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Q. The frequency of a tuning fork $A$ is $2\%$ more than the frequency of a standard tuning fork. The frequency of another fork $B$ is $3\%$ less than the frequency of a standard tuning fork. If $6 \, beats \, s^{- 1}$ are heard when the two tuning forks $A$ and $B$ are excited, the frequency of $A$ is

NTA AbhyasNTA Abhyas 2020Waves

Solution:

Let the frequency of standard fork $=x$
$\therefore n_{A}=\frac{102}{100}x$ , $n_{B}=\frac{97}{100}x$
Number of beats per second $=n_{A}-n_{B}=6$
$\frac{102}{100}x-\frac{97}{100}x=6$
$x=\frac{6 \times 100}{5}=120 \, Hz$
$\therefore n_{A}=\frac{102}{100}x=122.4Hz$