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Thread: 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 $\displaystyle V(x, y)=8(x^2+4y^2)$. If the position and velocity of the particle at time t = 0 are given by $\displaystyle r_{0}=2i-j$and $\displaystyle v_{0}=4i+8j$,
    (a) Find the position and velocity of the particle at any time t > 0.

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

    (b) Determine the period of the motion.

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

    (c) Find the total energy of the particle.

    $\displaystyle E=104$ How to get this?

    (d) Suppose that the potential is instead $\displaystyle V(x,y)=8(x^2+2y^2)$. 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; Aug 10th 2009 at 02:55 AM.
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  2. #2
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    (b) Use $\displaystyle r(t+T)-r(t)=0$ or $\displaystyle v(t+T)-v(t)=0$. Results should be the same. Or simply
    $\displaystyle T_x =\frac{2\pi}{4}=\frac{\pi}{2}$
    $\displaystyle T_y=\frac{2\pi}{8}=\frac{\pi}{4}$.
    $\displaystyle \frac{T_x}{T_y}=2\in \mathbb{Q}$
    So system period $\displaystyle T=T_x=\frac{\pi}{2}$

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

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

    Luobo
    Last edited by mr fantastic; Sep 19th 2009 at 12:58 AM. Reason: Restored original reply
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