How to parametrize motion of a pendulum in terms of Cartesian coordinates?

[zenterix]

Homework Statement

I'm finishing up a problem set from MIT OCW's 8.03 "Vibrations and Waves".

Relevant Equations

We are asked to parametrize the motion of the pendulum in terms of Cartesian coordinate $x$ in the coordinate system with "origin at the pendulum equilibrium position and $x$-axis horizontal in the plane of the pendulum. Find the exact equation of motion of the pendulum in terms of $x$".

Let the origin be where the pendulum string is attached to the ceiling.

$$\sin{\theta(t)}=\frac{x(t)}{L}\tag{1}$$

$$\cos{\theta(t)}=\frac{y(t)}{L}\tag{2}$$

$$\theta(t)=\sin^{-1}{\frac{x(t)}{L}}\tag{3}$$

$$\dot{\theta}(t)=\frac{\dot{x}(t)}{\sqrt{L^2-x^2(t)}}\tag{4}$$

$$\ddot{\theta}(t)=\frac{\ddot{x}(t)(L^2-x^2(t))+x(t)\dot{x}^2(t)}{(L^2-x^2)^{3/2}}=\frac{\ddot{x}(t)}{(L^2-x^2)^{1/2}}+\frac{x\dot{x}^2}{(L^2-x^2)^{3/2}}\tag{5}$$

From Newton's 2nd law,

$$\vec{F}_g=-mg\hat{j}=m\vec{a}=m(L\ddot{\theta}\hat{\theta}-L\dot{\theta}^2\hat{r})\tag{6}$$

$$=(y\ddot{\theta}-x\dot{\theta}^2)\hat{i}+(x\ddot{\theta}+y\dot{\theta}^2)\hat{j}\tag{7}$$

Equating the components we get the two equations

$$\dot{\theta}^2=\frac{y}{x}\ddot{\theta}\tag{8}$$

$$x\ddot{\theta}+y\dot{\theta}^2=-g\tag{9}$$

Thus

$$\ddot{\theta}=-\frac{x}{L^2}g\tag{10}$$

and using (5) we get

$$\frac{\ddot{x}(t)}{(L^2-x^2)^{1/2}}+\frac{x\dot{x}^2}{(L^2-x^2)^{3/2}}=-\frac{x}{L^2}g\tag{11}$$

This is a nonlinear differential equation.

Finally, let's change the coordinate system to have origin as prescribed in the problem statement (the position of the pendulum at the very bottom of its trajectory).

Let coordinate system 1 be the one used above (with origin at the ceiling) and coordinate system 2 be the new one with the origin at the bottom of the trajectory.

$$\vec{r}_{1}=\vec{r}_{1,2}+\vec{r}_2$$

where $\vec{r}_{1,2}=-L\hat{j}$ is the position of the origin of coordinate system 2's origin from the point of view of coordinate system 1. This is a fixed vector and so

$$\vec{v}_1=\vec{v}_2$$

$$\vec{a}_1=\vec{a}_2$$

Thus, in particular, for a point with coordinates $(x_1,y_1)$ in coordinate system 1 and $(x_2,y_2)$ in coordinate system 2 we have

$$x_1=x_2$$

$$\dot{x}_1=\dot{x}_2$$

$$\ddot{x}_1=\ddot{x}_2$$

If we sub these relationships into (11) there is no change in the differential equation.

Now, (11) looks quite complicated. I'm not sure if this is the differential equation that is being sought here. I obtained it essentially by taking derivatives of $\theta(t)=\arcsin{\left (\frac{x}{L}\right )}$. Is there another, better way?

 

[zenterix]

The next question is what to do if we can assume that $x$ is very small.

Recall that the equation of motion found in the OP is

$$\frac{\ddot{x}(t)}{(L^2-x^2)^{1/2}}+\frac{x\dot{x}^2}{(L^2-x^2)^{3/2}}=-\frac{x}{L^2}g\tag{11}$$

If $x$ is small I guess we can disregard the $x^2$ terms. If we do so we get

$$\frac{\ddot{x}}{L}+\frac{x\dot{x}^2}{L^3}=-\frac{g}{L^2}x$$

If we further eliminate the term $x\dot{x}^2$ then we are back to the SHM equation

$$\ddot{x}+\frac{g}{L}x=0$$

Assuming the calculations are correct, what justifies this last step. In other words, why do we know that we can just throw away the $x\dot{x}^2$ term?