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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`. |
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Answer» `(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. |
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