Rank Kinetic Energy Pendulum Problem: A-E-I < B-F < D-H < C-G

[sona1177]

Homework Statement

I have a pendulum problem that shows me the picture of the pendulum, which Physics Forums won't let me copy and paste. but anyway it is the picture of a pendulum that starts at point A, goes down to point B, then C (lowest point), then up to D (which looks the same height as B), and then up to E (which looks the same height as A). Then the pendulum comes back down to point (F), then down to G, which is again the lowest point, then back up to H, and finally I, which is the last point I see. If that image is clear, can someone help me rank the kinetic energies from smallest to largest?

Basically the pendulum swings back and forth, (once forward, and then back).

My answer was A,E, I < B,F < D,H < C,G but that is wrong. Can someone please help me?

Homework Equations

The Attempt at a Solution

 

[Stonebridge]

Without the diagram it's rather difficult.
You say D "looks the same height" as B.
If it is the same height then the kinetic energy at D will be the same as at B.
In which case the ranking is wrong because you have B<D [and F<H]

 

[sona1177]

Without the diagram it's rather difficult.
You say D "looks the same height" as B.
If it is the same height then the kinetic energy at D will be the same as at B.
In which case the ranking is wrong because you have B<D [and F<H]

Ok so would B, F, D, and H all be equal then?

So the order would be A, E, I <B, F, D, and H< C, G

Doesn't time make a difference? Wouldn't the pendulum swing slower when it's swinging for the second time (E ->I)?

 

[Stonebridge]

If at the same vertical height, then B,F,D and H will all be the same.
Yes, so will A,E and I.

Questions like this usually say something like "ignore friction".
If this is the case the pendulum will actually swing for ever. The ke at any point will be such that the sum of ke plus pe will be constant.
In this case, as the pe depends only on the vertical height of the mass (=mgh) then the ke also only depends on the vertical height. ke=total - pe

If the pendulum is slowing down, due to friction, then yes, it depends on time.
There is nothing in the question as I read it to suggest which is the case. I have assumed no friction.
Maybe if I could see the diagram I could be more helpful.
If the final point is at the same height as the initial, A and I, then in one swing there has been no loss in energy. So the frictionless case would be the correct one.