A tuning fork ' $A$ ' produces 5 beats per second with a tuning fork of frequency 480 $\mathrm{Hz}$. When a…
A tuning fork ' $A$ ' produces 5 beats per second with a tuning fork of frequency 480
$\mathrm{Hz}$. When a little wax is stuck to a prong of fork $\mathrm{A}$, the number of beats heard per
second becomes $2 .$ What is the frequency of tuning fork $\mathrm{A}$ before the wax is stuck
to it?
$485 \mathrm{~Hz}$
$478 \mathrm{~Hz}$
$475 \mathrm{~Hz}$
$482 \mathrm{~Hz}$
Solution
Frequency of tuning fork $\mathrm{B}=480 \mathrm{~Hz}$. It produces 5 beats with tuning fork $\mathrm{A}$. Therefore frequency of tuning fork $\mathrm{A}$ can be $485 \mathrm{~Hz}$ or $475 \mathrm{~Hz}$. On loading the tuning fork $\mathrm{A}$ with wax its frequency will decrease. Since the number of beats decreases, it means its frequency is higher than that of B. Hence its frequency is $485 \mathrm{~Hz}$.
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