An object of mass 0.2 kg executes simple harmonic oscillations along the X - axis with frequency of ( 25 ) Hz . At the position x=0.04 ~m , the object has kinetic energy 1 J and potential energy 0.6 J . The amplitude of oscillation is

Step-by-step solution

The total energy for simple harmonic motion is conserved and equals the sum of kinetic and potential energy at any point. Given KE = 1 J and PE = 0.6 J , E = KE + PE = 1 + 0.6 = 1.6 J . The angular frequency is derived from the frequency f = 25 Hz as = 2 f = 2 25 = 50 rad/s . The spring constant is k = m ^2 = 0.2 50^2 = 0.2 2500 = 500 N/m . Using the energy formula in terms of amplitude, E = 1 2 kA^2 , which rearranges to A = 2E k = 2 1.6 500 = 3.2 500 = 0.0064 = 0.08 m . The amplitude is 0.08 m .