# B SHM of Oscillating Pendulum

1. Jul 22, 2015

### andyrk

For SHM of oscillating pendulum, when the pendulum is at the extreme position, what is considered as the displacement? The curve/arc of the circle the bob is following or the straight line distance from the mean position?

2. Jul 22, 2015

### Dvorak

It is the curve or arc of the circle the bob is following.

3. Jul 23, 2015

### andyrk

But shouldn't it be the straight line displacement from the mean position?

4. Jul 23, 2015

### Qwertywerty

Technically , for SHM of a pendulum we use angular diplacement θ rather than straight line displacement . However as θ is a small we can still use straight line displacement . This is approximately equal to the length of the arc .

5. Jul 23, 2015

### andyrk

What if θ is not small?

6. Jul 23, 2015

### Qwertywerty

Then you wouldn't have SHM .

7. Jul 23, 2015

### sophiecentaur

What do your text books tell you about these questions you are asking on PF? Yet again, this topic is well covered on line and in books.

8. Jul 23, 2015

### andyrk

I don't know. They didn't mention if it was small or not. But they have taken the displacement to be $lθ$. So I think that must mean that θ is small. Am I right?

9. Jul 23, 2015

### Nathanael

Yes. As qwerty said, Lθ is just an approximation for Lsinθ, but this approximation is only accurate for small angles.
If you know about taylor series, you can look at the taylor series for sinθ and see where this approximation comes from (the smaller θ, the smaller the effect of the higher order terms).
If you don't know about taylor series then you can understand this approximation geometrically, too. But if that's the case then you should go study taylor series! Useful stuff..

10. Jul 24, 2015

### sophiecentaur

SHM is a specific kind of oscillation - the simplest there is, in fact. SHM requires the restoring force to be directly proportional to the displacement and, for a simple pendulum, it is not. Look at this link, which shows that the motion of a simple pendulum 'approaches' pure SHM as the amplitude approaches zero.

11. Jul 24, 2015

### Qwertywerty

Do you know the derivation for SHM ?

12. Jul 24, 2015

### Nidum

Some old clockmakers went to considerable trouble to make the motion of their pendulums follow SHM as near as possible at practical angles of swing . They did this by devising ways of making the effective length of the pendulum change slightly as it swung back and forth .

13. Jul 24, 2015

### nasu

You can use either one as a coordinate. You can also use the angle, as it was already mentioned.
In general, for a system, there is more than one way to choose the coordinates. Some may be more convenient but not in anyway "correct" or "wrong".
But even if you use the straight distance, the motions is SHM only approximately, for small angles.

And the length of the arc is Lθ for any value of the angle. Is not an approximation.
The straight distance though, is only approximately so, for small angles.

14. Jul 24, 2015

### sophiecentaur

Yes but, if you are using the arc in your equation, the restoring 'force' becomes a torque and that is not proportional to the angle in any case. It's still non linear with change of position.

15. Jul 24, 2015

### sophiecentaur

I don't know who the "they" you refer to were. If they were showing you the proper derivation of SHM, they would have included the limiting case for small angles.

16. Jul 24, 2015

### nasu

Did I say that it becomes linear if the angle is used?

The last part was a separate point. Someone said in a previous post that s=Lθ it is just an approximation for small theta, which is not true. I did not mean (and did not say)that the motion becomes SHM.

Last edited: Jul 24, 2015
17. Jul 24, 2015

### sophiecentaur

It bothered me that it may have been interpreted wrongly. You were discussing a change of co ordinates. The result is that the non linearity turns up somewhere else and that point needs to be made constantly, IMO.