If an object of mass 5 kilograms, suspended by a spring, is released from rest at a point 3.5 meters below its equilibrium position, and if the spring has force constant `forceConstant Newtons/meter, then what will be its average kinetic energy between times t = -.001 sec and t = .001 sec?
What is the average of the net forces on the object at the equilibrium point its point of release? Using this average, determine the work done on the object by the spring between its position of release and the equilibrium position.
Comment in your summary on the relationship between your results and the conservation of energy.
From the given information, we find that the angular velocity is
At t = -.001 second, the angular position is
resulting in a y position of
A similar calculation for t = .001 second yields y position
Thus the y position changes by
in .002 seconds, for an average velocity of
The object, having mass 5 kilograms, will therefore have a kinetic energy of
The force exerted by the spring at the y = 3.5 meter position is
Since the force changes linearly as the object returns to equilibrium, its average between 3.5 meters and equilibrium is
The object moves a distance of 3.5 meters with an average force of 4900 Newtons exerted on it in the direction of motion. The work done on it is therefore
This is very close to the kinetic energy attained by the object.
The average velocity near the equilibrium point, as seen in previous exercises, is near vAve = `sqrt(k/m) * A = `omega^2 * A. The KE near the equilibrium point is therefore near
approx KE near equilibrium = .5 m v^2 = .5 m * (`sqrt(k/m) * A) ^ 2 = .5 k A^2.
The maximum net restoring force experienced is at the maximum displacement A from equilibrium, and is
Fmax = k A.
The net force at the equilibrium point is zero, so the average force between equilibrium and extreme has magnitude
|Fave| = (kA + 0) / 2 = .5 k A.
The distance through which the force acts, between equilibrium and extreme, is A, so the work done has magnitude
| W | = | Fave | * A = .5 k A * A = .5 k A^2.
It is important to note that the expressions for KE and magnitude of work done are identical.
The figure below depicts an object in SHM at its equilibrium and extreme positions. The KE of the object and the net force on it are indicated at each point. When the work done on the object from extreme to equilibrium is calculated from average force and distance, the result is the same as the KE obtained from the mass and velocity of the object at equilibrium. This is as expected. The work done between extreme and equilibrium is the source of the KE at equilibrium.