A set of 14 tuning forks is arranged in a series of increasing frequencies. Each fork produces ' x ' beats per second with the preceding fork and the last fork is an octave of the first fork

Question

A set of 14 tuning forks is arranged in a series of increasing frequencies. Each fork produces ' x ' beats per second with the preceding fork and the last fork is an octave of the first fork. If the seventh fork has the frequency of 114 Hz, the value of x 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 and the frequencies are in increasing order, the frequencies form an arithmetic progression with a common difference of $x$ .

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

The frequency of the 14th tuning fork is $f_{14} = f_{1} + 13 x$ .

It is given that the last fork is an octave of the first fork, which means $f_{14} = 2 f_{1}$ .

Substituting this into the equation, we get $2 f_{1} = f_{1} + 13 x \Rightarrow f_{1} = 13 x$ .

The frequency of the 7th tuning fork is given as $114$ Hz.

Using the formula for the 7th fork, $f_{7} = f_{1} + 6 x$ .

Substituting $f_{1} = 13 x$ , we get $f_{7} = 13 x + 6 x = 19 x$ .

Equating this to the given frequency, $19 x = 114$ .

Solving for $x$ , we get $x = \frac{114}{19} = 6$ .

Answer: $6$

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