Question

A meter stick swinging in vertical plane about a fixed horizontal axis passing through its one end undergoes small oscillation of frequency $f_{0}$. If the bottom half of the stick were cut off, if its new frequency of small oscillation is found to be $\sqrt{x} f_{0}$. The value of $x$ is ____

Answer 1

We use $f_{0}=\frac{1}{2 \pi} \sqrt{\frac{m g l}{I}}$
where $l$ is distance between point of suspension and centre of mass of the body.
Thus, for the stick of length $L$ and mass $m$ frequency is
$f_{0}=\frac{1}{2 \pi} \sqrt{\frac{m \cdot g \cdot \frac{L}{2}}{\left(m L^{2} / 12\right)}}=\frac{1}{2 \pi} \sqrt{\frac{6 g}{L}}$
when bottom half of the stick is cut off we use
$f_{1}=\frac{1}{2 \pi} \sqrt{\frac{\frac{m}{2} \cdot g \cdot \frac{L}{4}}{\frac{m}{2} \frac{(L / 2)^{2}}{12}}}$
$=\frac{1}{2 \pi} \times \sqrt{\frac{12 g}{L}}=\sqrt{2} f_{0}$