Pendulum oscillating in an accelerating car

In summary, 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}}##.Equivalently,$$\ddot{\phi} + \frac{\sqrt{a^2 + g^2}}{
  • #1
saadhusayn
22
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.
 
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  • #2
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}) ##
 
  • #3
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
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
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'##.
 

What is a pendulum oscillating in an accelerating car?

A pendulum oscillating in an accelerating car refers to a pendulum that is attached to the ceiling of a car that is accelerating. The motion of the car causes the pendulum to move back and forth, creating a unique oscillation pattern.

What causes the pendulum to oscillate in an accelerating car?

The pendulum oscillates in an accelerating car due to the inertia of the pendulum. As the car accelerates, the pendulum tries to keep its original position, causing it to move back and forth.

What happens to the pendulum when the car stops accelerating?

When the car stops accelerating, the pendulum will continue to oscillate for a short period of time. Eventually, the pendulum will come to a stop and hang vertically due to the force of gravity.

Does the length of the pendulum affect its oscillation in an accelerating car?

Yes, the length of the pendulum does affect its oscillation in an accelerating car. The longer the pendulum, the slower the oscillation period. This is because the longer pendulum has a larger swing arc and takes longer to complete a cycle.

What is the relationship between the acceleration of the car and the amplitude of the pendulum oscillation?

The acceleration of the car and the amplitude of the pendulum oscillation have a direct relationship. This means that as the acceleration of the car increases, the amplitude of the pendulum oscillation also increases. Similarly, when the acceleration decreases, the amplitude of the pendulum oscillation decreases.

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