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Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m\text{ kg}$ and the spring constant is $k\text{ Nm}^{-1}$. At a given instant, the extension of the spring is $x\text{ meter}$ 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 :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$k=m$

A simple harmonic oscillator consists of a mass attached to a spring. The mass moves back and forth due to the restoring force exerted by the spring. The parameters given for this system are the mass \(m\) kg and the spring constant \(k\) Nm-1.

The equation for a simple harmonic oscillator is given by:

\(F = -kx = ma\)

The acceleration \(a\) can be linked to velocity \(v\) and position \(x\) using the energy conservation principle:

\(\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}\)

The equation above signifies that the sum of kinetic and potential energy remains constant, representing an ellipse in the \(x-v\) phase space.

For the graph of \(v\) as a function of \(x\) to be a circle, the equation should be of the form where the coefficient of \(x^2\) and \(v^2\) are equal. Thus:

\(\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = E \quad \Rightarrow \quad mv^2 = kx^2\)

For the coefficients of \(v^2\) and \(x^2\) to be equal, we require:

\(m = k\)

Therefore, for the graph to be a circle, the relationship \(k = m\) must hold true.

Hence, the correct option is:

$k=m$

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