A particle of mass 12 grams moves along the x axis. It has a restoring force F= -0.06 N/m. If it starts from x=10 cm with a speed of 20 cm/sec toward the equilibrium position, Find its amplitude, period, and frequency. Determine when the particle reaches the equilibrium point for the first time. Edit: I solved the question this way: F= -w^2*m -> w=1/2(sqroot) v=x*w 20=x*1/2(sqrt) x=28.2 -- T=2pi/w -> T=8.8 -- a(accelaration) = -w^2*x a= -14,1 Vf=Vi+a.t t=1,141 sec. -------------- 2nd way- and some one else solved the question with this way : Let w = angular frequency, A = amplitude, T = time period Acceleration = F / m = 0.06 / 0.012 = 5 m/s^2 Acc = w^2 * x w^2 * 0.10 = 5 Simplifying, w = 5 (sqrt 2) T = 2 pi / w m * v^2 + m *w^2* x^2 = m* w^2 * A^2 0.04 + 0.5 = 50 * A ^2 A = (sqrt 1.08) / 10 m = sqrt 1.08 * 10 cm -- and I'M confused :( Thanks,..