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In a set, 21 turning forks are arranged in a series of decreasing frequencies. Each tuning fork produces 4 beats per second with the preceding fork. If the first fork is an octave of the last fork, find the frequencies of the first and tenth fork. - Physics

In a set, 21 turning forks are arranged in a series of decreasing frequencies. Each tuning fork produces 4 beats per second with the preceding fork. If the first fork is an octave of the last fork, find the frequencies of the first and tenth fork.

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Solution

Given: N = 21, x = 4, nF = 2 nL

To find:

i. Frequency of first fork (nF)

ii. Frequency of tenth fork (n10)

Formula: nL = nF - (N – 1)x

Calculation:-

i. When tuning forks are arranged in the decreasing order of frequencies, the frequency of the pth tuning fork is,

nL = nF - (N - 1)x = n1 - (21 - 1) 4

∴ nL = nF - 80                           ............(1)

As frequency of first fork is an octave of last,

∴ nF = 2nL

∴ nL = nF/2

From equation (1),

nF/2 = nF - 80

∴ nF - (nF/2) = 80

∴ nF/2 = 80

∴ nF = 160 Hz

The frequency of the first fork is 160 Hz.

ii. For 10th fork,

n10 = n1 - (10 - 1)x

      = 160 - 9 * 4 = 160 - 36

n10 = 124Hz

The frequency of the tenth fork is 124 Hz.

Concept: Study of Vibrations of Air Columns
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