Question

A 3.0 kg object subject to a restoring force F is undergoing simple harmonic motion with a small amplitude. The potential energy U of the object as a function of distance x from its equilibrium position is shown. This particular object has a total energy E: of 0.4 J.

 

(a) What is the object's potential energy when its displacement is +4 cm from its equilibrium position? 

(b) What is the farthest the object moves along the x axis in the positive direction? Explain your reasoning. 

(c) Determine the object's kinetic energy when its displacement is – 7 cm. 

(d) What is the object's speed at x = 0?

(e) Suppose the object undergoes this motion because it is the bob of a simple pendulum as shown above. If the object breaks loose from the string at the instant the pendulum reaches its lowest point and hits the ground at point P shown, what is the horizontal distance d that it travels?

Answer 1

(a) From graph U = 0.05J

(b) Since the total energy is 0.4 J, the farthest position would be when all of that energy was potential spring energy. From the graph, when all of the spring potential is 0.4 J, the displacement is 10 cm

(c) At –7 cm we read the potential energy off the graph as 0.18 J. Now we use energy conservation. 

ME = U_sp + K 

0.4J = 0.18J + K 

→ K = 0.22J

(d) At x=0 all of the energy is kinetic energy K= ½mv² 

0.4 = ½ (3)v² 

v = 0.5m/s

(e) The object moves as a horizontally launched projectile when it leaves.