# Homework Help: Pendulum oscillating in an accelerating car

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1. May 11, 2017

We have a car accelerating at a uniform rate $a$ and a pendulum of length $l$ hanging from the ceiling ,inclined at an angle $\phi$ to the vertical . I need to find $\omega$ for small oscillations.

From the Lagrangian and Euler-Lagrange equations, the equation of motion is given by

$$l \ddot{\phi} + a \cos(\phi) + g \sin(\phi) = 0$$

Equivalently,

$$l \ddot{\phi} + \sqrt{a^2 + g^2} \sin(\phi + \arctan(\frac{a}{g})) = 0$$

I know that $\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}$ .

In order to make the small angle approximation for the sine, we must assume that $a << g$. So is it the case that $\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}$ only if $a << g$, because this assumption is not stated in the question.

2. May 11, 2017

### BvU

No. You want to assume $\phi + \arctan(\frac{a}{g}) << 1$ so that $\sin \phi' \approx \phi'$ (where $\phi'= \phi + \arctan(\frac{a}{g})$

3. May 11, 2017

### Dr.D

I would suggest that you first find the equilibrium position of the pendulum in the accelerating vehicle, then linearize the equation of motion about that position. The equilibrium position is not straight down if a>0.

4. May 11, 2017

The equilibrium position is $\phi_{0} = \arctan{(\frac{a}{g})}$. So if we linearize the equation, it becomes

$$\ddot{\phi} + \frac{\sqrt{a^2 + g^2}}{l} (\phi +\phi_{0}) = 0$$ if $\phi_{0}$ is small, or equivalently the acceleration is small.

Equivalently,
$$\ddot{\phi} + \frac{\sqrt{a^2 + g^2}}{l} \phi = - \frac{\sqrt{a^2 + g^2}}{l} \phi_{0}$$

Then the general solution has an angular speed of $$\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}$$

This expression for $\omega$ doesn't hold for large values of acceleration, does it?

5. May 12, 2017

### BvU

Again, no. There is a restoring force for $\phi'$ that is proportional to $\sin\phi'$. $a$ can be $+g$, $-g$ or anything and you still get the same equation, only with an different $\phi_0$ and a different $\omega$. The solution is a harmonic motion only for small $\phi'$.