Question

A wire of length $l$ having tension $T$ and radius $r$ vibrates with fundamental frequency $f$. Another wire of the same metal with length $2l$ having tension $2T$ and radius $2r$ will vibrate with fundamental frequency :

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

Answer 1

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

Given : length has doubled so $\lambda$ becomes double

$so,\frac{{\lambda }_1}{{\lambda }_2}=\frac{1}{2}-\left(1\right)$

frequency $\left(f\right)=\frac{v_1}{{\lambda }_1}=\frac{\sqrt{T/\rho \pi r^2}}{{\lambda }_1}$

frequency in another wire = $\frac{v_2}{{\lambda }_2}=\frac{\sqrt{\frac{2T}{\rho \pi (2r{)}^2}}}{{\lambda }_2}$

$so,\frac{f}{f^{,}}=\frac{\frac{v_1}{{\lambda }_1}}{\frac{v_2}{{\lambda }_2}}\frac{\sqrt{\frac{T}{\rho \pi (r{)}^2{\lambda }_1}}}{\sqrt{\frac{2T}{\rho \pi (2r{)}^2{\lambda }_2}}}$

$=\sqrt{\frac{2}{1}}\times \frac{{\lambda }_2}{{\lambda }_1}$

$=\sqrt{2}\times 2\left(From\right(1\left)\right)$

$so,f^{,}=\frac{f}{2\sqrt{2}}$