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) $k = m$
Concept. Energy conservation gives the $x$–$v$ curve: $\dfrac12mv^2 + \dfrac12kx^2 = \dfrac12kA^2$.
Step 1. Curve. $x^2 + \dfrac{m}{k}v^2 = A^2$, an ellipse in general.
Step 2. Circle condition. The coefficients must be equal: $\dfrac{m}{k} = 1$, i.e. $k = m$ (numerically, in SI units; $\omega = 1$ rad/s).
Answer. $k = m$ (Option B).
Solution by Sreeraj P, M.Sc Physics