Beginning of class quiz:

Class Vote:

Three ways of looking at this:

1. As the mass is pulled down, the net force is down and the displacement is down, so the net work on the mass must be positive.  This increases the total energy of the system.
2. If you release the mass at equilibrium, it doesn’t move. If you release the mass when it is pulled down, it starts to oscillate and in fact passes the equilibrium point with positive kinetic energy.  Since total energy is a constant of motion during oscillations, the total energy must be greater for the case when it is pulled down.
3. When the spring hangs at equilibrium it is stretched so that F_s =weight, or kx = mg.  We may set h = 0 at equilibrium, so the total energy when speed=0 is E = 0.5 k x² + mgh = 0.5 k x².

Next, let’s pull the mass down a distance, d, and hold it at rest.  The new x is x_new =  x+d, and the new height is –d.  So the total energy is E_new = 0.5 k x_new² + mgh = 0.5 k (x+d)² – mgd = 0.5 k x² + k x d + 0.5 k d² – mgd .  But since kx = mg (equilibrium condition) we may write this as E_new = 0.5 k x² + 0.5 k d² + mgd – mgd = 0.5 k x² + 0.5 k d², which is always larger than E = 0.5 k x².

Class Vote:

Period depends on frequency which depends on g and L only; not mass.

Class Vote:

Assuming the two people sit side-by-side, the length of the pendulum does not change, just the mass, so the frequency should be the same.

Class Vote:

The effective length, L, is the distance from the pivot to the centre of mass (com) of the mass.  This will be smaller if you stand up, making angular frequency larger.