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.
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.
