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Physics · PYQ NEET UG 2026
हिन्दीConsider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m$ kg and the spring constant is $k$ $\text{Nm}^{-1}$. At a given instant, the extension of the spring is $x$ meter and the speed of the particle is $v$ $\text{ms}^{-1}$. On the $x - v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is :
Correct answer · Option 2
$k = m$
Explanation
From energy conservation in SHM, $\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = E$. For the $x-v$ trajectory to be a circle, the coefficients of $x^2$ and $v^2$ must be equal, leading to $k = m$.
