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

[xiaozegu]

Homework 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?

 

[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.

 

[xiaozegu]

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

 

[voko]

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

 

[xiaozegu]

So it needs A+B<0?

 

[haruspex]

So it needs A+B<0?

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?

 

[xiaozegu]

I can get Be^2γt>A

 

[haruspex]

I can get Be^2γt>A

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?

 

[xiaozegu]

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?

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?

 

[haruspex]

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?

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.

 

[xiaozegu]

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.

If B is negative, is it OK?

 

[haruspex]

If B is negative, is it OK?

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

 

[xiaozegu]

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

So it needs A>0,B<0?

 

[haruspex]

So it needs A>0,B<0?

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?

 

[xiaozegu]

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?

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

 

[haruspex]

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

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?

 

[xiaozegu]

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?

I think it is A>0 and A>B

 

[haruspex]

I think it is A>0 and A>B

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?

 

[xiaozegu]

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?

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

 

[haruspex]

A equals to zero.

Yes!

 

[xiaozegu]

Yes!

thank you very much