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

  1. Nov 17, 2012 #1
    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. jcsd
  3. Nov 17, 2012 #2
    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.
     
  4. Nov 17, 2012 #3
    Is it a resistance force? If not, how can r decrease?
     
  5. Nov 17, 2012 #4
    Before you can answer the question "how can r decrease", you must obtain the acceleration of the particle.
     
  6. Nov 17, 2012 #5
    So it needs A+B<0?
     
  7. Nov 17, 2012 #6

    haruspex

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    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?
     
  8. Nov 17, 2012 #7
    I can get Be2γt>A
     
  9. Nov 17, 2012 #8

    haruspex

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    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?
     
  10. Nov 17, 2012 #9
    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?
     
  11. Nov 18, 2012 #10

    haruspex

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    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.
     
  12. Nov 18, 2012 #11
    If B is negative, is it OK?
     
  13. Nov 18, 2012 #12

    haruspex

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    With A being what? If A > 0, what will happen as t increases (regardless of B)? If A < 0 what will happen?
     
  14. Nov 18, 2012 #13
    So it needs A>0,B<0?
     
  15. Nov 18, 2012 #14

    haruspex

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    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?
     
  16. Nov 18, 2012 #15
    If A>0, r will decrease as time, If A<0, r will increase as time.
     
  17. Nov 18, 2012 #16

    haruspex

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    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?
     
  18. Nov 19, 2012 #17
    I think it is A>0 and A>B
     
  19. Nov 19, 2012 #18

    haruspex

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    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?
     
  20. Nov 19, 2012 #19
    A equals to zero. Because r cannot be negative, so A cannot <0. Plus, B>0.
     
  21. Nov 19, 2012 #20

    haruspex

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    Yes!
     
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