A pipe of length `85cm` is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below `1250 Hz`. The velocity of sound in air is `34m//s`.

Explanation

`(a)` When the tube is closed at one end
`v=(v)/( 41)(2n+1), `where` n=0,1,2,........`
`=(340)/( 4 xx 0.85)(2n+1)=100(2n+1)`
Thus for `n=0,1,2,3,4,5,6,....., ` we get
`n_(1)=1001 H_(z), n_(2)=300 1 H_(z), n_(3)=5001 H_(z), n_(4)=700 1 H_(z),`
`n_(5)=900 1 H_(z), n_(6)=1100 1 H_(z), n_(7)=1300 1 H_(z)`
Since `v` shoul be `lt v_(0)=12501 H_(z)`, we nedd not go beyond `n_(6)`.
`(b)` Organ pipe opened from both ends vibrates with all harmonics of the fundamental frequency. now, the fundamental mode frequency is given as
`v=v//lambda`
or, `v=v//2l`
Here, also, end correction has been neglected. So,the frequencies of higher modes of vibrations are given by
`v=n(v//2l) ...(1)`
or, `v_(1)=v//2l, v_(2)=2(v//2l), v_(3)=3(v//2l)`
If may be checked by patting the values of `n` in equation `(1)` that belox `1285 Hz`, there are a total of six possible natural oscillation requencies of air column in the open pipe.