A pipe 20cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open ? ( speed of sound in air is `340ms^(-1)`)
First (Fundamental), No
Length of the pipe, l = 20 cm = 0.2 m
Source frequency `= n^(“th”)` normal mode of frequency, `ν_(n)` = 430 Hz
Speed of sound, v = 340 m/s
In a closed pipe, the `n^(th)` normal mode of frequency is given by the relation:
`V_(n)=(2n-1)(v)/(4l)” “:n` is an integer=0,1,2,3…
`430=(2n-1)(340)/(4xx02)`
`2n-1=(430xx4xx0.2)/(340)=1.01`
`2n=2.01`
`n~1`
Hence, the first mode of vibration frequency is resonantly excited by the given source.
In a pipe open at both ends, the `n^(th)` mode of vibration frequency is given by the relation:
`v_(n)=(mv)/(2l)`
`n=(2lV_(n))/(v)`
`=(2xx0.2xx430)/(340)=0.5`
Since the number of the mode of vibration (n) has to be an integer, the given source does not produce a resonant vibration in an open pipe.
Here, `l= 20 cm=0.2m , v_(n) = 430 Hz,v= 340 m//s `
The frequency of `n^(th) ` normal mode of vibration of closed pipe s
`v_(n) =( 2n-1) (v)/( 4l)`
`:. 430 = ( 2n-1) ( 340)/( 4 xx 0.2)`
`2n-1= ( 430 xx 4xx 0.2)/( 340)= 1.02 `
`2n =2.02 , n = 1.01 `
Hence it will be the`1^(st)` normal mode of vibration . In a pipe, open at both ends we have
`v_(n) =n xx( v)/( 2l) = ( n xx 340 )/( 2 xx 0.2) = 430`
` :. n = ( 430 xx 2 xx 0.2)/( 340 ) =0.5 `
As n has to be an integer, therefore open organ pipe cannot be in resonance with the source.