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I just calculated the amplitude of a pendulum for a constant applied force (without friction)

$\left( \frac{\partial^2}{\partial t^2} + \omega^2 \right) x(t) = f(t)$

where $f(t) = F \theta(t)$

The solution yields

$x(t) = \frac{F}{\omega^2} ( 1 - \cos(\omega t) ) $ for $t>0$

I tried to find an experiment that shows this because I find it quite interesting that although a constant force is applied there is still a periodic amplitude which was counter intuitive for me at the first glance but made sense when I thought about the problem as a pendulum with an initial displacement that will freely swing around the new equilibrium point $x_0$. Does somebody know about such experiments? I tried with a pendulum in a capacitor with constant electric field between the walls but all experiments just deal with that case where the ball actually hits the capacitor walls and picks up charge to then get attracted by the opposing wall etc. and those experiments where a pendulum is inside a train with constant acceleration is always affected by friction so the amplitude will get damped too quickly to see the effect.

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  • $\begingroup$ There is constant gravity applied and it results in harmonic motion. Why is the application of an external force any different? It just changes the direction and magnitude of the weight applied. PS. In your question the force isnt constant as it changes magnitude with time. $\endgroup$ Feb 4, 2020 at 17:59
  • $\begingroup$ You can imagine small oscillations of a pendulum where you blow on the pendulum (say from above) at constant flow. The force is approximately constant. $\endgroup$
    – lcv
    Jun 4, 2020 at 3:03
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    $\begingroup$ You need to define $\theta (t)$. And what you propose isn't constant. A constant force requires $\frac{df}{dt}=0.$ Your example in the comment doesn't apply because the blowing is not in the $x$ direction. The differential equation has $f$ in the $x$ direction. $\endgroup$
    – Bill N
    Jun 1, 2021 at 22:17
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    $\begingroup$ Your force is manifestly not constant. Also, your solution suspiciously does not depend on the initial conditions. $\endgroup$
    – Roger V.
    Jun 14, 2023 at 14:13

2 Answers 2

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Where you write $f(t) = F \theta(t)$ this could come from $g' \sin \theta \approx g' \,\theta$ which is due to some extra resorting force for a simple pendulum undergoing small angle oscillations.
In effect you are changing the local value of the acceleration of free fall by introducing a constant vertical force on the bob in addition to the gravitational attraction.

That extra force could be achieved by you doing your experiment in a lift which is accelerating upwards at $g'$ or allowing the bob with charge $q$ oscillate in a constant electric field $E$ produced by a pair of charged parallel plates with $g' = \frac {qE}{m}$ where $m$ is the mass of the bob.

If that constant force is not vertical you then need to find the effective force on the bob by adding the gravitational field strength $g$ to $g'$ vectorially.
For example if the magnitude of $g'$ was the same as the magnitude of $g$ but it acted in a horizontal direction (charged parallel plates vertical, constant acceleration in a car along a horizontal road etc.) the $g$ in your simple pendulum experiment would be $\sqrt 2 g$ and the rest position of the bob would have the string inclined at $45^\circ$ to the horizontal.

A possible set up might be on a large turntable (roundabout/merry-go-round) rotating at constant speed?

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Here we can offer such an experiment: the ball on the spring moves horizontally, then we turn the entire system and set it upright. In this case, gravity will work as a constant force that changes the equilibrium position.

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