Question

Two wires are kept tight between the same pair of supports. The tensions in the wire are in the ratio $3:4$. Their radii are in the ratio $1:2$ and the density are in the ratio $3:4$. The ratio of their fundamental frequencies are

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Frequency of standing wave on a string fixed at both end
$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$
For fundamental frequency $\left(f_1\right)$
$f_1=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
So, where all have usual menning
$\frac{f_1}{f_2}=\frac{\frac{1}{2L_1}\sqrt{\frac{T_1}{{\mu }_1}}}{\frac{1}{2L_2}\sqrt{\frac{T_2}{{\mu }_2}}}=\sqrt{\frac{T_1}{T_2}.\frac{{\mu }_2}{{\mu }_1}}[\because L_1=L_2]$
Now
$\mu =\frac{m}{L}=\frac{\rho AL}{L}=\frac{\rho \frac{\pi }{4}{r}^{2}L}{L}\propto \rho \pi r^2$
$\left[\begin{array}{c}\rho =\text{density of wire}\\ r=\text{radius of wire}\\ L=\text{Length of wire}\end{array}\right]$
So
$\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}\frac{{\rho }_2{r}_2^2}{{\rho }_1{r}_1^2}}$
$=\sqrt{\frac{3}{4}\times \frac{4}{3}\times {\left(\frac{2}{1}\right)}^2}=2$