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प्रश्न
A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz, and 1152 Hz. Calculate the fundamental frequency.
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उत्तर १
The difference between the given overtone frequencies is 256 Hz. This implies that they are overtones that follow one another. Let nc represent the fundamental frequency of the closed pipe and nq, nq-1 represent the frequencies of the qth, (q + 1)th and (q + 2)th consecutive overtones, where q is an integer.
Data: nq = 640 Hz, nq-1 = 896 Hz, nq+2 = 1152 Hz Since only odd harmonics exist as overtones,,
nq = (2q + 1) nc
and nq+1 = [2(q + 1) + 1] nc = (2q + 3) nc
∴ `("n"_("q+1")/("n"_"q"))=(2"q" +3)/(2"q"+1)=896/640=7/5`
∴ `(2"q" + 3)/(2"q"+1)=7/5`
∴ 7(2q + 1) = 5(2q + 3)
∴ 14q + 7 = 10q + 15
∴ 4q = 8
∴ q = 2
As a result, the second, third, and fourth overtones, i.e. the fifth, seventh, and ninth harmonics, correspond to the three stated frequencies.
∴ 5nc = 640 ∴ bc = 128 Hz
उत्तर २
Given: Let frequency for pth overtones
∴ (2p - 1)n = 640 ...(1)
(2p + 1)n = 896 ...(2)
(2p + 3)n = 1152 ...(3)
To find: n = ?
Subtracting (1) from (2)
(2p + 1)n - (2p - 1)n = 896 - 640
2p.n + n - 2pn + n = 256
2n = 128 Hz
Also, subtracting equation (2) from equation (3)
(2p + 3)n - (2p + 1)n = 1152 - 896
n(2p + 3 - 2p - 1) = 256
∴ 2n = 256
∴ n = 128 Hz
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