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 :

Step-by-step solution

The total mechanical energy of a spring-mass simple harmonic oscillator is given by: E = 1 2 mv^2 + 1 2 kx^2 Rearranging the terms, we get: v^2 + k m x^2 = 2E 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: k m = 1 k = m Answer: k=m