# A Pendulum Experiment

1. Apr 22, 2013

### IkeB777

If you were to set up two pendulums, identical except that one pendulum had 500 grams of an element with atoms of low mass, and the other with 500 grams of atoms of higher mass, would the two pendulums have a different rate, even if infinitesimally? Has this been done? Thank you.

2. Apr 22, 2013

### Staff: Mentor

Welcome to PF.

First, the question is oddly worded. I think you mean either density or atomic mass when you say mass because 500 grams of mass is 500 grams of mass.

Anyway, the answer depends on how precise you want to be. The simplistic equation for pendulum swing you learn in high school says that the rate depends only on the length of the string. But a real-world pendulum swings in air, so a bob with a lower density will experience damping due to air resistance.

3. Apr 22, 2013

### IkeB777

Yes, but 500g of atoms with a higher atomic mass is going to have fewer atoms. I have a hunch that measuring the action from the reference of '500 gram object' is different from measuring the action on each individual atom's mass, and that the difference may be still imperceptibly small even with current instrumentation. Possibly have to measure the effects down to the individual sub-atomic parts.

Essentially, if these two pendulums were to be able to swing infinitely, would a difference emerge over time?

4. Apr 22, 2013

### WannabeNewton

If there is no air drag and no forcing, and if the identical pendulums are large enough so that the .5kg masses hanging from the respective pendulums can be treated as point particles, then the resulting equations of motion for this system (the simple pendulum) will only have the length of the identical pendulums show up as a parameter as russ_waters already stated.

Even if there is some damping constant $c$ (but still no forcing), the equations of motion will be $\ddot{\theta} + \frac{c}{m}\dot{\theta} + \frac{g}{L}\sin\theta = 0$ hence if .5kg of mass hangs from each of the identical pendulums, as you wanted, the motions will still be identical.

Last edited: Apr 22, 2013
5. Apr 22, 2013

### IkeB777

Or put another way, pretty much the same thing I'm asking in my other thread. Would rate of time slow due to a differing rate of gravity due to larger mass of the aggregate of the atoms within an object, not of the mass of the object itself. Still, to perceive a difference in time relative to gravity, you would never be able to do it comparing the scales of say, planets, but if you were measuring on something that is moving very fast and is very massive, a difference may be able to be seen.

This is just an idea, and I would love for anyone to tell me how and why I am completely off base. That is why I'm here. I'm going to ask a lot of questions. Thank you.

6. Apr 22, 2013

### IkeB777

And also I would be interested what the difference would be if this particular experiment could be replicated in a 3 dimensional gravitational environment as well as our own 2 dimensional gravity situation.

7. Apr 23, 2013

### Staff: Mentor

No.

What do you mean with "2-dimensional gravity situation"?

8. Apr 23, 2013

### IkeB777

When you live your entire life in a gravity field, your perception of the world conforms to it. Gravity is 2 dimensional on the surface of the planet, because we only see it's action in the vertical. This screws us intuitively and conceptually because gravity is a 3 dimensional force, but we can only directly observe that in space, and problematic to that is that the scales are imperceptibly small on the size of objects we have to work with there. If you were at the conceptual center of the Earth, you would be in a 3 dimensional gravitational environment. My question is, would you be weightless there?

9. Apr 23, 2013

### IkeB777

And see, here is where I think I might be going crazy:

I think that if you were to run these pendulums long enough, they would start to run out of phase, but the effect might be so small that you would possibly have to wait thousands of years to perceive a difference.

10. Apr 23, 2013

### TurtleMeister

If I'm understanding you correctly, I think what you're saying is that the passive gravitational mass of one pendulum will not equal the passive gravitational mass of the other pendulum (even though their inertial masses are equal), but that this inequality will be very small and undetected. This is actually an experiment which has been conducted many many times. I think Isaac Newton was the first to do it. Some of the latest experiments have been done by the Eot-Wash group with sensitivities reaching 10-13 level. Proposed satellite experiments will exceed that.

Edit:
Here are two links that might interest you. The first talks about Newton's experiments with pendulums and the second is a link to the Eot-Wash Group web page.

http://www.mathpages.com/home/kmath582/kmath582.htm
http://www.npl.washington.edu/eotwash/experiments [Broken]

Last edited by a moderator: May 6, 2017
11. Apr 23, 2013

### WannabeNewton

How are you coming to this conclusion? You can't just state things are true without physical reasoning behind them.

12. Apr 23, 2013

### IkeB777

I'm basing my conclusions on what I envision gravity to actually be. And I don't believe it is a repulsive or an attractive force, or a force at all for that matter. I think 'gravity' has been under our noses all along, and at one point almost conceptualized to the point of understanding, but the predominant thoughts prevailed, and still prevail today. I believe that when the true nature of gravity is realized, then, and only then, will the holes and discrepancies in our calculations make sense, will the confusing nature of the quantum world suddenly become apparent, and will no longer have to 'add the number to the other side' to cancel our confusion. It was never really necessary.

I'm still apprehensive about engaging in a full blown discussion about it, not because I'm afraid I'm wrong, but because I'm afraid I'm right.

13. Apr 24, 2013

### Staff: Mentor

That is not true. We live in a 3-dimensional environment. It does not matter where you are, the situation is the same everywhere. In space, you have no floor, of course, but gravity acts in the same way.

Relative to the mass around you: Sure. You would still be attracted by the sun (and the galactic center, and Andromeda, and ...), but that applies to everything around you, too.

Well, gravity does not care what we think it "should" be.
You will like the description of General Relativity (GR) then, where gravity is not a force.
Our concepts of gravity changed completely with GR.
What do you mean with that?

You are (probably) right that a new, more fundamental theory, combining quantum mechanics and GR, will solve some issues in physics. You are probably (>99.9%) wrong if you think you will contribute to that theory.

14. Apr 24, 2013

### pumila

Have you allowed for angular momentum? The lighter element will have a higher volume so you have to get the shape right (cylinders of identical diameter) to stop angular momentum having a differential effect. Or have a mounting arrangement that precludes rotation.