A set of 24 tuning forks is arranged in a ser | vedclass

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(D) Let the frequency of the first tuning fork be $f_1 = a$.The frequencies form an arithmetic progression $(AP)$ with common difference $d = 4 \, Hz$

Vedclass · Accepted Answer

$108$

Vedclass · Answer

(D) Let the frequency of the first tuning fork be $f_1 = a$.The frequencies form an arithmetic progression $(AP)$ with common difference $d = 4 \, Hz$.The number of tuning forks is $n = 24$.The frequency of the $n^{th}$ tuning fork is given by $f_n = a + (n - 1)d$.For the $24^{th}$ fork: $f_{24} = a + (24 - 1)4 = a + 92$.According to the problem,the frequency of the last fork is twice the frequency of the first fork: $f_{24} = 2f_1$.Substituting the values: $a + 92 = 2a$,which gives $a = 92 \, Hz$.The frequency of the $5^{th}$ tuning fork is $f_5 = a + (5 - 1)d$.$f_5 = 92 + 4(4) = 92 + 16 = 108 \, Hz$.

$A$ set of $24$ tuning forks is arranged in a series of increasing frequencies. If each fork gives $4 \, Hz$ beats per second with the preceding one and the frequency of the last tuning fork is two times that of the first fork,find the frequency of the $5^{th}$ tuning fork in $Hz$.

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