Question

A metre stick swinging in a vertical plane about a fixed horizontal axis passing through its one end undergoes small oscillations of frequency $f_0$. If the bottom half of the stick is cut off and glued to the upper half, then the new frequency for small oscillations is

• A. $f=\sqrt{2}{f}_0$
• B. $f=\frac{1}{\sqrt{2}}{f}_0$
• C. $f=\sqrt{3}{f}_0$
• D. $f=f_0$

Answer 1

The correct option is A $f=\sqrt{2}{f}_0$


The system behaves as a physical pendulum
$T=2\pi \sqrt{\frac{I}{mgd}}$
where,
$I=$ moment of inertia about support
$d=$ distance between support and center of mass of body
$\omega =$ angular frequency
We know
$T=\frac{2\pi }{\omega }\&\omega =2\pi f$
$\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{mgd}{I}}$
Initially,
$f_0=\frac{1}{2\pi }\sqrt{\frac{mg\frac{l}{2}}{\frac{m{l}^2}{3}}}=\frac{1}{2\pi }\sqrt{\frac{3g}{2l}}$
After cutting bottom half,


$f=\frac{1}{2\pi }\sqrt{\frac{mg\frac{l}{4}}{\frac{m(l/2{)}^2}{3}}}$
[mass $m$ remains same]
$\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{12g}{4l}}=\left(\frac{1}{2\pi }\sqrt{\frac{3g}{2l}}\right)\sqrt{2}$
$=\sqrt{2}{f}_0$
Hence, option $\left(a\right)$ is the correct answer.