Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kg and the spring constant is k Nm − 1 . At a given instant, the extension of the spring is x meter an

Question

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kg and the spring constant is k Nm − 1 . At a given instant, the extension of the spring is x meter and the speed of the particle is v 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 :NEET 2026 (Re-NEET)

TestHub · Accepted Answer

The total mechanical energy of a spring-mass simple harmonic oscillator is given by:

$E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}$

Rearranging the terms, we get:

$v^{2} + \frac{k}{m} x^{2} = \frac{2 E}{m}$

In the $x$ - $v$ plane, this equation generally represents an ellipse. For the graph to be a circle, the coefficients of $v^{2}$ and $x^{2}$ must be equal.

Therefore, we must have:

$\frac{k}{m} = 1$

$k = m$

Answer: $k = m$

Physics - SHM Question with Solution | TestHub

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