Two strings A and B of lengths L_A = 80 text{ | vedclass

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(C) Given: $T_A = T_B$, $f_A = f_B$, $L_A = 80 	ext{ cm}$, $L_B = x 	ext{ cm}$.Ratio of densities: $\frac{d_A}{d_B} = 0.81$.Ratio of diameters: $\fr

Vedclass · Accepted Answer

$144$

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(C) Given: $T_A = T_B$, $f_A = f_B$, $L_A = 80 	ext{ cm}$, $L_B = x 	ext{ cm}$.Ratio of densities: $\frac{d_A}{d_B} = 0.81$.Ratio of diameters: $\frac{D_A}{D_B} = 2$.Linear mass density $\mu = 	ext{Area} 	imes 	ext{density} = \frac{\pi D^2}{4} 	imes d$.Therefore, $\frac{\mu_A}{\mu_B} = \left(\frac{D_A}{D_B}ight)^2 	imes \frac{d_A}{d_B} = (2)^2 	imes 0.81 = 4 	imes 0.81 = 3.24$.The fundamental frequency of a string is $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since $f_A = f_B$ and $T_A = T_B$, we have $L_A \sqrt{\mu_A} = L_B \sqrt{\mu_B}$.$\frac{L_B}{L_A} = \sqrt{\frac{\mu_A}{\mu_B}} = \sqrt{3.24} = 1.8$.$x = 80 	imes 1.8 = 144 	ext{ cm}$.

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