Question

A particle is performing a linear simple harmonic motion. If the acceleration and the corresponding velocity of the particle are a and v respectively, which of the following graphs is correct?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

If a particle performs SHM with angular frequency '$\omega$' and amplitude 'A', then its displacement from mean position will be equal to $x=Asin\omega t$, if its initial phase is equal to zero.

Velocity will be equal to

$v=\frac{dx}{dt}=A\omega cos\omega t\to \left(1\right)$ and acceleration will be $\alpha =\frac{dv}{dt}=-A{\omega }^{2}sin\left(\omega t\right)\to$ (2)

From these equations, $cos\omega t=\frac{v}{A\omega },sin\omega t=\frac{\alpha }{-A{\omega }^2}$

Squaring and adding, $\frac{v^2}{A^2{\omega }^2}+\frac{{\alpha }^2}{A^2{\omega }^4}=1$

If v is taken on the y-axis and $\alpha$ on the x-axis, then it will be an ellipse