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Question
A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces Y beats per second with the previous one. The last is an octave of the first. The fifth fork
has a frequency of 90 Hz. Find Y and frequency of the first and the last tuning forks.
Solution
Solution :
Given : N = 12, nL = 2 nF , n5 = 90 Hz
To find: Number of beats (x),
Frequency of last fork (nL)
Formula : nL = nF + (N - 1) x
nL = nF + (5- 1) x
nF + 4x = 90 ….(1)
nL = nF + (12 - 1) x
nL = nF + 11x ….(2)
nL= 2nF
nF= 11x ….From (2)
Substituting in equation (1),
15x = 90 or x = 6 beat/s
nF = 11 x 6 = 66 Hz
nL = 2 x 66 = 132 Hz
The frequency of the first and last tuning fork is 66 Hz and 132 Hz respectively.
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