Pendulum Theory (small and large amplitude)

The picture below shows a mass swinging on a long string. The component of force in the direction of motion is mgsinq , so the tangential acceleration is gsinq . Substitute x/L for sinq , and note that when x is to the right, a is to the left, so a = -gx/L. Recall that a = -w ²x, so you can easily show that w = (g/L)^1/2.

The above formula for w is valid only for small angles. There is an error hiding in the above paragraph, an error which is negligible for small swings of the pendulum. See if you can find it yourself; do not read beyond this point.

Yeah, right. The problem is that it does not go exactly in a straight line. A small arc is very close to a straight line, so you can have minimal error, but remember that
a = -w ²x is exact for straight line motion only.

What if you want w for a large amplitude? Gotta devise a computer program to do the job. Here is one in Basic that might work if you make dq very small, but note that Basic does not permit Greek letters or subscripts, so change it a little. It is based on
v₂² = v₁² + 2aD x, D t = D x/(average v), and time to reach q = 0 is a quarter of a period.

g=9.8
v=0
t=0
q _o= (you choose a negative angle, 0 to -p /2 in radians)
L= (choose a length in meters)
for q = q _o to 0 step dq
a = -g*sin(q )
v1 = v
v = SQR(v^2+2*a*L*dq )
t = t + 2*L*dq /(v1+v)
next q
P=4*t
print "period = "P" seconds”

And if you want w , it is 2p /period. Here is a test to see if it is valid (besides the experimental test). See if v agrees with the v from conservation of energy. It started from height L-Lcosq , and mgh = ¹/₂mv² ---> v = (2gh)^1/2. So add lines at the end:

vf = SQR(2*g*L*(1-cos(q _o)))
print "final v by conservation of energy is "vf" m/s"
print "compare to v by the program, which is "v" m/s"

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