Q. Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is
A) 2:1
B) 3:4
C) 3:2
D) 1:3
Explanation
(d) Frequency of 1st overtone of A
$n_{1} = \frac{2}{2l_{1}}\sqrt{\frac{T}{m}} = \frac{2}{l_{1}D_{1}}\sqrt{\frac{T}{\pi\rho}}$
Frequency of 2nd overtone of B
$n_{2} = \frac{3}{2l_{2}}\sqrt{\frac{T}{m}} = \frac{2}{l_{2}D_{2}}\sqrt{\frac{T}{\pi\rho}}$
As $n_{1} = n_{2}$
$\therefore\frac{2}{l_{1}D_{1}} = \sqrt{\frac{T}{\pi\rho}} = \frac{3}{l_{2}D_{2}}\sqrt{\frac{T}{\pi\rho}}$
$\frac{l_{1}D_{1}}{l_{2}D_{2}} = \frac{2}{3}$
$\frac{l_{1}}{l_{2}} = \frac{2D_{2}}{D_{2}} = \frac{2}{3};\frac{l_{1}}{l_{2}} = 1:3
$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.