Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m$ kg and the spring constant is $k\ \text{N m}^{-1}$. At a given instant, the extension of the spring is $x$ meter and the speed of the particle is $v\ \text{m s}^{-1}$. On the $x$ - $v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is :
Answer: (B) $k = m$
Energy conservation: $\dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2 = \dfrac{1}{2}kA^2$, which gives
$$x^2 + \frac{m}{k}v^2 = A^2$$
This is an ellipse in general. It is a circle when the coefficients of $x^2$ and $v^2$ are equal:
$$\frac{m}{k} = 1 \;\Rightarrow\; k = m$$
(numerically, in SI units; this is the case $\omega = 1\ \text{rad s}^{-1}$).
Solution by Sreeraj P, M.Sc Physics