In Part 1 we looked at what is meant by Iterative Functions, we reviewed what is meant by angles in their standard position versus bearings, degrees versus radians, addition of displacement vectors and the use of spreadsheets. We also said that when you are dealing with non-uniform acceleration things can get complicated.
A First Look at the Pendulum
Here we see a pendulum. P is the 'bob'; usually a ball of something
heavy like lead. F is the frictionless fulcrum. The pendulum's
locus is an arc of constant radius.
In this discussion we will consider the half-cycle where the starting
point of the pendulum bob is at the 4 o'clock position of the
diagram and the swing is clockwise. g represents the gravitational
force.
During the first half of the cycle gravity acts at so as to increase the speed of the pendulum. It reaches maximum velocity at the lowest point. Then it slows down, coming to rest in this case apparently in the 8 o'clock position.
Another way of saying this is to say that the pendulum has a positive acceleration during the first half of the swing and a negative acceleration during the second half. At the half way point - at the 6 o'clock position - the acceleration is zero. Acceleration is greatest at the two extremes of the swing.
Velocity is zero at the two extremes
and maximum at the 6 o'clock position. Acceleration and velocity
are related as shown in this graph.
A Second Look at the Pendulum
Now let's get specific and assign some numbers. We start by assuming the pendulum is 1 metre long and is initially in the 4 o'clcock position, which corresponds to a bearing of 120 degrees from the fulcrum (F).
P0 is where the pendulum is at time 0. P1 is where it will be
at the at the end of the time slice and P is where it is half
way through the time slice.
If the time slice is short enough, then
we can make these assumptions:
- The distance travelled along the arc = length from P0 to P1
- the acceleration experienced will be as if the the gravitational
force vector was applied at P.
- the angles at P1, P and P0 are all right angles.
V(g) is the acceleration of gravity. Vc is the centripetal acceleration. The magnitude of the resultant, r = g*sin(60) = 8.487 m/sec^2.
Assuming a time slice of 0.01 seconds for now, we get:
velocity = acceleration * time
= 8.487 m/s^2 * 0.01 s
= 0.08487 m/s
distance = average velocity * time
= (v(i) + v(f))/2 * time
= 0.08487 m/s * 0.01 s
= 0.0004244 m
During the first time slice of 0.01 seconds the distance covered = distance from P0 to P1 = 0.0004244 m. The entire circumference of radius 1 m = 2 m * 3.14159 = 6.283 m, so the angle covered in the first time slice is
(0.0004244 m)/(6.283 m) * 360 degrees = 0.0243 degrees
The pendulum bob is now located at bearing 120.0243 degrees with respect to the fulcrum, and we can repeat the calculations for the next time slice.
Here we see the calculation set up in
a spreadsheet
The conversions from degrees to radians have been incorporated
into the formulas.
Note that with a starting angle of 120 degrees the time needed for the pendulum to reach 180 degrees to the nearest 0.01 second is 0.54 second; which means that the period is calculated to be 2.16 seconds, compared to 2.007 seconds.
Changing the starting bearing to 170
degrees gives us this result:
Here the angle is shown as 179.99060 degrees at time 0.500 seconds.
Ie: a 2 second period.
Further experimentation with this template allows us to demonstrate that with small initial deflections the pendulum period is indeed found by the formula discussed in Part 1. However with larger deflections the formula doesn't apply.
The reader should also consider experimenting with this template to determine the effect of larger and smaller time slices. 'Theoretically' the smaller the time slice the more 'accurate' the result... but of course there is a practical limit where the results obtained are 'close enough for all intents and purposes'.