A set of 28 turning forks is arranged in an increasing order of frequencies. Each fork produces ' x ' beats per second with the preceding fork and the last fork is an octave of the first. If

Question

A set of 28 turning forks is arranged in an increasing order of frequencies. Each fork produces ' x ' beats per second with the preceding fork and the last fork is an octave of the first. If the frequency of the 1 2 th fork is 152 Hz , the value of ' x ' (no. of beats per second) is

TestHub · Accepted Answer

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

Since each fork produces $x$ beats per second with the preceding fork, the frequencies are in an arithmetic progression with a common difference of $x$ .

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

For the $2 8^{th}$ fork, $f_{28} = f_{1} + 27 x$ .

It is given that the last fork is an octave of the first, which means $f_{28} = 2 f_{1}$ .

$2 f_{1} = f_{1} + 27 x \Rightarrow f_{1} = 27 x$

The frequency of the $1 2^{th}$ fork is given as $152 Hz$ .

$f_{12} = f_{1} + 11 x = 152$

Substituting $f_{1} = 27 x$ into the equation:

$27 x + 11 x = 152$

$38 x = 152$

$x = 4$

Answer: 4

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