Name: Superposition of Two or More SHMs
Uploaded: 2019-07-19
Duration: 9 min 11 s
Description: What happens when two or more simple harmonic motions are superimposed? Watch this video to learn what would be the result of the superposition of SHMs.

Question

A uniform disc of mass $m$ and radius $R$ is connected with two light springs $1$ and $2 .$ The springs are connected at the highest point $M$ and the $CM' N'$ of the disc. The other ends of the springs are rigidly attached with vertical walls. If we shift the CM in horizontal by a small distance, the disc oscillates simple harmonically. Assuming a perfect rolling of the disc on the horizontal surface, find the angular frequency of oscillation.

Accepted Answer

On displacing the CM towards right by the small distance $x$ , which results restoring forces in the both springs opposite direction of displacement.

Since CM of mass displaced through a small linear displacement, therefore corresponding angular displacement for spring $1$ will be $x^{,} = 2 R θ$ and for spring $2$ $x = R θ$ , where $θ$ is the angle of rotation of the rolling body.

Spring forces due to displacement,

$F_{1} {=k}_{1} x^{,} {=2k}_{1} Rθ$ and $F_{2} {=k}_{2} {x=k}_{2} Rθ$

These forces generate anticlockwise torque,

$T_{1 =} F_{1} {2R=4k}_{1} R^{2} θ$ and $T_{2} {=F}_{2} {R=k}_{2} R^{2} θ$

Net torque, $T_{net} = - (4 R^{2} k_{1} + R^{2} k_{2}) θ = - (4 k_{1} + k_{2}) R^{2} θ$

This torque produces anticlockwise angular acceleration $α$ , which is given by,

$α = \frac{T_{net}}{I_{}} = \frac{- (4 k_{1} + k_{2}) R^{2} θ}{I_{}}$ ,here $I$ is moment of inertia at contact point, ${⇒I}_{} = m R^{2} + I_{C} = m R^{2} + \frac{m R^{2}}{2} = \frac{3 m R^{2}}{2}$

$\Rightarrow α = \frac{- 2 (4 k_{1} + k_{2})}{3 m} θ$

Comparing the above equation with $α = - ω^{2} θ,$

$\Rightarrow ω = \sqrt{\frac{2 (4 k_{1} + k_{2})}{3 m}}$

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