A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Step-by-step solution

fundamental frequency L = λ 4 v 1 = v 4 L L = λ 2 + λ 4 = 3 λ 4 v 2 = 3 v 4 L ∴    v = odd v 4 L Now L = 8 5 cm = 0.85 m v = 3 4 0 m/s = odd 3 4 0 4 × 0.85 = odd × 1 0 0   Hz Hence possible frequencies below 1250 Hz are 100 Hz, 300 Hz, 500 Hz...........1100 Hz Hence 6