0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @MBarbosa The original question was a pendulum in a train. In that case, the acceleration is (more or less) perpendicular to the gravity, resulting in a g-force > 1. Your edit made it an undefined accelerating frame, where the direction of the acceleration relative to the force of gravity is unknown. The statement is no longer true, since the period will become longer when the pendulum accelerates downwards. $\endgroup$ – Previous Jun 12 '16 at 13:18
  • $\begingroup$ @Previous thanks, good catch. I edited again. Nonetheless, my original edit made it way more comprehensible! $\endgroup$ – M Barbosa Jun 12 '16 at 13:26
  • $\begingroup$ Sorry guys, I mistakenly deleted your edit. Should be fixed now. $\endgroup$ – levitopher Jun 12 '16 at 13:39
  • $\begingroup$ What thoughts do you have about this? What have you done to try to resolve the problem for yourself? $\endgroup$ – sammy gerbil Jun 12 '16 at 15:53
  • $\begingroup$ @JohnRennie In that case, the acceleration is in the same direction as acceleration due to gravity, which is a bit different to the problem here. $\endgroup$ – jim Jun 12 '16 at 19:35
2
$\begingroup$

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 $\mathbf{g}\perp \mathbf{g}_\mathrm{eff}$, we have $|\mathbf{g}_\mathrm{eff}|>|\mathbf{g}|$, and that's why $T_2<T_1$.

$\endgroup$
1
$\begingroup$

Look at the included figure: enter image description here

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 $\sqrt{g^2 + a^2}$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Yes, thanks for that. I think I was led along by my expectation that the acceleration would affect the motion of the bob symmetrically. $\endgroup$ – jim Jun 12 '16 at 21:49
  • 1
    $\begingroup$ Happy to have helped! It's nice to have the equation solved explicitly as you did. +1 $\endgroup$ – stafusa Jun 12 '16 at 22:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.