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
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)
wheree is the end correction.
Another resonance occurs at
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
wherev is velocity of sound,f is frequency andlambda 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