A segment of wire vibrates with a fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9…
A segment of wire vibrates with a fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg} \mathrm{wt}$. Then tension at which the fundamental frequency of the same wire becomes $900 \mathrm{~Hz}$ is
$36 \mathrm{~kg}-\mathrm{wt}$
$27 \mathrm{~kg}-\mathrm{wt}$
$18 \mathrm{~kg}-\mathrm{wt}$
$72 \mathrm{~kg}-\mathrm{wt}$
Solution
Fundamental frequency of wire
$\begin{array}{rlrl}f & =\frac{1}{2 \pi} \sqrt{\frac{T}{m}} \\ \text { or } & f & \propto \sqrt{T} \\ \text { or } & \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \\ \text { or } & \frac{900}{450} & =\sqrt{\frac{T_2}{9}} \\ \text { or } & T_2 & =4 \times 9=36 \mathrm{~kg}-\mathrm{wt}\end{array}$