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 \(12^{\text {th }}\) fork is 152 Hz , the value of ' \(x\) ' (no. of beats per second) is

Forks arranged in a series of increasing frequency from \(\mathrm{n}_1\) to \(\mathrm{n}_{28}\)
Each for produces ' \(x\) ' beats with preceding fork,
\(\therefore \quad\) For. \(12^{\text {th }}\) fork,
\(\mathrm{n}_{12}=\mathrm{n}_1+11 \mathrm{x}\)
\(\therefore \quad 152=\mathrm{n}_1+11 \mathrm{x}\)
Similarly,
\(\mathrm{n}_{28}=\mathrm{n}_1+27 \mathrm{x}\)
Given condition is \(\mathrm{n}_{28}=2 \mathrm{n}_1\)
\(2 \mathrm{n}_1=\mathrm{n}_1+27 \mathrm{x} \Rightarrow \mathrm{n}_1=27 \mathrm{x}\)
Substituting this value in equation (i),
\(\begin{array}{ll}
& 152=27 x+11 x \\
\therefore \quad & x=\frac{152}{38}=4
\end{array}...(i)\)