Given that the length of a pendulum - and not it’s release point, for example - is the only significant determinant of its period and its frequency, it seems to me that if I have two pendulums of the same length, they will have the same frequency and the same period. And I do have two such pendulums, about a meter long, taped to the side of a filing cabinet. And if I lift one up to, say, a 45 degree angle, and the other to, say, 22 degrees, and release them, they don’t hit at the same moment. Why not? It should take the same amount of time to swing from the apex to the other apex and back. Shouldn’t it take the same amount of time to swing from the apex to the directly vertical position?
T= 2pisqrt(l/g) is probably what you’re thinking of. But it only holds for small angles. 45 deg is not a small angle.
To elaborate, if you were to derive all the forces acting on the weight at any given time, you would have either sine or cosine terms in at least one equation. This makes it much harder to solv for the equation of motion of the object, as you would have arcsine and arccosine all over the place. Engineers simplify this by assuming that for small angles, cos(x) = 1 and sin(x) = 0.
Wait - if the frequency of this pendulum is, say, 30 cycles per minute when I release it from a high release point (say, 50 degrees), and it stays at 30 cpm when its arc has diminished to, say 20 degrees, then the height of the swing doesn’t affect the amount of time it takes to make one complete cycle. So one fourth of a cycle - from apex to the bottom of the swing - should take the same amount of time with a small displacement as it did with a large one, shouldn’t it? But it doesn’t.
Are they identical pendulums? Or just identical length? Maybe one has more friction at the fulcrum or something?
to reiterate what the others have said, pendulums are only approximately constant swing time. It holds well for small angles, less well for larger
That doesn’t happen, although the effect may be smaller than you can notice in a few swings. I calculate that a pendulum with a 50 deg amplitude has a period that is 4.18% longer than the same pendulum with a 20 deg amplitude and 4.98% longer than the same pendulum at the small angle limit.
I don’t know what sort of timing equipment you have available (stopwatch, etc.), but if you can measure time to better than a second, you’d need at least 12 full swings at 30 cpm to accumulate one second of difference between a 50-deg trial and a 20-deg trial. If your pendulum doesn’t damp out too fast for this, it might be neat to experimentally confirm the above numbers in your own home. (Of course, if your pendulum last for more than 12 swings with little damping, this will be easy to measure.)
For completeness: the period T of a pendulum released from rest at an angle [symbol]q[/symbol] is related to the small-angle-limit period T[sub]0[/sub] by
T = T[sub]0[/sub]*(2/[symbol]p[/symbol])*K(sin([symbol]q[/symbol]/2))
where K(k) is the complete elliptic integral of the first kind.
No, because the pendulum isn’t simply released from the point at 20 deg. The two equations are different, in that the pendulum released from 20 deg has no initial velocity. The one released at 50 deg has some initial velociity when it gets to 20 deg. Imagine two pendulums held at 20 deg. One you just let go, and the other one you push. They won’t have the same frequency.
That didn’t come out right. If you pushed one of the pendulums just a little, it would settle into its natural fequency after it reached the first apex, PROVIDED that apex was still a small angle. If it went out past the point where the small angle approximation is no longer valid, then you don’t have a simple pendulum anymore.