A closed organ pipe is vibrating in its air column. The difference in frequency between its 2^ nd overtone and 1^ st overtone is 165 Hz . If the velocity of sound in air is 330 m/s , the length of the closed organ pipe in cm is:

Step-by-step solution

For a closed organ pipe of length L , the frequencies of the harmonics are given by f_ n = (2n-1) v 4L , where n = 1, 2, 3, The 1^ st overtone corresponds to the 3^ rd harmonic ( n=2 ), so its frequency is f₁ = 3v 4L . The 2^ nd overtone corresponds to the 5^ th harmonic ( n=3 ), so its frequency is f₂ = 5v 4L . The difference in frequency is: f₂ - f₁ = 5v 4L - 3v 4L = 2v 4L = v 2L Given that this difference is 165 Hz and v = 330 m/s : 330 2L = 165 2L = 330 165 = 2 m L = 1 m = 100 cm . Answer: 100