20 [2] Fig. 2 shows a m mass of 0.5 0.9 173 Fig. 2. Mass and spring system 5 kg that is attached to a spring with a stiffness constant &= 200 N/m. At equilibrium, the length of the spring is 0.5 m. Let x be the vertical position of the mass with respect of its equilibrium position (r is positive downwards). If the spring is stretched downwards from its equilibrium position by a force F and released, the position of the mass will vary in time according to Newton's second law of motion dx F 172 773 +kx = 0 (external resisting forces due to air resistance and di friction are ignored). If the spring is stretched to a total length of 0.9 m and then released with no initial velocity, use the Taylor series of order 4 numerical method and two steps to find the position of the mass after 0.2 s