Two uniform stretched steel strings A and B a | vedclass

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(A) The frequency of a stretched string is given by $f = \frac{n}{2l} \sqrt{\frac{T}{m}}$,where $n$ is the harmonic number,$l$ is the length,$T$ is th

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$1/3$

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(A) The frequency of a stretched string is given by $f = \frac{n}{2l} \sqrt{\frac{T}{m}}$,where $n$ is the harmonic number,$l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.Since $m = \pi r^2 \rho$ (where $\rho$ is the density),the frequency formula becomes $f = \frac{n}{2l r} \sqrt{\frac{T}{\pi \rho}}$.For the first overtone of $A$,$n = 2$. Thus,$f_A = \frac{2}{2 l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.For the second overtone of $B$,$n = 3$. Thus,$f_B = \frac{3}{2 l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.Given $f_A = f_B$,we have $\frac{1}{l_A r_A} = \frac{3}{2 l_B r_B}$.Rearranging for the ratio of lengths: $\frac{l_A}{l_B} = \frac{2 r_B}{3 r_A}$.Given $r_A = 2 r_B$,we substitute to get $\frac{l_A}{l_B} = \frac{2 r_B}{3 (2 r_B)} = \frac{1}{3}$.

Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of $A$ is equal to the second overtone of $B$. If the radius of $A$ is twice that of $B$,then the ratio of the lengths of the strings $l_A/l_B$ is:

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