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Q.

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 $(g = 10 m / s^{2})$

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a

$π / \sqrt{3}$

b

$π / \sqrt{2}$

c

$π / \sqrt{6}$

d

$π / \sqrt{5}$

answer is D.

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Detailed Solution

$O P = a \sin 60 ° = \frac{\sqrt{3} a}{2} O C = l = \frac{2}{3} (O P) = \frac{a}{\sqrt{3}} I = I_{0} = 2 (\frac{m a^{2}}{3}) + m (O P)^{2} + \frac{m a^{2}}{12} = \frac{2}{3} m a^{2} + \frac{3}{4} m a^{2} + \frac{1}{12} m a^{2} = \frac{3}{2} m a^{2}$

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 Delta always

$T = 2 π \sqrt{\frac{I}{(3 m) g l}} = 2 π \sqrt{\frac{(3 / 2 m a^{2})}{(3 m) g (\frac{a}{\sqrt{3}})}} = 2 π \sqrt{\frac{\sqrt{3} \frac{a}{2}}{g}}$

Putting $a = \frac{1}{\sqrt{3}} m$ and $g = 10 m / s^{2}$

we get, $T = \frac{π}{\sqrt{5}} s$

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