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A set of 31 tuning forks is arranged in ...

A set of 31 tuning forks is arranged in series of decreasing frequency. Each fork gives 6 beats/sec. with the preceding one. The first fork is the octave of the last. The frequency of the last tuning fork is

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Given, frequency of last fork `(n_(1))=2xx` Frequency of first fork `(n_(F)),n_(5)=90Hz`
Let number of beats be x.
We known that,
`n_(1)=n_(F)+(N-1)x`
For N=5, `n_(L)=n_(F)+(5-1)x`
`:.n_(F)+4x=90" "......(i)`
For N=12, `n_(L)=n_(F)+(12-1)x`
`n_(L)=n_(F)+11x" ".......(ii)`
`:'n_(L)=2n_(F)`
`n_(F)=11x" "` [from equation]
Substituting in eyuation (i), we get
15x=90
x=6 beat/sec.
`n_(F)=11xx6=66Hz`
`n_(L)=2xx66=132Hz`
Therefore, Y=6 and the frequency of the first and last tuning fork is 66 Hz and 132 Hz respectivly.
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