Why does the period of a pendulum decrease in an accelerating frame? [duplicate]

Question

If there is a simple pendulum in a non-accelerating frame with period $T_1$, it will have period $T_2 < T_1$ when placed in a frame accelerating perpendicularly to the direction of gravity. Why?

Answer 1

A body in an accelerated reference frame (say, a train with acceleration $\mathbf{ A}$) will appear subjected to an inertial force, i.e., it's necessary to add $\mathbf{F}_i=-m\cdot\mathbf{A}$ to the 'real' forces acting on the body for Newton's second law to hold in this reference frame.

You can obtain this result from a change of reference frame, which, oversimplifying a bit, can be seen as a change of coordinates: $\mathbf{x}'=\mathbf{x}-\mathbf{X}$, where $\mathbf{x}'$ is the position of the body measured with respect to the train, $\mathbf{X}$ is the train position with respect to an inertial frame of reference, say, a station, and $\mathbf{x}$ is the position of the body measured from this station. Deriving that twice gives you $\mathbf{a}'=\mathbf{a}-\mathbf{A}$.

Thus, in the train's reference frame, this force is composed with gravity, so the resultant force is $\mathbf{F}=\mathbf{F}_g+\mathbf{F}_i=m\cdot(\mathbf{g}-\mathbf{A})\equiv m\cdot \mathbf{g}_{\mathrm{eff}}$.

The period of oscillation in the non-accelerating frame is given by $T_1=2\pi\sqrt{l/g}$ and, in the accelerating (non-inertial) one, by $T_2=2\pi\sqrt{l/g_{\mathrm{eff}}}$. Here, given that $\mathbf{g}\perp \mathbf{A}$, we have $|\mathbf{g}_\mathrm{eff}|>|\mathbf{g}|$, and that's why $T_2<T_1$.

Answer 2

Look at the included figure:

This shows a simple pendulum consisting of a bob of mass $m$ and length $l$ acted upon by two forces, namely $mg$ (force due to gravity, acting downwards) and the force $ma$ (due to the acceleration of the train) that is perpendicular to the force of gravity.The angular acceleration of the bob is $\frac{d^2 \theta}{dt^2}$ and can be determined by resolving forces acting on the bob. These are $mg \cos \alpha = mg \sin \theta$ and $ma \cos \theta$ and yield the equation $$m l \frac{d^2 \theta}{d t^2} = - mg \sin \theta + ma \cos \theta.$$ (The reason for the signs being the force due to gravity tends to make $\theta$ smaller, and the $mg$ term makes $\theta$ larger.)

As pointed out by stafusa, there is a new equilibrium position, at $\theta = \theta_0$ determined from $-g \sin \theta_0 + a \cos \theta_0 = 0$, so putting $\theta = \theta_0 + \Delta \theta$ in terms of $\Delta \theta$ have $$\frac{d^2 \Delta \theta}{dt^2} = -\sin \Delta \theta ( g \cos \theta_0 + a \sin \theta_0)$$

For simple harmonic motion assume $\Delta \theta$ is small which yields $$\frac{d^2 \Delta \theta}{d t^2} = - g' \Delta \theta, \,\,\, g' = g \cos \theta_0 + a \sin \theta_0$$ and $g$ changes to effectively $\sqrt{g^2 + a^2}$.

Answer 3

A comment on Jim's answer (I don't have enough reputation to properly 'comment' yet).

I think you make a mistake [edit: his answer's been corrected since] when you expand for small angles measured with respect to the vertical, because that is not any more the equilibrium position.

In the new, inclined equilibrium position, the effective $g$ is higher, but the expansion can be done as usual iff $\theta=0$ coincides with it.