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64 tuning forks are arranged such that each fork produces 4 beats per second next one. If the frequency of the last is octave of the first, then the frequency of 16th fork is……….Hz

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answer is 312.

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Detailed Solution

According to the question :

Frequency of the 64th tuning fork is

$\begin{array}{l} f_{1} + 63 β = 2 f_{1} \end{array}$

$\Rightarrow f_{1} = 63 \times 4$

Therefore, the frequency of the 16th tuning fork is

$f_{16} = f_{1} + 15 β = 312 Hz$

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