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Q. The frequencies of two tuning forks $A$ and $B$ are respectively $1.5\%$ more and $2.5\%$ less than that of the tuning fork $C$. When $A$ and $B$ are sounded together, $12$ beats are produced in $1\, \sec$. The frequency of the tuning fork $C$ is

KEAMKEAM 2010Waves

Solution:

Let the frequency of tuning fork $C$ is $x\, Hz$.
The frequency of tuning fork $A$ is
$ x + \frac{1.5}{100}x = 1.015\, x$
The frequency of tuning fork $B$ is $x - \frac{2.5}{100}x$
$= 0.975\, x$
Beat frequency of tuning fork $A$ and $B$ is
$1.015\,x - 0.975\, x = 0.04\, x$
Thus, $0.04\, x = 12$
$\Rightarrow x = \frac{12}{0.04} = 300 \, Hz$