Energy of the Simple Harmonic Oscillator?

Question

a) A particle executes simple harmonic motion with an amplitude of 3 cm. At what position des its speed equal one half of its maximum speed?

b) A 65kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the cord is 11m. The jumper reaches the bottom of her motion 36m below the bridge before bouncing back. Her motion can be separated into an 11m free fall and a 25m section of simple harmonic osciilation.
- For what time interval is she in free fall?
- Use the principle of conservation of energy to find the spring constant of the bungee cord.

Answer 1

a) use (kx^2)/2 = (mv^2)/2

b) use (kx^2)/2 = mgh
mgh = 65*9.8*11
x = 25
calculate k from here