Question

A uniform square plate of side $a$ is hinged at one of its corners as shown. It is suspended such that it can rotate about horizontal axis. Find out its time period for small oscillations about its equilibrium position.

• A. $2\pi \sqrt{\frac{\sqrt{2}a}{6g}}$
• B. $2\pi \sqrt{\frac{2\sqrt{2}a}{3g}}$
• C. $2\pi \sqrt{\frac{2a}{g}}$
• D. $2\pi \sqrt{\frac{a}{2g}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{2\sqrt{2}a}{3g}}$



Moment of inertia of a square plate about one of its corners and perpendicular to its plane,
$I=I_{Cm}+m{d}^2$
$I=m\frac{a^2}{6}+m{\left(\frac{a}{\sqrt{2}}\right)}^2$
Time period of the physical pendulum is
$T=2\pi \sqrt{\frac{I}{mgd}}$
$T=2\pi \sqrt{\frac{m\frac{a^2}{6}+m{\left(\frac{a}{\sqrt{2}}\right)}^2}{mg.\frac{a}{\sqrt{2}}}}$
$T=2\pi \sqrt{\frac{(\frac{a}{6}+\frac{a}{2}).\sqrt{2}}{g}}=2\pi \sqrt{\frac{2\sqrt{2}a}{3g}}$