Two identical strings of length $50.0 \mathrm{~cm}$ and $51.0 \mathrm{~cm}$ are tensioned by the sam

Question

Two identical strings of length $50.0 \mathrm{~cm}$ and $51.0 \mathrm{~cm}$ are tensioned by the same force $T$. The mass per unit length of both strings is $1 \mathrm{~g/m}$. When both strings vibrate simultaneously in their fundamental modes, they produce 4 beats per second. The tension $T$ is approximately

Professor PC Thomas Classes · Answer

(a) : We are given $L_1 = 0.50 \mathrm{~m}$, $L_2 = 0.51 \mathrm{~m}$, and linear density $\mu = 0.001 \mathrm{~kg/m}$. Let the velocity of the transverse wave be $v$. The fundamental frequencies are $f_1 = \frac{v}{2L_1}$ and $f_2 = \frac{v}{2L_2}$. The beat frequency is $f_1 - f_2 = 4$. So, $\frac{v}{2(0.50)} - \frac{v}{2(0.51)} = 4$. This simplifies to $v (1 - \frac{1}{1.02}) = 4 \Rightarrow v (\frac{0.02}{1.02}) = 4$, which gives wave speed $v = \frac{4 	imes 1.02}{0.02} = 204 \mathrm{~m/s}$. Since $v = \sqrt{\frac{T}{\mu}}$, the tension $T = \mu v^2 = 0.001 	imes (204)^2 = 41.616 \mathrm{~N} \approx 41.6 \mathrm{~N}$.