Question 119 of 142

A set of $20$ tuning forks is arranged in a series of increasing frequencies. If each fork gives $4$ 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 last fork is _____ $Hz$.

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Answer: 152

Each fork produces $4$ beats per second with the previous means each fork has frequency $4Hz$ more than the previous.
Using relation, ${f}_{last}={f}_{first}+\left(N-1\right)x$, here, $N$ is the number of tuning fork in series and $x$ is beat frequency between two successive forks.
$2{f}_{0}={f}_{0}+\left(20-1\right)4$
$\Rightarrow 2{f}_{0}={f}_{0}+76 \Rightarrow {f}_{0}=76Hz$
Frequency of last tuning fork is $152Hz$.