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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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  1. 24 July, 12:41
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    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
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