Particle of Mass M Moving in XY-Plane: Potential Energy & Orbit Analysis

[Reshma]

A particle of mass M is free to move in the horizontal plane(xy-planne here). It is subjected to force \vec F = -k\left(x\hat i + y\hat j\right), where 'k' is a positive constant.
There are two questions that have been asked here:
1] Find the potential energy of the particle.

\vec \nabla \times \vec F = 0
The given force is conservative and hence a potential energy function exists.
Let it be U.
F_x = -\frac{\partial U}{\partial x} = -kx

F_y = -\frac{\partial U}{\partial y} = -ky

U(x,y) = \frac{k}{2}\left(x^2 + y^2) + C

2]If the particle never passes through the origin, what is the nature of the orbit of the particle?

I am not sure what the PE function tells about the trajectory of the particle. Explanation needed...

 

[George Jones]

The potential energy can be used to find the Lagrangian, and then Lagrange's equation can be used to find the motion.

Alternatively, m \ddot{x} = F_x = -kx and m \ddot{y} = F_y = -ky[/tex] can be solved directly.<br /> <br /> These equations should look very familiar.<br /> <br /> Regards,<br /> George

 

[Reshma]

The potential energy can be used to find the Lagrangian, and then Lagrange's equation can be used to find the motion.

Alternatively, m \ddot{x} = F_x = -kx and m \ddot{y} = F_y = -ky[/tex] can be solved directly.<br /> <br /> These equations should look very familiar.<br /> <br /> Regards,<br /> George

<br /> <br /> Thank you for replying.<br /> <br /> So this is a 2-dimensional harmonic oscillator. The general solution would be:<br /> x = A\cos(\omega_0 t - \alpha) &amp; y = B\cos(\omega_0 t - \beta)<br /> <br /> So, the resultant path of these two SHMs would be an ellipse, right?

 

[George Jones]

So this is a 2-dimensional harmonic oscillator. The general solution would be:
x = A\cos(\omega_0 t - \alpha) & y = B\cos(\omega_0 t - \beta)

So, the resultant path of these two SHMs would be an ellipse, right?

Yes.

Regards,
George