If an organ pipe is sounded with a tuning fork of frequency 256 Hz, the resonance occured at 35 cm and 105 cm, then the velocity of sound is?

1 Answer
Jun 23, 2017

It is an organ pipe with one closed end.

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For a close ended organ pipe: The fundamental (first harmonic) needs to have an node at the close end since the air cannot move and an antinode at the open end.

λ/4 = L_1 + e …... (1)
where e is the end correction.

Another resonance occurs at L_2=105cm
For a closed end organ pipe this is the 3rd harmonic resonating as there is no 2nd harmonic for closed ended pipes.

We get

(3λ)/4 = L_2 + e …... (2)

Subtracting equations (1) from (2) we get

λ/2 = (L_2 – L_1) …... (3)

Using the expression

v=flambda
where v is velocity of sound, f is frequency and lambda is the wavelength.

we get

v = 2f (L_2 - L_1)

Inserting given values we get

v=2xx256(105-35)
=>v=2xx256(105-35)
=>v=35840cms^-1