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I took the case where the simple pendulum has a fixed length and the frame is accelerating perpendicular to gravity. I found the following differential equation:

$$\ddot{\theta} = - \frac{g}{l} \theta – \frac{a}{l}(1 – \theta^2)^{1/2} ;$$ assuming small vibration

how to proceed?

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2 Answers 2

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If you are sitting in the accelerating frame (acceleration $\vec a$ relative to the laboratory) and measured the acceleration of free fall $\vec g_{\rm accelerated}$ what value would you get?

The answer is that $\vec g_{\rm accelerated} = \vec g - \vec a$ where $\vec g$ is the acceleration of free fall in the laboratory frame.

You can think of it as a "fictitious' force $-m\;\vec a$ acting in the accelerated frame where $m$ is the mass of the bob.

If the pendulum was not oscillating then it would orientate itself along the direction of $\vec g_{\rm accelerated}$ and this would be the position about which the bob would oscillated with shm as though the gravitational field strength was $\vec g_{\rm accelerated}$.

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Rotate the coordinates so that y axis is along the equilibrium direction (acceleration + gravity). The solve it the same as regularly but the angle $\theta$ measured not from vertical but from the y axis. Also the new gravity is the combined gravity and base acceleration.

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