Exercise 2
Equations of Motion
Newton’s law applies: \[\vec{F}(\vec{r}) = - \vec{\nabla} V(\vec{r}) = m \ddot{\vec{r}}\]
or written with components: \[F_x = - \frac{\partial V}{\partial x} = m \ddot{x}\] \[F_y = - \frac{\partial V}{\partial y} = m \ddot{y}\]
Using \(V(x, y) = \frac{1}{2}k(x^2+y^2)\) we get: \[\ddot{x}(t) = -\frac{k}{m}x(t)\] \[\ddot{y}(t) = -\frac{k}{m}y(t)\]
These are the differential equations for the harmonic oscillator in two dimensions. The general solution for these are (see Exercise 4, Lecture 3): \[x(t) = A \cos \omega t + B \sin \omega t\] \[y(t) = C \cos \omega t + D \sin \omega t\]
with \(\omega = \sqrt{k/m}\) and the initial conditions \(A = x(0), B=v_x(0)/\omega, C = y(0), D = v_y(0)/\omega\).
These are the equations of motion.
Circular orbits
The following applies to a circular motion: \(r = \sqrt{x^2 +y^2} = \mathrm{const}\). The radius has to be independent of time t. \[x^2 +y^2 = \cos^2 \omega t (A^2 + C^2) + \sin^2 \omega t (B^2 +D^2) + \cos \omega t \sin \omega t (2AB+ 2CD) = \mathrm{const.}\]
To achieve this, the following equations are to be fulfilled: \[A^2 + C^2 = B^2 +D^2\] (with \(\cos^2 \omega t + \sin^2 \omega t = 1\)) \[AB = -CD\] The last expression in the equation cancels out.
Let \[A=D \Rightarrow B = -C\] or \[A=-D \Rightarrow B = C\]
This fulfills the requirements.
It follows: \[ v_y(0) = \omega x(0) \] \[ v_x(0) = -\omega y(0) \] or \[ v_y(0) =- \omega x(0) \] \[ v_x(0) = \omega y(0) \]
Other values for A, B, C and D lead to elliptical orbits.
After one period of time \(t=T\) the system is again in the initial state at \(t=0\). In the equations of motion there are sums of cosine and sine, these functions have the period \(2\pi\). The period \(T\) does not depend on the radius (see Exercise 6, Lecture 2): \[ T = \frac{2\pi}{\omega} = \frac{2\pi \sqrt{m}}{\sqrt{k}} \]
Animation
red: position vector
blue: velocity vector
green: orbit of particle
Adjust the sliders to check the inital conditions for a circular orbit. \(\omega = 1/s\), one period takes about 6.28 s.
The values for A, B, C and D can be adjusted in the range of -200 … 200.A:
B:
C:
D:
Conservation of Energy
For the total energy \(E = T + V\) the following applies: \[E = \frac{1}{2}m (\dot{x}^2 + \dot{y}^2) + \frac{1}{2}k (x^2 + y^2)\]
For conservation of energy, \(E\) has to be independent of \(t\):
\[\frac{\mathrm{d}E}{\mathrm{d}t} = 0\]
\[=\frac{1}{2}m (2 \dot{x} \ddot{x} + 2 \dot{y} \ddot{y}) + \frac{1}{2}k (2 x \dot{x} + 2 y \dot{y} )\]
\[= m \dot{x} \ddot{x} + m \dot{y} \ddot{y} +k x \dot{x} + k y \dot{y}\] \[= m( \dot{x} \ddot{x} +\dot{y} \ddot{y} + \frac{k}{m} x \dot{x} +\frac{k}{m} y \dot{y})\] \[= m \left( \dot{x} \underbrace{(\ddot{x} + \frac{k}{m} x )}_{=0} +\dot{y} \underbrace{(\ddot{y} + \frac{k}{m} y )}_{=0} \right)\] \[=0\]