In an open organ pipe _ 3 and _ 6 are 3^ rd and 6^ th harmonic frequencies, respectively. If _ 6 - _ 3 =2200 ~Hz then length of the pipe is \_\_\_\_ mm . (Take velocity of sound in air is 330 ~m / s .)

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Quantrex Academy · Accepted Answer

Correct answer: C. 225

For an open organ pipe, the n -th harmonic frequency is f_n = n · (v)/(2L) . The 3rd harmonic is v_3 = 3 · (330)/(2L) and the 6th harmonic is v_6 = 6 · (330)/(2L) . Taking the difference: v_6 - v_3 = (6-3) · (330)/(2L) = 3 · (330)/(2L) = 2200 . Solving for L : L = (3 × 330)/(2 × 2200) = (990)/(4400) = 0.225 m = 225 mm.

In an open organ pipe _ 3 and _ 6 are 3^ rd and 6^ th harmonic...

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