A particle executes simple harmonic motion. When its velocity is plotted against its displacement (assuming SI units for both axes), the resulting graph is a perfect circle. The time period of the particle's oscillation is

Step-by-step solution

The equation relating velocity v and displacement x in simple harmonic motion is given by: v = A^2 - x^2 Squaring and rearranging the terms, we get: x^2 A^2 + v^2 A^2 ^2 = 1 This represents an ellipse. For the graph to be a perfect circle, the semi-major and semi-minor axes must be equal. Thus, we must have: A = A = 1 rad/s The time period T of the oscillation is: T = 2 = 2 1 = 2 s Answer: 2 s