The period of oscillation of a bob in an accelerating frame

In summary: 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: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.I'm not quite sure I follow what you're doing. Try this.
  • #1
Soffie
10
0
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)
 
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  • #2
Soffie said:
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?)
 
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  • #3
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).
 
  • #4
osilmag said:
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".

osilmag said:
I think it still would be 2*pi*sqrt(l/g).
That is not correct.
 
  • #5
Doc Al said:
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)
206628-4e0ca3aac3a018d12cddbf1a1e339fa6.jpg

Now here's the diagram for the displacement:
206629-ecf3704eee6272619b8401bae6a43f20.jpg

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.
 

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  • #6
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!)
 

1. What is the definition of oscillation in a physics context?

Oscillation is the repetitive motion of an object around a fixed point or equilibrium position.

2. How is the period of oscillation affected by acceleration?

The period of oscillation is inversely proportional to the acceleration. This means that as acceleration increases, the period decreases and vice versa.

3. What is the equation for calculating the period of oscillation in an accelerating frame?

The equation is T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

4. How does the period of oscillation change as the length of the pendulum changes?

The period of oscillation is directly proportional to the square root of the length of the pendulum. This means that as the length increases, the period also increases and vice versa.

5. Does the mass of the bob affect the period of oscillation in an accelerating frame?

No, the mass of the bob does not affect the period of oscillation in an accelerating frame. The period is only dependent on the length of the pendulum and the acceleration due to gravity.

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