A uniform narrow tube 1.80 m long is open at both ends. It resonates at two successive harmonics of frequencies 280 Hz and 300 Hz. (a) What is the fundamental frequency (Hz) ? (b) What is the speed of sound in the gas in the tube (m/s) ?

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Physics

Susan Rowe

24 July, 09:27

A uniform narrow tube 1.80 m long is open at both ends. It resonates at two successive harmonics of frequencies 280 Hz and 300 Hz. (a) What is the fundamental frequency (Hz) ? (b) What is the speed of sound in the gas in the tube (m/s) ?

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Osvaldo Cortez · Answer 1

Harmonics are integer multiples of the fundamental frequency.

I found the greatest common factor of 280 Hz and 420 Hz, and that was 140 Hz. Therefore, I knew that this was the fundamental frequency. (280 Hz and 420 Hz are the second and third harmonics, respectively)

The formula for the fundamental frequency of an open pipe is

f = (v/2L)

where v is the velocity of the gas inside the tube and L is the length of the tube.

I assume that I'm ignoring end corrections in this calculation.

f = 140 Hz

v = unknown

L = 1.85 meters

140 = (v) / (2*1.85)

v = (140) (2) (1.85) = 518 m/s