Question

A rod of mass $m$, length $l$ is held horizontal, using a vertical string through its centre. If it is turned a little, the frequency of oscillation will be proportional to- [$C$-torsional constant of the string]

• A. $\frac{1}{2\pi }\sqrt{\frac{3C}{m{l}^2}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{12C}{m}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{12C}{m{l}^2}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{m}{12C}}$

Answer 1

The correct option is C $\frac{1}{2\pi }\sqrt{\frac{12C}{m{l}^2}}$

Time period of torsional pendulum is $T=2\pi \sqrt{\frac{I}{C}}$
Here $I=\frac{m{l}^2}{12},$ so
$T=2\pi \sqrt{\frac{m{l}^2}{12C}}$
Frequency of pendulum is
$f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{12C}{m{l}^2}}$