Each of the two strings of length 51.6 , cm a | vedclass

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(A) Given: Lengths $l_1 = 0.516 \, m$ and $l_2 = 0.491 \, m$. Tension $T = 20 \, N$. Mass per unit length $\mu = 1 \, g/m = 0.001 \, kg/m$.The fundame

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$7$

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(A) Given: Lengths $l_1 = 0.516 \, m$ and $l_2 = 0.491 \, m$. Tension $T = 20 \, N$. Mass per unit length $\mu = 1 \, g/m = 0.001 \, kg/m$.The fundamental frequency of a vibrating string is given by $v = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Calculate the wave speed $v_w = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{0.001}} = \sqrt{20000} = 100\sqrt{2} \approx 141.42 \, m/s$.Frequency $v_1 = \frac{141.42}{2 	imes 0.516} \approx 137.04 \, Hz$.Frequency $v_2 = \frac{141.42}{2 	imes 0.491} \approx 144.02 \, Hz$.The number of beats is the difference in frequencies: $|v_2 - v_1| = 144.02 - 137.04 = 6.98 \approx 7 \, Hz$.

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

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