Question

A metre stick swinging in vertical plane about a fixed horizontal axis passing through its one end undergoes small oscillation of frequency $1\text{Hz}$. If the bottom half of the stick is cut off, then its new frequency for small oscillation (in $\text{Hz}$) would become

Answer 1

$f_0=\frac{1}{2\pi }\sqrt{\frac{mgl}{I}}$
where $l$ is distance between point of suspension and centre of mass of the body.
Thus, for the stick of length $L$ and mass $m$
$f_0=\frac{1}{2\pi }\sqrt{\frac{m.g\frac{L}{2}}{\left(\frac{m{L}^2}{3}\right)}}=\frac{1}{2\pi }\sqrt{\frac{3g}{2L}}$
When bottom half of the stick is cut off, its mass becomes $\frac{m}{2}$ and length $\frac{L}{2}$.
Hence new frequency is
$f_0^{\prime }=\frac{1}{2}\sqrt{\frac{\frac{m}{2}.g.\frac{L}{4}}{\frac{m}{2}\frac{{\left(\frac{L}{2}\right)}^2}{3}}}=\frac{1}{2\pi }\times \sqrt{\frac{3g}{L}}=\sqrt{2}{f}_0$
$f_0^{\prime }=1.414\text{Hz}$

Aliter,

$f\propto \frac{1}{\sqrt{L}}$
$\therefore \frac{f_0^{\prime }}{f_0}=\frac{\sqrt{L}}{\sqrt{L/2}}$
$\therefore f_0^{\prime }=\sqrt{2}{f}_0$

Why this Question? Tip: The result is independent of mass and is inversely proportional to square root of L.