Question

A wire (fixed at both ends) of length $l$ having tension $T$ and radius $r$ vibrates with natural frequency $f$. Another wire of same metal with length $2l$ having tension $2T$ and radius $2r$ will vibrate with natural frequency

• A. $f$
• B. $2f$
• C. $2\sqrt{2}f$
• D. $\frac{f}{2\sqrt{2}}$

Answer 1

The correct option is D $\frac{f}{2\sqrt{2}}$

Natural frequency for a wire with both ends fixed.
$f=\frac{v}{2L}=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
Mass per unit length $\left(\mu \right)=\rho s$
where $\rho \to$ density of wire and $s\to$ cross - section
$\Rightarrow f=\frac{1}{2L}\sqrt{\frac{T}{\pi r^2\rho }}$
$\Rightarrow f\propto \frac{\sqrt{T}}{Lr}$
Let $f_2$ be frequency of second string, then
$\therefore \frac{f}{f_2}=\sqrt{\frac{T}{2T}}\times \frac{2L}{L}\times \frac{2r}{r}$
$\Rightarrow \frac{f}{f_2}=\frac{4}{\sqrt{2}}$
$\therefore f_2=\frac{\sqrt{2}}{4}f=\frac{f}{2\sqrt{2}}$