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 $\frac{1}{\sqrt{3}}m$. The time period in seconds of small oscillations of the frame will be :

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

Answer 1

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

G = C.M of system

Time period $T=2\pi \sqrt{\frac{I}{mgd}}$

$I_{CM}$ moment of inertial of the shape about C.M.

$=\frac{1}{16}m{l}^2$ ( Here $l$= length of side d = distance of C.M. from point of contact)

$=\frac{2}{3}\sqrt{l^2-\frac{l^2}{4}}=\frac{2}{3}\times \frac{\sqrt{3}}{2}l=\frac{l}{\sqrt{3}}$

I = moment of oscillation about point of contact O

$=\frac{1}{6}m{l}^2+m{d}^2=\frac{1}{6}m{l}^2+\frac{m{l}^2}{3}=\frac{1}{2}m{l}^2$ (by applying parallel axis theorem)

So,$T=2\pi \sqrt{\frac{\frac{1}{2}m{l}^2}{mg.\frac{l}{\sqrt{3}}}}=2\pi \sqrt{\frac{\sqrt{3}l}{2g}}$

Now $l=\frac{1}{\sqrt{3}}mT=2\pi \sqrt{\frac{1}{2g}}=\frac{\pi }{\sqrt{5}}$

I have taken g = 10 m$/s^2$