Expert: Ahmed Salami - 10/27/2009

Question
I hope you can help cause I've been banging my head on this one...

Ok, I have a problem where we are trying to duplicate the motion of a pendulum (ignoring the effects of gravity). The pendulum is one foot long and has a period of 2 seconds. At t=0 the pendulum is at point A (which is 30 degrees from the centerline). Point O is the point where A would be in rest. Point C is the opposing coord. to point A. Use point O as the axis origin and compute the X-coord. vs Time and the Y-coord. vs Time for every 5 seconds. Write the Sine or Cosine functions to represent the dataset.

I've reverse engineered the functions, and I've found the Amplitude and the Period. But my Period in the functions I created (which will produce all data points accurately) are different then the period found by converting the degrees to pi. Let me try to show you...

Knowing that point A is 30 degrees from the axis point I can use trig functions to find the X-value: Since the X value is opp of the 30 then the X-value is half of the Hyp which is the length of string. So X is .5 and is going to be the same on both sides of the Y-Axis. Which gives you X= +/-.5 (sign dependent of side of axis)

Y-Value you find by taking l(the length of pendulum) and subtracting the cos(30) (Height of right triangle) Which gives you: Y= .13

Treating these as constants we can then draw them into a table to see the values over the entire period:

Time(t)   X-Value    Y-Value
0.0      -0.5        0.13
0.5       0.00       0.00
1.0       0.5        0.13
1.5       0.0        0.00
2.0      -0.5        0.13

Let's do X First...
Amplitude of the X would then be calculated by taking the (.5-(-.5)/2) => (.5+.5)/2 => .5 So, Amplitude of the X-Function is .5, it does not start on the axis, and starts as a negative so it is a (-A Cos(Bx-D)+C) function. At this point the function should theoretically look like this: x(t)= -.5Cos(b(t)). Then we figure out the period which is normally 2pi... This is where I get lost, The formula that I found that gives the right data set is
x(t) = .5Cos(pi(t)) So some how I had to divide my Period by 2? Is it because the period is completed in 2 seconds or (1/2) periods per second?

Funny thing is the normal period of 2pi works with Y-Function,
With Y, my Amplitude and vertical shift are the same .065 (half of .13) it's positive and doesn't start and origin so it's a(
ACos(Bx-D)+C) function. y(t)=.065Cos(2pi(t))+.065 This function works!

If I used the degrees that are traversed over the period (angular velocity) I get a period of 2pi/3... That doesn't plug into the formula's at all...

what am I doing wrong cause I know it's something lol. Thank You!!!!!!!

Answer
Hi Thomas,
The delay is terribly regretted.
The general equation of simple harmonic motion is written
x = Asin(wt + φ)
where A is the amplitude, w is the angular velocity and φ represents phase shift. If T is the period, then
w = 2π/T
Applying this to your problem,
x = 0.5sin(wt - π/2)
 = 0.5cos(wt)
But T = 2 and w = 2π/2 = π
x = 0.5cos(πt)
In the y situation, you have to realise that there are two complete vertical oscillations in the space of one horizontal oscillations. Think about it.
So for y, T = 1 and w = 2π

I hope its now understandable (and still useful, the delay is once again regretted).

Regards