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

Answer: (C)

If a particle performs SHM with angular frequency ${}^{\prime }{\omega }^{\prime }$ and amplitude ' $A$ ' then its displacement from mean position will be equal to $x=a\sin \omega t$. If its initial phase is equal to zero, velocity will be
$v=\frac{dx}{dt}=A\omega \cos \omega t$ .......(i)
and acceleration will be
$a=\frac{dv}{dt}=-A{\omega }^2\sin (\omega t)$ ......(ii)
From these equations,
$\cos \omega t=\frac{v}{A\omega },\sin \omega t=\frac{a}{-A{\omega }^2}$
Squaring and adding,
$\frac{v^2}{A^2{\omega }^2}+\frac{a^2}{A^2{\omega }^4}=1$ ......(iii)
If $v$ taken on $y$ -axis and $a$ on $x$ -axis, then it will be an ellipse.