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 [itex]m \ddot{y} = F_y = -ky[/tex] can be solved directly.

These equations should look very familiar.

Regards,
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 [itex]m \ddot{y} = F_y = -ky[/tex] can be solved directly.

These equations should look very familiar.

Regards,
George

Thank you for replying.

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?

 

[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