Free Undamped Motion in Spring Mass System

Assume that a mass m is fastened to the free end of a flexible spring that is hanging vertically from a rigid support [5]. Naturally, the mass of the spring will determine how much it stretches or elongates; different weight masses will stretch the spring in different ways. The spring itself exerts a restoring force F that is proportional to the amount of elongation s and opposed to the direction of elongation, according to Hooke's law. Put simply. When a mass m is connected to a spring, the mass extends the spring by a certain amount, reaching an equilibrium where the restoring force ks balances the mass W. Remember that [6] defines weight. w = mg (0.3.1) the condition of equilibrium is mg = −ks (0.3.2) or mg − ks = 0 (0.3.3) In the event that the mass deviates from its equilibrium position by an amount x, the spring's restoring force is equal to k(x + s). We can associate Newton's second rule with the net, or resultant, force of the restoring force and the weight W is balanced by the restoring force ks, provided that there are no retarding forces operating on the system and that[6] the mass vibrates free of other external forces—free motion. Remember that [6] defines weight. (ⅆ^2 x)/(ⅆt^2 )= -k(s + x) + mg (0.3.5) = −kx + mg − ks = −kx (0.3.5) The spring's restoring force works in the opposite direction of motion, as shown by the negative sign. Additionally, we follow the tradition that displacements recorded in positive values below the equilibrium position [6].