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Find out the frequency of the first ove...

Find out the frequency of the first overtone emitted by a 1 . 25 m long organ pipe closed at one end. Given , velocity of sound in air ` = 320 m * s^(-1)`

Text Solution

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For the 1st overtone, a node-antinode pair is formed in between a node at the closed end and an antinode at the open end .
So, distance between the node at the closed end and the 2nd antinode from it ` = (3lambda)/(4)= l = ` length of the pipe .
` :. lambda = (4l)/(3)`
So, the frequency of this 1st overtone is
` n_(1) = (V)/(lambda) = (3V)/(4l) = (3 xx 320)/(4 xx 1.25) = 192` Hz .
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