Two successive resonance frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air is 324 m s-1.

Sol. Let the length of the longer tube be L base 2 and smaller will be L base 1. According to the data 440 = 3 * 330/4 * L base 2 …..(1) (first over tone) and 440 = 330/4 * L base 1 ….(2) (fundamental) solving equation we get L base 2 = 56.3 cm and L base 1= 18.8 cm.