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Grade: upto college level
        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.
5 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							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.

						
5 years ago
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