A tuning fork gives 3 beats with $50 \mathrm{~cm}$ length of sonometer wire. If the length of the wire is…
A tuning fork gives 3 beats with $50 \mathrm{~cm}$ length of sonometer wire. If the length of the wire is shortened by $1 \mathrm{~cm}$, the number of beats is still the same. The frequency of the fork is
$256 \mathrm{~Hz}$
$288 \mathrm{~Hz}$
$297 \mathrm{~Hz}$
$320 \mathrm{~Hz}$
Solution
The frequency of a vibrating wire $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
$\therefore \quad \mathrm{f} \propto \frac{1}{l} \Rightarrow \mathrm{fl}=\text { constant }$
Let the frequency of the fork be $f$ and the initial and final frequencies of the wire be $f_1$ and $f_2$.
The number of beats heard before decreasing the length is $f-f_1=3$
The number of beats after decreasing the length is $\mathrm{f}_2-\mathrm{f}=3$
$\begin{array}{ll}
\therefore & \mathrm{f}_1 l_1=\mathrm{f}_2 l_2 \\
\therefore & (\mathrm{f}-3) 50=(\mathrm{f}+3) 49 \\
& 50 \mathrm{f}-49 \mathrm{f}=147+150 \\
\therefore \quad & \mathrm{f}=297 \mathrm{~Hz}
\end{array}$