Pendulum oscillating in an accelerating car

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SUMMARY

The discussion focuses on the dynamics of a pendulum oscillating in a car that accelerates at a uniform rate, denoted as ## a ##. The equation of motion derived from the Lagrangian and Euler-Lagrange equations is given by $$l \ddot{\phi} + a \cos(\phi) + g \sin(\phi) = 0$$. The angular frequency for small oscillations is expressed as ##\omega = \sqrt{\frac{\sqrt{a^2 + g^2}}{l}}##, contingent upon the small angle approximation, specifically when ##\phi + \arctan(\frac{a}{g}) << 1##. The equilibrium position of the pendulum is established as ##\phi_{0} = \arctan{(\frac{a}{g})##, which alters the dynamics depending on the acceleration.

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saadhusayn
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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.
 
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saadhusayn said:
we must assume that a<<ga<<ga
No. You want to assume ##
\phi + \arctan(\frac{a}{g}) << 1## so that ##\sin \phi' \approx \phi'## (where ##\phi'= \phi + \arctan(\frac{a}{g}) ##
 
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.
 
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?
 
saadhusayn said:
if ##\phi_{0}## is small, or equivalently the acceleration is small.
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'##.
 

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