Question

In Fig. 1.6 the 1 kg mass is released when the spring is unstretched (the spring constant k = 400 N/m). Neglecting the inertia and friction of the pulley, find (a) the amplitude of the resulting oscillation, (b) its centre point of oscillation, and (c) the expressions for the potential energy and the kinetic energy of the system at a distance y downward from the centre point of oscillation.

Answer 1

(a) Suppose the mass falls a distance h before stopping. The spring is elongated by h. At this moment the gravitational potential energy (mgh) the mass lost is stored in the spring

After falling a distance h the mass stops momentarily, its kinetic energy T = 0 at that moment and the PE of the system V = 1/2 kh², and then it starts moving up. The mass will stop in its upward motion when the energy of the system is recovered as the gravitational PE (mgh). Therefore, it will rise 0.049 m above its lowest position. The amplitude of oscillation is thus 0.049/2 = 0.0245 m. 

(b) The centre point of motion is at a distance h/2 = 0.0245 m below the point from where the mass was released. (c) Total energy of the system

At a distance y downward from the centre point of oscillation, the spring is elongated by (h/2 + y) and the total potential energy of the system is

and the kinetic energy