1. Write the standard SHM relations

For a particle in simple harmonic motion, $x = A\cos(\omega t + \phi)$

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

2. Eliminate the time parameter

From $x = A\cos\theta$ and $v = -A\omega \sin\theta$ where $\theta = \omega t + \phi$.

Now square both equations: $x^2 = A^2\cos^2\theta$ $v^2 = A^2\omega^2\sin^2\theta$

Divide appropriately: $\frac{x^2}{A^2} = \cos^2\theta$ $\frac{v^2}{A^2\omega^2} = \sin^2\theta$

Adding, $\frac{x^2}{A^2} + \frac{v^2}{A^2\omega^2} = \cos^2\theta + \sin^2\theta = 1$

3. Identify the graph

This is the equation of an ellipse in the $x$-$v$ plane: $\frac{x^2}{A^2} + \frac{v^2}{(A\omega)^2} = 1$

So the graph of velocity as a function of displacement is elliptical.

4. Check options

• A: Circular — Incorrect in general. It becomes a circle only in specially scaled axes, not in the usual $v$ vs $x$ graph.
• B: Elliptical — Correct.
• C: Sinusoidal — Incorrect. Sinusoidal variation occurs with time, not with displacement.
• D: Straight line — Incorrect.

Therefore, the correct answer is B.