A pipe of length 85cm is closed from one end . find the number of possible natural oscillation of air column in the pipe whose frequencies lie below 1250 Hz . the velocity of sound in air 340m/s [JEE main 2014]
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Answered by
2
In fundamental mode
λ/4=0.85
λ=4×0.85
f=v/λ
=340/4×0.85
=100Hz.
∴ Possible frequencies = nf (n is odd) = 100 Hz, 300 Hz,
500 Hz, 700 Hz, 900 Hz, 1100 Hz below 1250 Hz.
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Answered by
4
Explanation:- For closed organ pipe = (2n+1)v/4L
(n = 0,1 ,2,3, ......)
(2n+1)v/4L < 1250
(2n+1) < 1250 × 4×0.85/340
- ⟹ (2n+1)<12.5
- ⟹ 2n < 11.50
- ⟹n < 5
so, n ,= 0, 1, 2, 3, ....5
so we have 6 possibilities .
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Alternative Method
In closed organ pipe , fundamental node
λ/4 = 0.85
λ = 0.85 × 4
As we know that , f = c/λ
340/4× 0.85 = 100Hz
∵ possible frequency (n-1)v/4L number will be odd ∴ 100Hz, 300Hz , 500Hz , 700Hz , 900Hz , 1100Hz below 1250
∴ number of possible natural oscillation is 6
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