For a certain organ pipe, three successive resonance frequencies are observer at `425`, `595` and `765 H_(Z)` respectively. Taking the speed of sound in air to be `340 m//s`, (a) explain whether the pipe is closed at one or open at boyh ends. (b) determine the fundamental frequency and length of the pipe.

Explanation

(a) The given frequencies are in the are
`425 : 595 : 765 ` i.e., `5 : 7 : 9`
And as clearly these are odd integers so the given pipe is closed pipe.
(b) From part (b) it is clear that the frequency of 5th harmonic (which is third overtone) is 425 Hz
So `425 = 5fc rArr f_(C) = 85 Hz` Further as `f_(C) = (v)/(4L), L = (v)/(4f_(C)) = (340)/(4 xx 85) = 1 m`