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1 December, 09:51

The contrabassoon is the wind instrument capable of sounding the lowest pitch in an orchestra. It is folded over several times to fit its impressive 18 ft length into a reasonable size instrument.

a. If we model the instrument as an open-closed tube, what is its fundamental frequency? The sound speed inside is 350 m/s because the air is warmed by the player's breath.

b. The actual fundamental frequency of the contrabassoon is 27.5 Hz, which should be different from your answer in part a. This means the model of the instrument as an open-closed tube is a bit too simple. But if you insist on using that model, what is the "effective length" of the instrument?

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  1. 1 December, 13:09
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    Answer: a) f = 15.95Hz effective length (L) = 50.91m

    Explanation: The musical instrument is an open-closed tube thus it length (L) is related to the wavelength of sound passing through it is given below as

    L = λ/4

    Where λ = wavelength of sound wave

    L = 18ft = 5.49m (1m = 3.28ft)

    Hence, λ = 4L

    λ = 5.49 * 4

    λ = 21.95m

    v = fλ where v = 350m/s, λ = 21.95m

    f = v/λ = 350 / 21.95

    f = 15.95Hz.

    b) the actual fundamental frequency is 27.5 Hz

    From v = fλ where v = 350m/s, we have the wavelength as

    λ = v/f

    λ = 350 / 27.5

    λ = 12.73m

    Assuming the instrument is an open-closed tube, then the length is given as

    L = λ/4, but

    λ = 12.73

    L = 12.73 * 4

    L = 50.91m
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