Small Oscillations around equilibrium

AI Thread Summary
The discussion focuses on deriving the equations of motion for a point pendulum under constant acceleration. The equation of motion is established as \ddot{\theta} + \frac{a\cos\theta + g\sin\theta}{L} = 0, with the equilibrium point identified as \theta_0 = -\arctan(a/g). A Taylor expansion around the equilibrium point leads to a simplified form for small oscillations. The correct frequency of these oscillations is given by \omega = L^{-1/2} (a^2 + g^2)^{-1/4}. Participants suggest expressing \theta_0 in terms of a and g for further simplification.
LiorE
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Homework Statement



The problem is: A point pendulum is being accelerated at a constant acceleration of a. Basically what's required is to find the equations of motion, the equilibrium point, and to show that the frequency of small oscillations about the e.p. is: \omega=L^{-1/2} (a^2+g^2)^{-1/4}

Homework Equations



The equation of motion I've arrived at is:

\ddot{\theta} + \frac{a\cos\theta + g\sin\theta}{L} = 0

So the e.p is:

\theta_0 = -\arctan(a/g)

The Attempt at a Solution



If we do a Taylor expansion for \eta << 1 around \theta_0:

\cos(\theta_0+\eta) = \cos\theta_0 - \eta \sin\theta_0+\ldots
\sin(\theta_0+\eta) = \sin\theta_0 + \eta \cos\theta_0+\ldots

We end up with:

\ddot{\eta} + \frac{\cos\theta_0}{L}\frac{g^2+a^2}{g} \eta = 0

Then what am I missing here?
 

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Correction:

<br /> \omega=L^{-1/2} (a^2+g^2)^{1/4}<br />
 
Anyone?
 
Seriously, can no one say anything?
 
Just saw your question.

LiorE said:
We end up with:

\ddot{\eta} + \frac{\cos\theta_0}{L}\frac{g^2+a^2}{g} \eta = 0

Then what am I missing here?

You're almost there.

Express θ0 in terms of a and g, using the arctan relation you came up with earlier. Things will simplify.
 
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