A set of 28 turning forks is arranged in an increasing order of frequencies. Each fork produces ' $x$ '…
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
2
4
6
8
Solution
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),
$\begin{array}{ll}
& 152=27 x+11 x \\
\therefore \quad & x=\frac{152}{38}=4
\end{array}...(i)$