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Homework Help: Simple Harmonic Motion of a particle

  1. Aug 24, 2007 #1

    danago

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    A particle moves in a straight line so that its position, x cm, from a fixed point O, at time t seconds is given by [tex]x=3+12sin(2\pi t)+5cos(2 \pi t)[/tex].

    Prove that the particle is undergoing simple harmonic motion, and find x when its speed is 26[tex]\pi[/tex] cm/s.


    I was fine with the first part of the question...

    [tex]
    \begin{array}{l}
    x = 3 + 12\sin (2\pi t) + 5\cos (2\pi t) \\
    v = \dot x = 24\pi \cos (2\pi t) - 10\pi \sin (2\pi t) \\
    a = \ddot x = - 48\pi ^2 \sin (2\pi t) - 20\pi ^2 \cos (2\pi t) = - 4\pi ^2 (x - 3) = 12\pi ^2 - 4\pi ^2 x \\
    \end{array}
    [/tex]

    With the second part, i can easily set the velocity equal to 26[tex]\pi[/tex] and solve it on my calculator, but how can i solve it algebraically?

    Thanks in advance,
    Dan.
     
    Last edited: Aug 24, 2007
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  3. Aug 24, 2007 #2

    Dick

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    If you want to solve it exactly, you can write an expression of the form A*sin(x)+B*cos(x) as a single trig function. See the wikipedia page of trig identities under 'Linear combinations'.
     
  4. Aug 25, 2007 #3
    Algebraically, the equation simplifies to [tex]x=3+13sin(2\pi t+\theta) [/tex] where [tex]\theta=sin^{-1}\frac{5}{13}[/tex].
     
  5. Aug 25, 2007 #4

    danago

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    Ah ofcourse, i forgot about solving it like that. Thanks :smile:
     
  6. Sep 19, 2007 #5
    I have a wooden log floating in the water and it is being pushed towards the water. I am to show that the force that is pushing the log back to the equilibrium position is harmonic, and i just have no idea how to do it... There are no equations attached.
     
  7. Sep 19, 2007 #6

    Dick

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    If you have a question, don't piggy-back it on someone else's thread. It won't get the notice it deserves. You don't need to show the 'force is harmonic', to show something is a harmonic oscillator you just need to show the restoring force is proportional to displacement. If this isn't clear, DON'T REPLY HERE. Start a new thread.
     
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