Question
A stretched sonometer wire is in unison with a tuning fork. When the length is increased by $4\%,$ the number of beats heard per second is $6.$ Find the frequency of the fork.

Answer

Given:
$ I _2= I _1+0.04 I _1=1.04 I _1$
$n _1- n _2=6$
To find: Initial frequency of tuning fork $\left( n _1\right)$
Formula: $\left. n _1\right|_1= n _2 I _2$
Calculation:
From formula,
$ n _1 I _1= n _2\left(1.04 I _1\right)$
$\therefore n _1=1.04 n _2$
$\therefore n _2=\frac{1}{1.04} n _1 $
But $n _1- n _2=6$
$ \therefore n _1-\frac{ n _2}{1.04}=6$
$\therefore 0.04 n _1=6.24$
$\therefore n _1=156 Hz $
The initial frequency of the tuning fork is $156 H z$.

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