Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below `f_(0) = 1250 Hz`. The length of the pipe is `l = 85 cm`. The velocity of sound is `v = 340 m//s`. Consider two cases
the pipe is closed from one end .

Answer

(a) `v_(n) = (v)/(4l)(2n + 1);` six oscillations;
(b) `v_(n) = (v)/(2l)(n + 1)`, also six oscillations.
Here `n = 0, 1, 2,`…..

Explanation

`(lambda)/(4) = l` `(3lambda)/(4) = l` , and for `n^(th)` overtone `(2n + 1)(lambda)/(4) = l`
`(2n + 1) (lambda)/(4) = l , lambda = (4l)/((2n + 1))`
`f = (V(2n + 1))/(4l) lt 1250 Hz`
`= (340(2n + 1)xx 100)/(4 xx 85) lt 1250 = (2n + 1) lt 12.5` , `n-`obertone
`2n lt 11.5`
`n lt (11.5)/(2)`
`n lt 5.75`
`n = 0, 1, 2, 3, 4, 5`
simillarly for open organ pipe
`f = (V)/(2l) (n + 1) lt 1250`
`(340 xx 100)/(2 xx 85)(n + 1) lt 1250`
`n + 1 lt (12.5)/(2)`
`n + 1 lt 6.25`
`n lt 5.25`
`n =` oberione
`n = 0, 1, 2, 3, 4, 5`
number of oscillation `(6)`