A set of 8 tuning forks is arranged in ascending order of frequencies. Each fork produces 4 beats with the next one and frequency of last fork is octave of first. The frequency of 5 th fork is

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Let the frequency of the first tuning fork be $f_{1}$ .

Since each tuning fork produces $4$ beats with the next one and they are arranged in ascending order of frequencies, their frequencies form an arithmetic progression with a common difference $d = 4$ Hz.

The frequency of the $n^{th}$ tuning fork is given by $f_{n} = f_{1} + (n - 1) d$ .

For the $8^{th}$ tuning fork, $f_{8} = f_{1} + (8 - 1) \times 4 = f_{1} + 28$ .

It is given that the frequency of the last fork is the octave of the first, which means $f_{8} = 2 f_{1}$ .

Substituting the value of $f_{8}$ , we get:

$2 f_{1} = f_{1} + 28$

$f_{1} = 28$ Hz

The frequency of the $5^{th}$ tuning fork is:

$f_{5} = f_{1} + (5 - 1) \times 4$

$f_{5} = 28 + 16 = 44$ Hz.

Answer: $44$ Hz

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