Frequency of an Accelerating Pendulum

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Problem. A simple pendulum oscillates with frequency [itex]f[/itex]. What is the frequency if it accelerates at [itex]1/2 g[/itex] (a) upward, and (b) downward?

Let [itex]m[/itex] be the mass of the bob at the end of the cord of length [itex]l[/itex] that is making an angle [itex]\theta[/itex] with the line through the point of equilibrium.

Given (a), the total net force on the bob in the vertical direction is [itex]\frac{1}{2} mg[/itex] and the component of this net force tanget to the motion of the bob is [itex]\frac{1}{2} mg \sin \theta[/itex] which is roughly [itex]\frac{1}{2} mg \theta[/itex] for small [itex]\theta[/itex]. As far is I can tell, this component is the net force tangent to the motion of the bob and since it isn't akin to Hooke's law, the pendulum is not undergoing simple harmonic motion so it has no frequency. Somehow, I feel that my analysis is wrong here. Did I overlook something?
 

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  • #2
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There are equations we use for simple pendulums basically undergoing SHM for small values of theta. Here the pendulum can be said to obey Hookes law because the restoring force is proportional to theta
 
  • #3
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There are equations we use for simple pendulums basically undergoing SHM for small values of theta. Here the pendulum can be said to obey Hookes law because the restoring force is proportional to theta
Yes, but apart from that, the restoring force must be negative which in this case, is not. For (b), however, it is.
 
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I was thinking: for any point in time in which the bob is accelerating upward, there exists a point in time where it is accelerating at the same rate downward. This means the pendulum bob in (a) is the same one as in (b), just at two different points in time. Hence, in both cases, the pendulum has the same frequency right?
 
  • #5
Doc Al
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Viewed from the noninertial frame of the elevator, the pendulum bob has an additional force on it equal to [itex]\frac{1}{2} mg[/itex] downward. This adds to the restoring force, making it [itex]\frac{3}{2} mg \sin\theta[/itex] instead of [itex]mg \sin\theta[/itex].

Another way to view it (equivalent to the above) is that the effective acceleration due to gravity in the elevator is 3g/2.
 
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Viewed from the noninertial frame of the elevator, the pendulum bob has an additional force on it equal to [itex]\frac{1}{2} mg[/itex] downward. This adds to the restoring force, making it [itex]\frac{3}{2} mg \sin\theta[/itex] instead of [itex]mg \sin\theta[/itex].

Another way to view it (equivalent to the above) is that the effective acceleration due to gravity in the elevator is 3g/2.
I see. So the problem states that the whole pendulum is moving at 1/2 g upward or downward. I was under the impression that only the pendulum bob is moving at 1/2 g upward or downward. I should have know better.
 

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