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Waves, solving Energy with position?

  1. Sep 8, 2006 #1
    Howdy,

    This is only for a second year course, but i didn't think it would get answered in the intro forums...

    Ok, the wording of this question doesn't make much sense to me, but I assure you that this is what it says...
    ********************************************
    Demonstrate how

    [tex]x=A \cos (\omega t+\phi)[/tex]

    can be the answer for

    [tex]E=\frac{1}{2}m\frac{dx}{dt}^2 + \frac{1}{2}kx^2 [/tex]

    ********************************************

    I just don't see how this could possibly work. Any ideas would be helpful

    -----------

    As for my thoughts, I figured that either the question was written wrong, or that using the simplified energy eq'n (E=kA^2) would help...but I get nowheres using that still.
     
    Last edited: Sep 8, 2006
  2. jcsd
  3. Sep 8, 2006 #2

    Doc Al

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    Staff: Mentor

    There's a typo in that last equation. d^2x/dt^2 is the acceleration, not the speed.
     
  4. Sep 8, 2006 #3
    thanks for pointing that out, it was supposed to be velocity so i edited it to say such. I was mixing two questions together for a second...i still don't know what to do though
     
  5. Sep 8, 2006 #4

    Doc Al

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    Staff: Mentor

    Just plug your equation for x into the energy equation and see if it makes sense as a solution.
     
  6. Sep 8, 2006 #5
    no, that would just complicate things. when i actually subsititute the derivative in the first eq'n i can simplify it easily to [tex]E= \frac{1}{2} k A^2[/tex] but that doesn't answer the question...what it seems to want is for you to show how [tex]x=A \cos (\omega t+\phi)[/tex] actually IS the answer for [tex]E=\frac{1}{2}m\frac{dx}{dt}^2 + \frac{1}{2}kx^2 [/tex]

    which just sounds absurd to me.
     
    Last edited: Sep 8, 2006
  7. Sep 8, 2006 #6
    hmmmm 40+ views and nothing. I personally think that the question must be wrong, but I'll continue to check back in case anyone sees a relationship that I'm missing between the two expressions
     
  8. Sep 8, 2006 #7
    Solve it as a differential equation? You have [tex]\frac{dx}{dt}^2[/tex] and [tex]x^2[/tex], so if you try [tex]x=A'e^{\lambda t}[/tex] it should work. Just a suggestion, I'm not 100% on this.
     
  9. Sep 8, 2006 #8

    Doc Al

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    Staff: Mentor

    I'm a bit puzzled why you ignored my advice to just plug it in and see if it satisfies the equation. That's the simplest way to verify a proposed solution to any equation! And that seems to be all you are asked to do: Just demonstrate that the given function of x is a solution to the energy equation. And it is!
     
  10. Sep 9, 2006 #9
    I'm sorry, I guess I just misunderstood what the question was asking really...
    if you just plug it in and work though it you get the [tex]e=\frac{1}{2}kA^2[/tex] equation that one would expect, I just didn't think that THAT was what the question wanted.

    Either way, sorry for ignorning your advise, but what you said is the only thing that I myself though about doing, and I wasn't convinced that was right.
     
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