A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period T 1 and (ii) back and forth in a direction perpendicular to its plane, with a period T 2 . T

Question

A ring is hung on a nail. It can oscillate, without slipping or sliding (i) in its plane with a time period T 1 and (ii) back and forth in a direction perpendicular to its plane, with a period T 2 . The ratio T 1 T 2 will be :

TestHub · Accepted Answer

The time period of oscillation of physical pendulum is given by

$T = 2 π \sqrt{\frac{I}{M g l}} . . . (1)$

Where, $I$ is the moment of inertia of ring about given axis and $l$ is the distance of centre of mass of ring from the axis of rotation.

Moment of inertia of ring in its plane is given by

$I_{1} = I_{c} + M d^{2} = M R^{2} + M R^{2} = 2 M R^{2}$

Moment of inertia of ring perpendicular to its plane is given by

$I_{2} = I_{c} + M d^{2} = \frac{M R^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2}$

Now the ratio of time period is given by

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1}}{I_{2}}}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{2 M R^{2}}{\frac{3}{2} M R^{2}}}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$

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