1. For a particle in simple harmonic motion (SHM), the displacement can be written as $x = A\sin \omega t$ where $A$ is amplitude and $\omega$ is angular frequency.

2. The velocity is $v = \frac{dx}{dt} = A\omega \cos \omega t$

3. Eliminate $t$ between $x$ and $v$. From $x = A\sin \omega t \implies \sin \omega t = \frac{x}{A}$ and $v = A\omega \cos \omega t \implies \cos \omega t = \frac{v}{A\omega}$

4. Using the identity $\sin^2 \omega t + \cos^2 \omega t = 1$ we get $\left(\frac{x}{A}\right)^2 + \left(\frac{v}{A\omega}\right)^2 = 1$

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

6. Hence, the graph of velocity as a function of displacement for SHM is an ellipse.

7. Checking options:

   • A: parabola $\rightarrow$ incorrect
   • B: helix $\rightarrow$ incorrect
   • C: ellipse $\rightarrow$ correct
   • D: circle $\rightarrow$ incorrect in general

Therefore, the correct answer is C.