Question

A thin rod of length $l$ in the shape of a semicircle is pivoted at one of its ends such that it is free to oscillate in its own plane. The frequency $f$ of small oscillations of the semicircular rod is :

• A. $\frac{1}{2\pi }\sqrt{\frac{g\pi }{2l}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{g\sqrt{{\pi }^2+4}}{2l}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{g\sqrt{\pi +2}}{l}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{g\sqrt{{\pi }^2+1}}{2\pi l}}$

Answer 1

Answer: (B)

$\frac{1}{2\pi }\sqrt{\frac{g\sqrt{{\pi }^2+4}}{2l}}$

Given that a thin rod of length $l$ in the shape of a semicircle is pivoted at one of its ends such that it is free to oscillate in its own plane.

We have to find the frequency $f$ of small oscillations of the semicircular rod.

The pictorial representation of the problem is shown below.

Length of the rod $l=\pi R$ and $I=2m{R}^2$

Since $f=\frac{1}{2\pi }\sqrt{\frac{mg{l}_0}{I}}$

From the figure, we have

$l_0=\sqrt{R^2+{\left(\frac{2R}{\pi }\right)}^2}$

Therefore $f=\frac{1}{2\pi }\sqrt{\frac{mg\times \sqrt{R^2+{\left(\frac{2R}{\pi }\right)}^2}}{2m{R}^2}}$

$=\frac{1}{2\pi }\sqrt{\frac{g\times \sqrt{1+\left(\frac{4}{{\pi }^2}\right)}}{2R}}$

$=\frac{1}{2\pi }\sqrt{\frac{g\times \sqrt{{\pi }^2+4}}{2\pi R}}$

$=\frac{1}{2\pi }\sqrt{\frac{g\sqrt{{\pi }^2+4}}{2l}}$

Thus $f=\frac{1}{2\pi }\sqrt{\frac{g\sqrt{{\pi }^2+4}}{2l}}$.