Given V=343 m/s

length of pipe= 0.7795

What would be the fundamental frequency that resonates from the closed organ pipe?

I got 110 Hz but i'm afraid I may be wrong..

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Given V=343 m/s

length of pipe= 0.7795

What would be the fundamental frequency that resonates from the closed organ pipe?

I got 110 Hz but i'm afraid I may be wrong..

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## Answers (1)

Fundamental frequency of closed organ pipe = V/(4*L) as (lambda/4) = L or lambda= 4*L. Or required frequency = 343/(4*0.7795) = 110 Hz. You are right! But you must have confidence that comes from understanding.