Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below f 0 = 1250 Hz. The length of the pipe is λ = 85 cm. The velocity of sound is v = 340 m/s.
Consider the two cases:
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
v n = 
(2n + 1); six oscillations;
v n = 
(n + 1), also six oscillations.
Here n = 0, 1, 2, ...
Sol. 
= λ 
= λ
and for n th overtone n th 
(2n + 1) 
= λ , λ= 
f = 
< 1250 Hz
= 
< 1250
= (2n + 1) < 12.5 n – obertone
2n < 11.5
n < 
n < 5.75
n number of oscillation = 6
n = 0, 1, 2, 3, 4, 5
similarly for open organ pipe
f = 
(n + 1) < 1250
(n + 1) < 1250
n + 1 < 
n + 1 < 6.25
n < 5.25
n = obertone
n = 0, 1, 2, 3, 4, 5
number of oscillation
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