
(119)

which yields

(120)

Thus we see that the acceleration is proportional to the negative of the displacement. In order to obtain the velocity at any point, recall that (119) is an example of a conservative force, so that it has a potential energy function
From conservation of energy, we then find that
the energy at any point is

(121)

where E is a constant. We can relate the total energy E to the amplitude by noting that at the point where the amplitude is greatest, the velocity is zero and so (121) reduces to
Substituting this into the right hand side of
(121) and solving for v yields

(122)

We can see this relationship easier by looking at a plot of energy verses position
The potential energy yields a parabola centered on the origin, and the total energy yields a line at a constant height above the axis. The region contained inside these two curves is the physically allowed region for the oscillator. The amplitude can be read off directly by noting the positions where the two curves intersect, and the difference in energy at any point between the total energy and the potential energy is just the kinetic energy at that point.

(123)

where A and d_{0}
are arbitrary constants. Calculus shows us that this is the correct solution.
We can see immediately that the constant A is just the amplitude
of the cycle. d_{0}
is called the phase angle. Its purpose is to describe the position of the
oscillator at t = 0. What is the period of (123)? Recalling the
definition of the period, we see that

(124)

Similarly, we find that the frequency is

(125)

It is usually more convenient to work with the
angular frequency, w.
This is defined to be w
= 2pf so

(126)

Thus we can rewrite (123) as

(127)

Notice that the velocity is

(128)

Example:
A block of mass 5 kg is connected
to a spring. A force of 150 N is required to stretch it 1.5 m, at which
time it is released to undergo simple harmonic motion. Assuming no friction,
find the period, frequency and angular frequency.
First we must calculate the spring constant k. Recall that
then we have that
The forces acting on the pendulum are the tension,
T, and gravity. Decomposing gravity into a radial and tangential
component, we see that the restoring force is

(129)

This is not simple harmonic motion, since the force is proportional to the sine of the angle and not to the angle itself. Recall, however, that we can expand sinq as
so, if the angle is small, we can approximate
sinq as q.
Then (129) becomes

(130)

This is simple harmonic motion, so we immediately see that
and

(131)

Similarly,

(132)

and

(133)

Notice that all these are
independent of mass. This observation was first made by Galileo. We must
now ask, how small must the angle be for this approximation to be valid?
(129) can be solved using special functions called elliptical integrals,
and then we find that the period is given by

(134)

where now q is the maximum angular displacement. By comparing this with (133), we can determine the difference in the periods and find the accuracy of the approximation. For example, for 15 degrees, the true period differs from the approximate one by less that 0.5%.