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(C) The fundamental frequency of a string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \rho \pi r^2$.For string $B$: $n_B = \frac{

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(C) The fundamental frequency of a string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = ho \pi r^2$.For string $B$: $n_B = \frac{1}{2l_B} \sqrt{\frac{T_B}{\mu_B}}$.Given for string $A$: $l_A = 2l_B$,$d_A = 2d_B \Rightarrow r_A = 2r_B$,$T_A = 2T_B$,and $ho_A = 2ho_B$.Calculating $\mu_A$: $\mu_A = ho_A \pi r_A^2 = (2ho_B) \pi (2r_B)^2 = 8 ho_B \pi r_B^2 = 8\mu_B$.Now,the fundamental frequency of string $A$ is: $n_A = \frac{1}{2l_A} \sqrt{\frac{T_A}{\mu_A}} = \frac{1}{2(2l_B)} \sqrt{\frac{2T_B}{8\mu_B}} = \frac{1}{4} \left( \frac{1}{2l_B} \sqrt{\frac{T_B}{\mu_B}} ight) = \frac{1}{4} n_B$.The frequency of the $p^{	ext{th}}$ overtone of string $A$ is $f_p = (p+1)n_A$.We want $f_p = n_B$,so $(p+1) \frac{n_B}{4} = n_B$.This gives $p+1 = 4$,or $p = 3$.Thus,the $3^{	ext{rd}}$ overtone of string $A$ has the same frequency as the fundamental frequency of string $B$.

$A$ string $A$ has twice the length,twice the diameter,twice the tension,and twice the density of another string $B$. The overtone of $A$ which will have the same fundamental frequency as that of $B$ is:

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