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 ratio 425 : 595 : 765, i.e., 5:7:9 and clearly these are oddintegers so the given pipe is closed at one end.
b) From part (a) it is clear that the frequency of 5th harmonic (which is 2nd overtone) is 425 Hz
So `425 = 5f_c (or) f_c = 85Hz`.
Further as `f_C= v/(4L) ,L =v/(4f_C) =340/(94 xx85) =1m`