Waves in Closed Organ Pipe

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Question

Find the fundamental frequency and the frequencies of the first two overtones of a pipe 45.0 cm long if the pipe is closed at one end. Use $v = 344 m / s$ .

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Solution

We have , $v = 344 m / s, L = 45 c m = 0.45 m$

Fundamental frequency of a closed organ pipe is given by ,
$f_{1} = v / 4 L = 344 / (4 \times 0.45) = 191.1 H z$

therefore ,frequency of first overtone ,
$f_{2} = 3 f_{1} = 3 \times 191.1 = 573.3 H z$

Frequency of second overtone ,
$f_{3} = 5 f_{1} = 5 \times 191.1 = 955.5 H z$

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