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Wave Speed expression

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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 open at both ends.

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Solution

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

Fundamental frequency of an open organ pipe is given by ,
$f_{1} = v / 2 L = 344 / (2 \times 0.45) = 382.2 H z$ ,

Therefore ,frequency of first overtone
$f_{2} = 2 f_{1} = 2 \times 382.2 = 764.4 H z$

Frequency of second overtone
$f_{3} = 3 f_{1} = 3 \times 382.2 = 1146.6 H z$

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