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What is the second lowest resonant frequency for a 0.7 m long organ pipe closed at one end? e) 515 Hz d) 368 Hz c) 405 Hz b) 429 Hz a) 322 Hz

1 Answer

5 votes

Answer:

(d) 368 Hz

Step-by-step explanation:

The resonance frequency of one closed end pipe is given by
f=(nv)/(4L) where n =1,3,5,7------------

Here we are talking about second lowest resonant frequency so n=3

The speed of sound v = 343 m/sec

Length of pipe L =0.7 m

So
f=(nv)/(4L)=(3* 343)/(4* 0.7)=367.5\ Hz

Which is closest to 368 Hz so option (d) will be the correct answer

User Markus Kuhn
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