# Period of simple pendulum accelerated horizontally

I'm confused about simple pendulum problems where the pendulum is accelerated horizontally of anyway not vertically with acceleration $\vec{A}$.

$m\vec{g} + \vec{T}-m \vec{A} =m \vec{a}$

So

$\begin{cases} m l \ddot{\theta} = - mg sin(\theta)+m A cos(\theta) \\ m \dot{\theta} ^2 l = T - mg cos(\theta)-m A sin(\theta) \end{cases}$

From the first equation, on the tangential coordinate,

$l \ddot{\theta} = - g sin(\theta)+ A cos(\theta)$

Which is for small angles

$l \ddot{\theta} = - g \theta+ A$

And therefore the period of small oscillations should still be $\tau=\sqrt{\frac{l}{g}} 2\pi$

While of course it is different, but I don't see the mistake in what I wrote here.

• You can't assume your $\theta$ to be small anymore, since the pendulum oscillates about a new equilibrium. – JoDraX Mar 9 '16 at 18:53

Since your pendulum oscillates about a new angle, call it $\theta_0$, your Taylor series approximation should be about $\theta_0$. So, \begin{align*} l \ddot{\theta} & = -g \sin{\theta} + A \cos{\theta} \\ & \approx -g ( \sin{\theta_0} + (\theta - \theta_0) \cos{\theta_0} ) + A ( \cos{\theta_0} - (\theta - \theta_0) \sin{\theta_0}) \text{ for small deviations from } \theta_0 \\ & = A \cos{\theta_0} - g \sin{\theta_0} + \theta_0 (g \cos{\theta_0} + A \sin{\theta_0}) - \theta (g \cos{\theta_0} + A \sin{\theta_0}) \end{align*} Solving that should give you the period you're looking for.
Personal choice, I would change the frame such that $\vec{g}' = \vec{g} - \vec{A}$. Doing this means gravity is larger and the frame is a little rotated, so choose a frame that lines up with gravity and say it's all the same. Then choose $\theta=0$ to be parallel to $\vec{g'}$, you get the same equations as usual, where $g' = \big|\vec{g}'\big|$.
$$\ddot{\theta}+\frac{g'}{l}sin\theta = 0$$
due to $$\vec{g}=-g\cos(\theta )\hat{\rho} - g\sin(\theta )\hat{\theta}$$ and $$\vec{A} = A\hat{x} = A\cos(\theta )\hat{\rho} - A\sin(\theta )\hat{\theta}$$