Problem: What is the centripetal acceleration, in m/s ^ 2, of the reference point on the circle which models the simple harmonic motion of a mass of 2 kilograms, released from rest at a distance of 3 meters from its equilibrium position, when a restoring force of 5 Newtons/meter acts to pull the object back to its equilibrium point? What acceleration would you expect for the object at the instant it reaches and extreme point? Explain in your summary why you would expect the object itself to undergo this acceleration at the extreme positions in its cycle.
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Solution: The angular velocity of the reference point is `sqrt(k/m) = 1.581 radians / second.
At this angular velocity, a point on the circle will be moving at 1.581 radians/second ( 3 meters) = 4.743 meters/second.
At this velocity, the centripetal acceleration of a point on a circle of radius 3 meters will be ( 4.743 meters/second) ^ 2/( 3 meters) = 7.498 meters/second ^ 2.
Since the centripetal acceleration is directed vertically at the instant the object passes through its equilibrium point, the acceleration of the object, which has the same vertical motion as the reference point, must have the same acceleration.
In any event, the acceleration of the object can be found from Newton's Second Law as a = F/m, where F is the force at the extreme position and m the mass of the object.
At the extreme position, 3 meters from equilibrium, the force is ( 3 m)( 5 N/m) = 15 Newtons. Dividing this by the 2 kilogram mass we obtain the same result as above, 7.498 m/s ^ 2.
Generalized Response: In general the centripetal acceleration of the point on the reference circle is a = v^2 / r; since v = r `omega = A `omega, we have
a = (A `omega) ^ 2 / A = A `omega^2.
Since the acceleration function associated with position function y = A sin (`omega t) is y = - `omega^2 A sin(`omega t), we see that the maximum magnitude of the acceleration (which occurs when the value of the sine function is -1) is
aMax = `omega^2 * A,
the same as the centripetal acceleration on the reference circle.
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Figure Description: The figure shows the green object at its point of maximum force, and therefore maximum acceleration, at an extreme point of its motion. Since the acceleration of the point on the reference circle is in the same direction as the acceleration of the object whose position follows the y coordinate of the reference point, the two accelerations are equal.
