An essentially massless spring is connected to the ceiling. A 5.00 g mass is attached to the spring so that the spring is 2.00 cm longer than its unstretched length. The mass is the pulled downward so that it is 20 cm below the equilibriun position and released from rest. After 10s the amplitude of oscillation has been reduce to 15 cm, dut to the viscous drag of the air. The equation of motion is x(t) = A0e ^-bt/2m cos (omega)ot.
(a) Determine the value of b k and omega
Derive an expression for the net force on the mass as a function of time.
First off, keep in mind Hooke's law for massless springs: F = -kx, where x is the stretch in the spring from its equilibrium position (or length) and k is the spring constant, in N*m. It is a restoring force (that is, it points against the displacement of the end of the spring, back towards the equilibrium position), hence the minus sign.
So, in this case, a force of -mg (.005*9.8, negative because the mass pulls the spring down) stretches the spring .02 m (or 2.00 cm -- note ...
The explanations are very detailed as used in solving the problem.