2D harmonic oscillator Problem

zorop · #1 $Old$ August 6th, 2009, 07:41 PM

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 $r_{0}=2i-j$ and $v_{0}=4i+8j$ ,
(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

luobo · #2 $Old$ August 7th, 2009, 08:11 PM

(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