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(A) Let the frequency of the first tuning fork be $f_1 = f$.The number of tuning forks is $n = 56$.The beat frequency is $d = 4 	ext{ Hz}$.The freque

Vedclass · Accepted Answer

$292$

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(A) Let the frequency of the first tuning fork be $f_1 = f$.The number of tuning forks is $n = 56$.The beat frequency is $d = 4 	ext{ Hz}$.The frequency of the $n^{	ext{th}}$ fork is given by $f_n = f_1 + (n - 1)d$.So, $f_{56} = f + (56 - 1) 	imes 4 = f + 55 	imes 4 = f + 220$.Given that the frequency of the last fork is twice that of the first, we have $f_{56} = 2f$.Equating the two expressions: $2f = f + 220$, which gives $f = 220 	ext{ Hz}$.The frequency of the $19^{	ext{th}}$ fork is $f_{19} = f_1 + (19 - 1)d$.$f_{19} = 220 + 18 	imes 4 = 220 + 72 = 292 	ext{ Hz}$.

$56$ tuning forks are arranged such that each fork produces $4$ beats per second with its previous one. If the frequency of the last fork is twice that of the first, the frequency of the $19^{\text{th}}$ fork is . . . . . . (in $\text{Hz}$)

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