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?

If a particle performs SHM with angular frequency '$\mathrm{\omega }$' and amplitude 'A' then its displacement from mean position will be equal to x = a sin $\mathrm{\omega }$t. If its initial phase is equal to zero, velocity will be
$\mathrm{v} = \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{A\omega } \mathrm{cos\omega t}------\left(\mathrm{i}\right)$
and acceleration will be
$\mathrm{a}= \frac{\mathrm{dv}}{\mathrm{dt}} = -{\mathrm{A\omega }}^2\sin \left(\mathrm{\omega t}\right)------\left(\mathrm{ii}\right)$

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

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

If ‘v’ taken on y-axis and a on x-axis, then it will be an ellipse.