# I do not why the particle does the simple harmonic motion.

1. Nov 17, 2012

### xiaozegu

2. Relevant equations

I do not why the particle does the simple harmonic motion. And how to find such innitial condition to satify r decreases continually in time.

3. The attempt at a solution[/b

Is it need to take derivative of r?

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2. Nov 17, 2012

### voko

The equation given is NOT one of harmonic motion. It contains exponential non-periodic terms.

As for how to obtain the equation, recall the formula for acceleration in circular motion.

3. Nov 17, 2012

### xiaozegu

Is it a resistance force? If not, how can r decrease?

4. Nov 17, 2012

### voko

Before you can answer the question "how can r decrease", you must obtain the acceleration of the particle.

5. Nov 17, 2012

### xiaozegu

So it needs A+B<0?

6. Nov 17, 2012

### haruspex

Assuming γ>0 (which you can, without loss of generality), how do the individual terms e-γt and e+γt vary as t increases? How could you make a linear combination of them that always decreases?

7. Nov 17, 2012

### xiaozegu

I can get Be2γt>A

8. Nov 17, 2012

### haruspex

Do you mean that for any A and B the above will be true for all sufficiently large t? Not so.
If that's not what you meant, what do you mean?
Please answer the specific questions I asked before: how does each of the two terms change as t increases?

9. Nov 17, 2012

### xiaozegu

If γ >0, e-γt decreases as time goes up, e+γt increase. So I need find the derivative of these two items make de+γt/dt>de-γt/dt?

10. Nov 18, 2012

### haruspex

No, you need to find a linear combination of e+γt and e-γt, i.e. something of the form Ae+γt + Be-γt, which decreases (without going negative) as t increases. You have correctly worked out that e-γt does that. There's a very easy solution for A and B.

11. Nov 18, 2012

### xiaozegu

If B is negative, is it OK?

12. Nov 18, 2012

### haruspex

With A being what? If A > 0, what will happen as t increases (regardless of B)? If A < 0 what will happen?

13. Nov 18, 2012

### xiaozegu

So it needs A>0,B<0?

14. Nov 18, 2012

### haruspex

Instead of firing off guesses, please try to answer my questions: If A > 0, what will happen as t increases (regardless of B)? If A < 0 what will happen?

15. Nov 18, 2012

### xiaozegu

If A>0, r will decrease as time, If A<0, r will increase as time.

16. Nov 18, 2012

### haruspex

We may be at cross purposes here. In my posts I have A as the coefficient for the +γ term and B as that for the -γ term. Do you have it the other way around?
Anyway, what I'm trying to steer you to is that if A > 0 then Ae+γt will tend to +∞ as t→+∞, and completely dominate over the e-γt term, making it irrelevant (regardless of the value of B). Can you see that?
Conversely, if A < 0 then Ae+γt will tend to -∞ as t→+∞, again completely dominating the e-γt term, making it irrelevant.
So if you need the function to decrease as t→+∞ but without going negative, what does that leave as a possible value for A?

17. Nov 19, 2012

### xiaozegu

I think it is A>0 and A>B

18. Nov 19, 2012

### haruspex

No, if A>0 then Ae+γt+Be-γt →+∞ as t→+∞, for all B. So we can absolutely rule out A>0.
Similarly, if A<0 then Ae+γt+Be-γt →-∞ as t→+∞, for all B. So we can absolutely rule out A<0.
What is left?

19. Nov 19, 2012

### xiaozegu

A equals to zero. Because r cannot be negative, so A cannot <0. Plus, B>0.

20. Nov 19, 2012

Yes!