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Worked Examples · Example 6.1

Q.An air column is of length 17 cm long. Calculate the frequency of the 5th overtone if the air column is

(a) closed at one end and
(b) open at both ends. (Velocity of sound in air = 340 m s⁻¹.)
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Closed: nᶜ = 340/(4×0.17) = 500 Hz; 5th overtone = (2×5+1)×500 = 5500 Hz. Open: nᵒ = 340/(2×0.17) = 1000 Hz; 5th overtone = (5+1)×1000 = 6000 Hz.

  1. Closed at one end. The fundamental is

    nc=v4L=3404×0.17=500 Hzn^c = \frac{v}{4L} = \frac{340}{4\times0.17} = 500\ \mathrm{Hz}

    The p-th overtone is nᶜₚ = (2p + 1)nᶜ (only odd harmonics). For the 5th overtone (p = 5):

    n5c=(2×5+1)×500=11×500=5500 Hzn^c_5 = (2\times5 + 1)\times 500 = 11\times 500 = 5500\ \mathrm{Hz}

  2. Open at both ends. The fundamental is

    no=v2L=3402×0.17=1000 Hzn^o = \frac{v}{2L} = \frac{340}{2\times0.17} = 1000\ \mathrm{Hz}

    The p-th overtone is nᵒₚ = (p + 1)nᵒ (all harmonics). For the 5th overtone:

    n5o=(5+1)×1000=6000 Hzn^o_5 = (5 + 1)\times 1000 = 6000\ \mathrm{Hz}

    ✓Final answer

    (a) Closed: fundamental 500 Hz, 5th overtone 5500 Hz. (b) Open: fundamental 1000 Hz, 5th overtone 6000 Hz.

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