Pendulum oscillating in an accelerating car

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Homework Help Overview

The discussion revolves around a pendulum oscillating in a car that is accelerating uniformly. The participants are exploring the dynamics of the pendulum, particularly focusing on the conditions under which the angular frequency, denoted as ##\omega##, can be accurately described using the derived equations of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivation of the equation of motion for the pendulum and the implications of the small angle approximation. Questions arise regarding the assumptions needed for the approximation to hold, particularly the relationship between the acceleration of the car and gravitational acceleration.

Discussion Status

The discussion is active, with participants offering different perspectives on the assumptions regarding the small angle approximation and the equilibrium position of the pendulum. There is an exploration of how the equilibrium position changes with acceleration and how this affects the derived expressions for ##\omega##.

Contextual Notes

Participants note that the assumption of small angles may not hold under certain conditions, and they emphasize the need to consider the equilibrium position in the context of the car's acceleration. The discussion highlights the complexity of the problem and the need for careful consideration of the assumptions involved.

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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