Problem: What is the speed, in meters/second, of the reference point on the circle which models the simple harmonic motion of a mass of 7 kilograms, released at a distance of 2 meters from its equilibrium position, when a restoring force of 2 Newtons/meter acts to pull the object back to its equilibrium point? Find the average speed of the object between t = -.001 and t = .001 seconds, if the object moves upward through its equilibrium position at t = 0. Explain in your summary why the two velocities should be related as they are.
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Solution: The angular velocity of the reference point is `sqrt(k/m) = .5345 radians / second. At this angular velocity, a point on the circle will be moving at
ref circle velocity = .5345 radians/second ( 2 meters) = 1.069 meters / second
(note that on a circle of radius 2 meters, each radian corresponds to a distance of 2 meters, so that .5345 radians corresponds to .5345 meters).
Using the y = A sin(`omega t) function to calculate the average velocity between the t = -.001 and t = .001 times, we obtain
vAve = 1.069 meters/second.
The near equality of these two velocities should not be surprising. As the object moves vertically true equilibrium, the point on the reference circle moves vertically through the x axis, parallel to the object. At this instant of object and reference point will be moving in parallel directions at the same speed.
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Figure Description: The figure below depicts a reference circle and the object whose motion it models. The red reference point is at the x axis when the green object is at its equilibrium position. In this position both objects are moving in the y direction and their velocities must match.
