Question

A closed organ pipe can vibrate at a minimum frequency of 500 Hz. Find the length of the tube. Speed of sound in air = 340 m s−1.

Answer 1

The minimum frequency of the vibration is the fundamental frequency of the vibration. In a closed organ pipe, the fundamental frequency of vibration is given as,  ν = V/4L, where V = speed of the sound in air = 340 m/s and ν = 500 Hz. L = length of the pipe =? Thus, L =V/4ν =340/(4*500) m =340/2000 m  →L = 0.17 m = 17 cm.