Question 26 of 194

When a particle executes simple Harmonic motion, the nature of graph of velocity as function of displacement will be

ACircularBEllipiticalCSinusoidalDStraight line

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Answer: (B) Ellipitical

The displacement equation of a particle is given by
$x=A\sin \left(\omega t+\phi \right) \Rightarrow \sin \left(\omega t+\phi \right)=\dfrac{x}{A}$
The velocity equation is given by,
$v=\omega A\cos \left(\omega t+\phi \right) \Rightarrow \cos \left(\omega t+\phi \right)=\dfrac{v}{\omega A}$
Since, ${\sin }^{2}\theta +{\cos }^{2}\theta =1$
$\Rightarrow \dfrac{{x}^{2}}{{A}^{2}}+\dfrac{{v}^{2}}{{\omega }^{2}{A}^{2}}=1$
The above equation represents an ellipse.

JEE Main Physics practice question on Oscillations (Simple Harmonic Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.