Problem: Suppose that an object on a spring has mass m and the force constant of the spring is k. If the object is displaced a distance A from its equilibrium position and released, the y coordinate of its position at time t will be the same as that of a point moving with angular velocity
angular velocity = `omega = `sqrt(k/m)
on a circle of radius A (the y direction is taken as the direction in which the object moves). Here k is the force
constant (generally measured in Newtons / meter) of the spring and m the mass of the object.
If an object of mass 5 kilograms suspended from a spring is released from rest at a point 9 meters below its equilibrium position, and if the spring has force constant 3 Newtons/meter, then how long does it take the object to move through one complete cycle?
If t = 0 when the object first passes upward through the equilibrium position, then what will be its position at t = 2 seconds?
pWhat will be its average velocity and kinetic energy between t = -.001 sec and t = .001 sec?
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Solution: From the given information, we find that the angular velocity is
angular velocity = `sqrt(k/m) = `sqrt(k/m) = `sqrt[ 4 Newtons/meter) / ( 5 kilograms)] = `sqrt( .8000 / sec ^ 2) = .8944 rad/sec.
{You are probably wondering where the the radian came from in that calculation. For now you can probably get away with regarding the radian as a unit that appears when needed and disappears when it doesn't belong. It is needed here, since it is part of the unit for angular velocity}.
To determine the time required for a complete cycle, the model makes it clear that the cycle will take place as the point makes one complete revolution, which requires 2 `pi rad / ( .8944 rad/sec) = 7.025 second.
Since the y coordinate will be 0 at the equilibrium position, the equilibrium position will correspond to either of the times when the point on the circle crosses the x axis. When the point passes through the positive x axis, the y coordinate is increasing, so the object is moving upward. When the point passes through the negative x axis, the y coordinate is going from positive to negative, which corresponds to downward motion. When the object, which was originally pulled downward, first passes through the equilibrium position it is moving upward. Since this happens when t = 0, we see that t = 0 when the point is moving through the positive x axis.
At t = 4 sec, the angular position of the point on the circle (called the reference point) is
ang. Pos. = `theta = `omega * t = .8944 radian / second ( 4 sec) = 3.577 radians.
This corresponds to a y position of 5 meters * SIN( 3.577 radians) = -.1085 meters.
At t = -.001 second, the angular position is
ang pos at -.001 sec = ( .8944 radian / second) (-.001 second) = -.0008944 radian,
resulting in a y position of
y(-.001 sec) = 5 meters * SIN( -.0008944 radian) = -.004472 meters.
A similar calculation for t = .001 second yields y position
y(.001 sec ) = .004472 meters.
Thus the y position changes by
`dy = .004472 meters - -.004472 meters = .008944 meters
in .002 seconds, for an average velocity of
vAve = `dy / `dt = 4.472 m/s. The object, having mass 5 kilograms, will therefore have a kinetic energy of .5 ( 5 kg)( 4.472 m/s) ^ 2 = 49.99 Joules.
Generalized Solution
Generalized Response: The motion of an object subjected to a linear net restoring force is modeled by the motion of a point moving around a reference circle whose radius its equal to the amplitude of motion (i.e., the maximum displacement from the equilibrium point). The angular velocity of the point is
angular velocity = `sqrt(k/m),
where k is the restoring force constant and m the mass of the object.
If clock time starts when the point is moving upward through the x axis, the angular position of the point on the reference circle at clock time t is
angular position of reference point = `omega * t.
If we let A stand for the amplitude of motion (equal to the radius of the circle), then the coordinates of the point relative to the center of the circle are
coord of ref circle point: (x, y) = (A cos(`omega * t), A sin(`omega * t) ) = (A cos(`sqrt(k/m) t), A sin(`sqrt(k/m) t) ).
If we orient the coordinate system so that the motion of the actual object is along the y axis, then the position of the object at time t is given by the y coordinate
position of object = y(t) = A sin(`sqrt(k/m) t).
The average velocity of an object with this position function between any two given clock times can be determined. If the clock times are t1 and t2, then
position change = `dy = y(t2) - y(t1) = A sin(`sqrt(k/m) t2) - A sin(`sqrt(k/m) t1),
and the corresponding average velocity is
vAve = `dy / `dt = (y(t2) - y(t1) ) / (t2 - t1) = A [ sin(`sqrt(k/m) t2) - sin(`sqrt(k/m) t1) ] / (t2 - t1).
It turns out that if t1 and t2 are near 0 and do not differ by much, this average velocity is approximately
approx. vAve near t = 0: vAve = `sqrt(k/m) * A = `omega * A (approx.).
A numerical calculation with t1 and t2 both near 0, and differing by a small fraction of the period of motion, will yield an average velocity close to this value.
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Explanation in terms of Figure(s), Extension
Figure Description: The figure below depicts an object of mass m which, when pulled a distance `dy, is pulled back by an elastic object (a spring or some other object for which the restoring force is a linear function of the distance through which it is stretched) with a force increasing from F1 to F2. If the restoring force is linear, then we have F = -ky, and the object will undergo simple harmonic motion with angular frequency
`omega = `sqrt(k/m).
This means that at clock time t, with clock time starting at the 0-degree position on the circle (the positive x axis), the position of the object will be modeled by the y coordinate of the point on the circle whose angular position is
angular position at clock time t: `omega * t.
The corresponding coordinates relative to the center of the circle are
(x,y) = (A cos(`omega * t), A sin(`omega * t) ).
The y coordinate in particular is modeled by the function
y(t) = A sin(`omega I t) = A sin ( `sqrt(k/m) * t).
Figure(s)