1.

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`.A. 25cmB. 100cmC. 50cmD. 12.5cm

Answer» Correct Answer - A
`n_2-n_1 = 2592 - 1944 = 648`
`n=(v)/(2l)`
`therefore l=(v)/(2n) = (324)/(2xx648)`
0.25cm


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