A set of 20 tuning forks is arranged in a ser | vedclass

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Question

(B) Let the frequency of the first tuning fork be $f_1 = f$.Since each fork gives $4 \; Hz$ beats with the preceding one,the frequencies form an arith

Vedclass · Accepted Answer

$152$

Vedclass · Answer

(B) Let the frequency of the first tuning fork be $f_1 = f$.Since each fork gives $4 \; Hz$ beats with the preceding one,the frequencies form an arithmetic progression with a common difference $d = 4 \; Hz$.The frequency of the $n$-th fork is given by $f_n = f_1 + (n - 1)d$.For the $20$-th fork $(n = 20)$:$f_{20} = f + (20 - 1) 	imes 4 = f + 19 	imes 4 = f + 76$.According to the problem,the frequency of the last fork is twice the frequency of the first:$f_{20} = 2f_1$.Substituting the values:$f + 76 = 2f$.Solving for $f$:$f = 76 \; Hz$.The frequency of the last fork is $f_{20} = 2f = 2 	imes 76 = 152 \; Hz$.

$A$ set of $20$ tuning forks is arranged in a series of increasing frequencies. If each fork gives $4 \; Hz$ beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first,then the frequency of the last fork is $\dots \; Hz$.

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