Question

Two wires are kept tight between the same pair supports. The tensions in the wires are in the ration $2:1$, the radii are in the ratio $3:1$ and the densities are in the ratio $1:2$. The ratio of their fundamental frequencies is

• A. $2:3$
• B. $2:4$
• C. $2:5$
• D. $2:6$

Answer 1

Answer: (A)

$2:3$

Let $\rho$ be the density of wire, and $\pi r^2$ be the area of the wire.

Thus, mass per unit length of wire $=\rho \times \pi r^2$

Thus, the ratio of mass per unit length of the wires is: $\frac{{\rho }_1}{{\rho }_2}\times \frac{r_1^2}{r_2^2}=\frac{1}{2}\times \frac{9}{1}=\frac{9}{2}$

Fundamental frequency of wire is given by $\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

Since they are kept between same pair of supports, their lengths are equal.

Thus, $\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}\times \frac{{\mu }_2}{{\mu }_1}}$ $=\sqrt{\frac{2}{1}\times \frac{2}{9}}$ $=2:3$