1. Write the standard SHM relations

For a particle executing simple harmonic motion with amplitude $A$ and angular frequency $\omega$,

$x = A\cos(\omega t + \phi)$

Velocity is

$v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi)$

2. Eliminate time to get relation between $v$ and $x$

Using

$\sin^2\theta + \cos^2\theta = 1$

we have

$\left(\frac{x}{A}\right)^2 = \cos^2(\omega t + \phi)$

and

$\left(\frac{v}{A\omega}\right)^2 = \sin^2(\omega t + \phi)$

Adding,

$\left(\frac{x}{A}\right)^2 + \left(\frac{v}{A\omega}\right)^2 = 1$

3. Identify the graph

The equation

$\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = 1$

is the standard equation of an ellipse in the $x$-$v$ plane.

So, the graphical representation of velocity as a function of displacement is elliptical.

4. Check options

• A: circular $\rightarrow$ Incorrect in general. It would be a circle only if the axes were scaled appropriately, not on ordinary $v$ vs $x$ axes.
• B: straight line $\rightarrow$ Incorrect.
• C: parabolic $\rightarrow$ Incorrect.
• D: elliptical $\rightarrow$ Correct.

Therefore, the correct option is D.