When a particle executes simple Harmonic motion, the nature of graph of velocity as function of displacement will be

We need to determine the nature of the graph of velocity as a function of displacement for a particle executing Simple Harmonic Motion (SHM).

For a particle in SHM with amplitude $$A$$ and angular frequency $$\omega$$, the velocity at displacement $$x$$ is:

$$v = \omega\sqrt{A^2 - x^2}$$

$$v^2 = \omega^2(A^2 - x^2)$$

$$v^2 = \omega^2 A^2 - \omega^2 x^2$$

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

This can be written as:

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

This is the equation of an ellipse with semi-major axis $$A$$ along the $$x$$-axis and semi-major axis $$\omega A$$ along the $$v$$-axis. Since $$\omega \neq 1$$ in general, the two axes are different, making it an ellipse (not a circle).

The correct answer is Option B: Elliptical.