Question

A wire frame in the shape of an equilateral triangle is hinged at one vertex so that it can swing freely in a vertical plane, with the plane of the triangle always remaining vertical. The side of the frame is $1/\sqrt{3}m$. The time period in seconds of small oscillations of the frame will be $(g=10m/s^2)$

• A. $\pi /\sqrt{2}$
• B. $\pi /\sqrt{3}$
• C. $\pi /\sqrt{6}$
• D. $\pi /\sqrt{5}$

Answer 1

The correct option is D $\pi /\sqrt{5}$

$OP=a\sin {60}^{\circ }=\frac{\sqrt{3}a}{2}$
$OC=l=\frac{2}{2}\left(OP\right)=\frac{a}{\sqrt{3}}$
$I=I_0=2\left(\frac{m{a}^2}{3}\right)+m{\left(OP\right)}^2+\frac{m{a}^2}{12}$
$=\frac{2}{3}m{a}^2+\frac{3}{4}m{a}^2+\frac{1}{12}m{a}^2$
$=\frac{3}{2}m{a}^2$
$T=2\pi \sqrt{\frac{I}{\left(3m\right)gl}}$
$=2\pi \sqrt{\frac{\left(3/2m{a}^2\right)}{\left(3m\right)g\left(\frac{a}{\sqrt{3}}\right)}}$
$=2\pi \sqrt{\frac{\sqrt{3}}{2}\frac{a}{g}}$
Putting $a=\frac{1}{\sqrt{3}}$ m and g=10 m/s${}^2$
We get, $T=\frac{\pi }{\sqrt{5}}$ sec