The period of oscillation of a bob in an accelerating frame

[Soffie]

If a suspended pendulum bob is accelerated (in a car, for example), if you're in the accelerating frame of reference, you will observe the fictitious force which appears to act on the bob (as you're in the accelerating frame, the bob is not 'moving' so to speak, so to establish equilibrium you introduce the fictitious force.
The bob is thus at an angle to the vertical, due to the fictious force in the accelerating frame, OR due to the acceleration in the inertial frame. If the bob performs small oscillations about the line the angle makes to the vertical, how would you go about finding the time period? Presumably it'd not just sqrt(l/g)

 

[Doc Al]

Presumably it'd not just sqrt(l/g)

In that accelerating frame, what would the effective value of "g" be? (If you drop something, how would it accelerate as measured in that frame?)

 

[osilmag]

Sum the forces on the bob. The y acceleration would still be g. And the other force would be a reaction from the car accelerating. I think it still would be 2*pi*sqrt(l/g).

 

[Doc Al]

Sum the forces on the bob. The y acceleration would still be g. And the other force would be a reaction from the car accelerating.

Only two "real" forces act on the bob: gravity and the tension in the cord. But viewed from the accelerating frame there is an additional "fictitious" force that must be added. The effect of that added force can be viewed as a change in the effective "g".

I think it still would be 2*pi*sqrt(l/g).

That is not correct.

 

[Soffie]

Only two "real" forces act on the bob: gravity and the tension in the cord. But viewed from the accelerating frame there is an additional "fictitious" force that must be added. The effect of that added force can be viewed as a change in the effective "g".That is not correct.

Ok so I've drawn some diagrams which may help solve the problem- I've oriented the axes so the tension is pointing up- what the bob looks like in its effective equilibrium position, and resolved the fictitious force and the weight in terms of the angle phi, which was the angle made due to the acceleration: (sorry I think you have to view the images full size down below)

Now here's the diagram for the displacement:

am I right in thinking the horizontal components of mg and Ffict still cancel out, so we're just left with tension and the downward force due to F fict and mg? Then you can just solve using SHM equations.

 

[Doc Al]

I'm not quite sure I follow what you're doing. Try this. Imagine the pendulum hanging from the ceiling of a train car. First, assume no acceleration. What forces act? Apply Newton's 2nd law. I'm sure you understand how that situation can lead to SHM about the equilibrium position.

Next, have the car accelerate. Viewed from within the accelerating car, what forces act? Apply Newton's 2nd law. (You'll need to add the fictitious force to apply Newton.) How can you rearrange the resulting equation to make it look similar to the previous one? (And thus convince yourself that it will also lead to SHM about the equilibrium position. The same equations have the same solutions!)