Question

Two uniform 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 radius of the first wire is twice the radius of the second wire then the ratio of the lengths of the first wire to second wire is

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{5}$
• D. $\frac{1}{6}$

Answer 1

Answer: (A)

Fundamental frequency of the first wire
$f=\frac{1}{2 L_{1}} \, \sqrt{\frac{T}{m} \, }=\frac{1}{2 L_{1}}\sqrt{\frac{T}{\pi r_{1 \, }^{2} \rho } \, }=\frac{1}{2 L_{1} r_{1}}\sqrt{\frac{T}{\pi \rho } \, }$
The first overtone $f_{1}=2f=\frac{2}{2 L_{1} r_{1}}\sqrt{\frac{T}{\pi \rho } \, }=\frac{1}{L_{1} r_{1}}\sqrt{\frac{T}{\pi \rho } \, }$
The second overtone of the second wire
$f_{2}=\frac{3}{2 L_{2} r_{2}}\sqrt{\frac{T}{\pi \rho } \, }$
$f_{1}=f_{2}$
$\frac{1}{L_{1} r_{1}}\sqrt{\frac{T}{\pi \rho } \, }=\frac{3}{2 L_{2} \, r_{2}} \, \sqrt{\frac{T}{\pi \rho } \, }$
$\therefore \, \, \, 3L_{1}r_{1}=2L_{2}r_{2}$
$\frac{L_{1}}{L_{2}}=\frac{2}{3}.\frac{r_{2}}{r_{1}}$
$=\frac{2}{3}.\frac{r_{2}}{2 r_{2}}$ $\left[\right.\therefore \, \, \, r_{1}=2r_{2}\left]\right.$
$=\frac{1}{3}$