# Simple harmonic motion of clock

1. Feb 25, 2006

### gandharva_23

let us consider a bob suspended by means of a thread from a fixed point (wrt ground and i m considering ground to be an inertial frame in this problem ) . What will be the time period of small oscillations of this pandulum as seen by a man accelerating with an acceleration a along the horizontal . will it be 2pi(root(l/g)) or will it be 2pi(root(l/root(a^2+g^2)) .
logic1 : if we solve the problem in ground frame we get T=2pi(root(l/g)) . now at small speeds this time is not frame dependent , hence the time period should be same in mans frame also .
logic2 : We want to calculate T as seen by man so lets solve the problem in mans frame . Since its a noninertial frame a pseudo force will act on the bob and the time period will come out to be
2pi(root(l/root(a^2+g^2)) .
which one is correct ?

2. Feb 25, 2006

### Tom Mattson

Staff Emeritus
If the pendulum itself is inertial then it acts as a clock. In Galilean relativity all clocks keep the same time in all frames. In Special Relativity they don't.

Which one are you using?

3. Feb 25, 2006

### gandharva_23

i am not getting into special relativity and i totally agree that the time period should be 2piroot(l/g) but what is the flaw in logic 2 ?

4. Feb 25, 2006

### topsquark

I believe the problem is that the pedulum will have the same period, but look as if it were pushed off to one side, as if it were not hanging straight down. The pseudo-force will act only in the horizontal direction and thus will not affect the period of motion.

-Dan

5. Feb 25, 2006

### gandharva_23

If pseudo force acts in the horizontal direction then it would definitely alter the time period . if we make the fbd and calculate the time period the time period after application of pseudo force comes out to be 2pi(l/((a^2+g^2)^1/2))^1/2

6. Feb 26, 2006

### topsquark

Also:
I was thinking in the shower this morning. (Always a good place to meditate! ) I think we have both been laboring under a particular assumption, me tacitly and you (gandharva) specifically, that the pendulum bob experiences an extra constant acceleration from the pseudo-force. That a freely falling mass will be "shunted" off to one side by the pseudo-force is true, but the pendulum mass is certainly NOT freely falling. It is constrained to move in a circle, and that circle will still be a circle in the (non-relativisitic) accelerating frame.

I am no longer going to blithely state what the motion of the pendulum is because the simple answer I gave before could easily be wrong. Certainly my previous post is incorrect in regards to detail.

The whole solution to the problem rests with Galilean Relativity. In Galilean Relativity time in an invariant. So whatever time we calculate in one reference frame is going to be the same time in any other reference frame. I am not prepared to say that the pendulum experiences nothing due to the action of the pseudo-force, but the period MUST be the same in both reference frames. I suspect this means the pseudo-force is not constant on the pendulum mass. (After all, it's a PSEUDO-force, so F=ma need not apply!)

-Dan

7. Feb 26, 2006

### gandharva_23

you seem to have meditated intensely :-) on i dont know what ;-) but the solution does not seem to be convincing ....... the period has to be 2piroot(l/g) but the logic that you give to defy the second argument that i gave does not seem to be too convincing ..... neways keep meditating on it... and i ll do the same

8. Feb 26, 2006

### topsquark

Convincing or not, the Galilean Relativity transforms leave t alone. Remember we aren't working in Einsteinian Relativity where there are no preferred reference frames. There ARE preferred reference frames in Galilean Relativity, called "inertial frames." Your observer isn't in one, so a calculation of the period of a pendulum (which really isn't a simple problem in general) might not work out to be so easy.

-Dan

9. Feb 26, 2006

### gandharva_23

Galilean Relativity transforms leave t alone ..... thats exactly i want to say . the time period that should appear to the observer should be same as the time period of pandulum in the ground frame which is 2 pi root(l/g) . but try solving the problem in thet non inertial frame and u 'll get a different result . i know it isnt really simple but i did calculated that and that came out to be equal to 2pi(l/((a^2+g^2)^1/2))^1/2 .

10. Feb 27, 2006

### topsquark

That was exactly my point when I said I suspected the pseudo-force was not acting in such a way to provide a constant acceleration to the pendulum mass. In order to figure out what the pseudo-force is doing in that frame we need to specify that the period is $$2 \pi \sqrt{l/g}$$ and work backward to figure out what the pseudo-force is doing. It doesn't sound like a pretty problem doing it that way :yuck: so, frankly, I'm not going to do it.

-Dan

11. Feb 27, 2006

### qtp

why can't u just add the acceleration into the calculation of the EOM for the pendulum and see what you get? i would think that it would change the measured period in the same way that things on the earth experience a coriolis effect.

12. Mar 2, 2006

### topsquark

gandharva_23 I owe you an apology for two reasons. First, I have made two errors in logic on this thread, but more importantly that I had convinced myself that I knew the Physics of the situation when in retrospect I clearly didn't. I can excuse myself for making a mistake (everyone does) but I can't excuse myself for posting an answer when I didn't completely understand the problem. (I'm a Quantum Physicist, not a Relativist, so I don't usually do accelerating frames.)

As to my error, in my defense, no one corrected it including the mentor-teachers on the forum. It's apparently an easy mistake to make.

Let me demonstrate the problem that's been bugging me for a few days since I made my last post. I concluded that the action of the pseudo-force was not constant because the pendulum period was the same in both an accelerating frame and an inertial frame. This is an obvious mistake as the following experiment would show. Consider the pendulum, not in motion, but as hanging at its equilibrium point. The accelerating observer will NOT see the pendulum hanging off to one side. The accelerating observer will be able to measure the point where the string meets the ceiling to support the pendulum and will be able to tell that it is directly over the pendulum.

So if there is a pseudo-force acting on the pendulum, why isn't it hanging off to one side? The answer is quite simple: EVERYTHING in the reference frame is accelerated by the pseudo-force! Our error was to assume that it applied merely to the pendulum mass.

In retrospect this should be obvious. According to the Galilean transformations, any and all objects in the stationary frame will be "accelerated" by the same amount according to an accelerating observer. The Galilean transforms apply to ANY measurement made in the stationary frame. The error is easy to make since we often only look at the object under study when we use Galilean Relativity. An Einsteinian Relativist might not be so prone to the error since is it quite normal to consider the time transformation at several points in the reference frame. (A similar comment should hold true for Cosmologists as well.)

I hope this has cleared up the confusion generated by my previous posts as well as answering what went wrong in logical argument 2 in your original post.

-Dan

13. Mar 3, 2006

### gandharva_23

What we can say is that if we get into the frame of observer the pivot will also be accelerating in the backward direction . thats what i thought about 2 days ago .