Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass p

Question

Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to $1 \mathrm{~g} / \mathrm{m}$. When both the strings vibrate simultaneously the number of beats is\

Professor PC Thomas Classes · Answer

(a) : $l_{1}=0.516 \mathrm{~m}, l_{2}=0.491 \mathrm{~m}, T=20 \mathrm{~N}$. Mass per unit length, $\mu=0.001 \mathrm{~kg} / \mathrm{m}$. Frequency, $v=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}$ 
$ \begin{aligned} & v_{1}=\frac{1}{2 	imes 0.516} \sqrt{\frac{20}{0.001}} \mathrm{~Hz}=137 \mathrm{~Hz} \ & v_{2}=\frac{1}{2 	imes 0.491} \sqrt{\frac{20}{0.001}} \mathrm{~Hz}=144 \mathrm{~Hz} \end{aligned} $
 $	herefore \quad$ Number of beats $=v_{1}-v_{2}=7 \mathrm{~Hz}$\