A particle of mass 2 kg is attached to a spring of force constant 18 N m ⁻¹ and executes simple harmonic motion. If the velocity v of the particle is plotted against its displacement x , the resulting graph is an ellipse. What is the magnitude of the ratio of the intercept on the velocity axis to the intercept on the displacement axis?

Step-by-step solution

The equation relating velocity v and displacement x for a simple harmonic oscillator is: v^2 = ^2(A^2 - x^2) Rearranging this into the standard equation of an ellipse: v^2 ( A)^2 + x^2 A^2 = 1 The intercept on the velocity axis (where x = 0 ) is the maximum velocity, v_ max = A . The intercept on the displacement axis (where v = 0 ) is the amplitude, x_ max = A . The ratio of the velocity intercept to the displacement intercept is: A A = The angular frequency is given by: = k m = 18 2 = 9 = 3 s ⁻¹ Answer: 3 s ⁻¹