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 m 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:

Solution & Explanation

Principle: in SHM the relation between speed and displacement comes from energy conservation, v = ω√(A² − x²). Squaring: v² = ω²(A² − x²), which rearranges to x² + v²/ω² = A². On the x–v plane this is an ellipse with semi-axes A (along x) and ωA (along v). It becomes a circle only when both semi-axes are equal, i.e. when the coefficients of x² and v² match, meaning ω² = 1. For a spring–mass oscillator ω² = k/m. Setting ω² = 1 gives k/m = 1, so k = m. Key point: "graph is a circle" forces the angular-frequency condition ω = 1, not an arbitrary spring constant — this matches answer (B), k = m.