When two tuning forks are sounded together, 5 beats per second are heard. One of the forks is in unison with…
When two tuning forks are sounded together, 5 beats per second are heard. One of the forks is in unison with $0.97 \mathrm{~m}$ length of sonometer wire and the other is in unison with $0.96 \mathrm{~m}$ length of the same wire. The frequencies of the two tuning forks are
$383 \mathrm{~Hz}, 388 \mathrm{~Hz}$
$388 \mathrm{~Hz}, 392 \mathrm{~Hz}$
$475 \mathrm{~Hz}, 480 \mathrm{~Hz}$
$480 \mathrm{~Hz}, 485 \mathrm{~Hz}$
Solution
Given number of beats $=5, l_1=0.97 \mathrm{~m}$, $l_2=0.96 \mathrm{~m}$
The frequencies of the given sonometer are as follows:
$\mathrm{f}_1=\frac{1}{2 \mathrm{l}_1} \sqrt{\frac{\mathrm{TL}}{\mathrm{m}}}$
$\mathrm{f}_2=\frac{1}{2 \mathrm{l}_2} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$
$\mathrm{f}_2-\mathrm{f}_1=5$
$\frac{1}{21_2} \sqrt{\frac{T}{m}}-\frac{1}{21_1} \sqrt{\frac{T}{m}}=5$
$\left(\frac{1}{2(0.96)}-\frac{1}{2(0.97)}\right) \sqrt{\frac{T}{m}}=5$
$\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=931.2$
$\therefore \quad \mathrm{f}_1=480 \mathrm{~Hz}$ and $\mathrm{f}_2=485 \mathrm{~Hz}$