Pendulum oscillating in an accelerating car

AI Thread Summary
The discussion focuses on analyzing the motion of a pendulum in an accelerating car using Lagrangian mechanics. The equation of motion derived is l \ddot{\phi} + a \cos(\phi) + g \sin(\phi) = 0, leading to the angular speed formula ω = √(√(a² + g²)/l). It is clarified that the small angle approximation requires the condition φ + arctan(a/g) << 1, not just a << g. The equilibrium position of the pendulum is determined to be φ₀ = arctan(a/g), and the linearized equation of motion shows that the solution remains harmonic for small deviations. The relationship between acceleration and angular speed holds true as long as the small angle condition is satisfied.
saadhusayn
Messages
17
Reaction score
1
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.
 
Physics news on Phys.org
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'##.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top