Two wires of the same material are vibrating | vedclass

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(A) The frequency of the $p$-th overtone for a string fixed at both ends is given by $f = (p+1) \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $\mu = \p

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(A) The frequency of the $p$-th overtone for a string fixed at both ends is given by $f = (p+1) \frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.Since the material is the same,density $\rho$ is constant. The tension $T$ is also constant.For the first wire,the first overtone $(p=1)$ is $f_1 = 2 \cdot \frac{1}{2 \ell_1} \sqrt{\frac{T}{\pi r_1^2 \rho}} = \frac{1}{\ell_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.For the second wire,the second overtone $(p=2)$ is $f_2 = 3 \cdot \frac{1}{2 \ell_2} \sqrt{\frac{T}{\pi r_2^2 \rho}} = \frac{3}{2 \ell_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.Given $f_1 = f_2$,we have $\frac{1}{\ell_1 r_1} = \frac{3}{2 \ell_2 r_2}$.Given $r_1 = 2 r_2$,substituting this into the equation: $\frac{1}{\ell_1 (2 r_2)} = \frac{3}{2 \ell_2 r_2}$.Simplifying,$\frac{1}{2 \ell_1} = \frac{3}{2 \ell_2}$,which gives $\frac{\ell_1}{\ell_2} = \frac{1}{3}$.

Two wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and the radius of the first wire is twice the radius of the second,then the ratio of the length of the first wire to the second wire is:

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