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Math Help - 2D harmonic oscillator Problem

  1. #1
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    2D harmonic oscillator Problem

    A particle of mass 1 moves in a 2D harmonic oscillator potential V(x, y)=8(x^2+4y^2). If the position and velocity of the particle at time t = 0 are given by r_{0}=2i-jand v_{0}=4i+8j,
    (a) Find the position and velocity of the particle at any time t > 0.

    r(t)=(2cos4t+sin4t)i+(-cos8t+sin8t)j
    v(t)=(-8sin4t+4cos4t)i+(8sin8t+8cos8t)j

    (b) Determine the period of the motion.

    T=\frac{\pi}{2} The period depends on r(t) or v(t) or both?
    How to get this? Why?

    (c) Find the total energy of the particle.

    E=104 How to get this?

    (d) Suppose that the potential is instead V(x,y)=8(x^2+2y^2). Is there a period defined for the motion in this case? Explain why or why not.

    Any advice and comments will be helpful, thanks a lot, guys
    Last edited by zorop; August 10th 2009 at 03:55 AM.
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  2. #2
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    (b) Use  r(t+T)-r(t)=0 or  v(t+T)-v(t)=0. Results should be the same. Or simply
    T_x =\frac{2\pi}{4}=\frac{\pi}{2}
    T_y=\frac{2\pi}{8}=\frac{\pi}{4}.
    \frac{T_x}{T_y}=2\in \mathbb{Q}
    So system period T=T_x=\frac{\pi}{2}

    (c) Total energy is the sum of kinetic energy and the potential, and total energy is conserved.
    E=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+V(x,y)=\frac{1 }{2}\times1\times(4^2+8^2)+8\times[2^2+4 (-1)^2]=104W

    (d) E=\frac{1}{2}\times1\times(\dot{x}^2+\dot{y}^2)+8( x^2+2y^2)
    \frac{\partial{E}}{\partial{x}}=\ddot{x}+16x=0
    \frac{\partial{E}}{\partial{y}}=\ddot{y}+32y=0
    So,
    \omega_x =4\, T_x=\frac{2\pi}{\omega_x}=\frac{\pi}{2}
    \omega_y=4\sqrt{2}, T_y=\frac{2\pi}{\omega_y}=\frac{\pi}{2\sqrt{2}}
    \frac{T_x}{T_y}=\sqrt{2}\notin\mathbb{Q}

    Luobo
    Last edited by mr fantastic; September 19th 2009 at 01:58 AM. Reason: Restored original reply
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