Q. A tuning fork of frequency $392\ Hz,$ resonates with $50\ cm$ length of a string under tension $(T)$. If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is
A) 4
B) 6
C) 8
D) 12
Explanation
(c) $n \propto \frac{1}{l} \Rightarrow \frac{\mathrm{\Delta}n}{n} = - \frac{\mathrm{\Delta}l}{l}$
If length is decreased by 2% then frequency increases by 2%
$i.e.,\ \frac{n_{2} - n_{1}}{n_{1}} = \frac{2}{100}$
$\Rightarrow n_{2} - n_{1} = \frac{2}{100} \times n_{1} = \frac{2}{100} \times 392 = 7.8 \approx 8$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.