Two strings A and B of lengths L_A = 80 cm a | vedclass

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(C) Given: $T_A = T_B$,$f_A = f_B$,$L_A = 80 \ cm$,$L_B = x \ cm$,$d_A / d_B = 0.81$,and $D_B = D_A / 2$.Linear mass density $\mu$ is given by $\mu =

Vedclass · Accepted Answer

$144$

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(C) Given: $T_A = T_B$,$f_A = f_B$,$L_A = 80 \ cm$,$L_B = x \ cm$,$d_A / d_B = 0.81$,and $D_B = D_A / 2$.Linear mass density $\mu$ is given by $\mu = 	ext{Area} 	imes 	ext{density} = \pi (D/2)^2 	imes d = \frac{\pi D^2 d}{4}$.Therefore,the ratio of linear densities is $\frac{\mu_A}{\mu_B} = \left(\frac{D_A}{D_B}ight)^2 	imes \frac{d_A}{d_B} = (2)^2 	imes 0.81 = 4 	imes 0.81 = 3.24$.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 $\frac{1}{L_A \sqrt{\mu_A}} = \frac{1}{L_B \sqrt{\mu_B}}$,which implies $\frac{L_B}{L_A} = \sqrt{\frac{\mu_B}{\mu_A}}$.Substituting the values: $\frac{x}{80} = \sqrt{\frac{1}{3.24}} = \frac{1}{1.8}$.$x = \frac{80}{1.8} = \frac{800}{18} \approx 44.44 \ cm$. Wait,re-evaluating the ratio: $\frac{L_B}{L_A} = \sqrt{\frac{\mu_B}{\mu_A}} \implies \frac{x}{80} = \sqrt{\frac{1}{3.24}} = \frac{1}{1.8}$.Actually,$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \implies L \propto \frac{1}{\sqrt{\mu}}$. So $\frac{L_B}{L_A} = \sqrt{\frac{\mu_A}{\mu_B}} = \sqrt{3.24} = 1.8$.$x = 80 	imes 1.8 = 144 \ cm$.

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