A rod of mass m and length l is suspended fro | vedclass

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(B) The rod forms an equilateral triangle with the two strings,so the distance from the suspension point to the rod is $d = l \sin(60^\circ) = \frac{l

Vedclass · Accepted Answer

$\frac{10}{9}$

Vedclass · Answer

(B) The rod forms an equilateral triangle with the two strings,so the distance from the suspension point to the rod is $d = l \sin(60^\circ) = \frac{l\sqrt{3}}{2}$.For oscillation in the plane of the page (physical pendulum),the moment of inertia about the suspension point is $I = I_{cm} + md^2 = \frac{ml^2}{12} + m\left(\frac{l\sqrt{3}}{2}ight)^2 = ml^2\left(\frac{1}{12} + \frac{3}{4}ight) = ml^2\left(\frac{10}{12}ight) = \frac{5ml^2}{6}$.The time period is $T_1 = 2\pi \sqrt{\frac{I}{mgd}} = 2\pi \sqrt{\frac{5ml^2/6}{mg(l\sqrt{3}/2)}} = 2\pi \sqrt{\frac{5l}{3\sqrt{3}g}}$.For oscillation perpendicular to the plane,the rod acts as a simple pendulum of length $d = \frac{l\sqrt{3}}{2}$.The time period is $T_2 = 2\pi \sqrt{\frac{d}{g}} = 2\pi \sqrt{\frac{l\sqrt{3}}{2g}}$.Calculating the ratio: $\frac{T_1^2}{T_2^2} = \frac{5l / (3\sqrt{3}g)}{l\sqrt{3} / (2g)} = \frac{5}{3\sqrt{3}} 	imes \frac{2}{\sqrt{3}} = \frac{10}{3 	imes 3} = \frac{10}{9}$.

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