# Infinite acceleration of bob in pendulum with no friction or air resistance

So i'm a bit confused about something. If we take a mathematical pendulum and we apply a force to it. We ignore all friction and air resistance and only consider gravity and the force applied to bob. My question is and what i've figured out so far: If the force applied to the bob is less than the acceleration due to gravity, the bob will have a forever constant period of swings. However, if the force applied to the bob is greater than the acceleration due to gravity - does that mean that the bob will accelerate infinitely? Or does it only mean that the bob will have an infinite swing in one direction? I can't figure out if the acceleration will decay due to gravity when the initial force applied to the bob is greater than the acceleration due to gravity.

Anyone who care to help? :)

• It can move in a vertical circle. Commented Apr 14, 2016 at 17:31
• It sounds as if you are thinking about forced harmonic oscillators aka driven harmonic oscillators, where a sinusoidally varying force is applied to the oscillator. If there is no damping then energy is continually being supplied and the amplitude of the oscillations increases without limit. In practice of course there is always damping. Commented Apr 14, 2016 at 17:31
• A force can be applied in many ways. To me at least it is unclear what your force is (which direction) and how you apply it (always the same? Periodically? Commented Apr 14, 2016 at 17:48
• @Martin as far as i understood it's when you apply the force at an initial angular displacement. So that you apply some force to the bob at the angle -Pi/2, e.g. give it a push with a force which is bigger than the gravitational pull. Will it then keep accelerating? Commented Apr 14, 2016 at 17:58
• Your description of the scenario is unclear in several ways. Maybe you could include a drawing and some formulas. Since you have come to the conclusion that the bob will accelerate infinitely, I assume that you have used a formula of some kind.
– jkej
Commented Apr 14, 2016 at 18:41

Whenever a pendulum moves, it accelerates since the velocity vector is constantly changing. So any pendulum that 'pendulates' forever will 'accelerate' forever. However, I think the acceleration you are talking about is better described as angular acceleration. Perhaps a better way to describe it is if it will complete a full rotation with increasing frequency. The answer to that question is no. The pendulum will have a constant oscillation frequency because the gravity that accelerates it on the way down is exactly equal to the gravity that decelerates it on the way up.

An pendulum released from rest will swing from side to side, going exactly as high as its initial position, repeating forever.

A pendulum given an initial "kick" in the form of some impulse has enough energy to rise higher than its initial position - if this height is higher than the pendulum string can reach, the pendulum will swing in a complete circle all the way around repeatedly, rather than oscillating side-to-side. If there is no force imparted after time 0, the overall pattern of the swing will not change from swing to swing.

A pendulum that has some force applied (like if the bob had a rocket motor attached) could remain stationary if the force balances gravity, or do an odd lopsided swing if it's weaker, or do vertical loops at ever-faster speeds. Different things may happen if the force has constant direction and magnitude, or if the force changes depending on the position of the bob. But in general, if there is a force applied over time, the period of the oscillations from swing to swig may change.

The freely moving mathematical pendulum has three different modes of motion. With potential energy gauged to zero in the upper turning point of zero force, and minimal value of $$-2 r m g$$ in the lower turning position.

The modes are

1: oscillaton, negative total energy if the kinetic energy $$1/2 m v^2 = 1/2 m r^2 \dot \phi ^2$$ in the potential minimum at $$\phi=0$$ is less than the the potential energy difference to the top.

2: rotation: positive total energy, kinetic energy in the lower extreme of the potential is larger than than the difference to the top position, the top position is passed with positive kinetic energy, the motion is a wobbling rotation $$\phi(t)= 2 \text(am)(\frac{\omega}{2} t, \frac{4}{\omega^2}), \omega > 2$$ described by the elliptic Jacobi amplitude $$\text{am}$$.

1. The single loop, top to top, in an infinite time. This on has an easy special solution $$\phi(t)= 4 \ \text{atan}(e^{\pm \omega\ t})$$ solving the equation of motion $$\ddot \phi =-\omega ^2 \ \sin(\phi)$$

An experiment is always performed by selecting a start position angle and preparing the system in a state of absolute rest.

Then you transfer an exactly defined amount of momentum $$p$$ to the mass, resulting in a momentarily change from $$v=0$$ to $$v=p/m$$, as any golfer knows.

Applied force is difficult to control, but the momentum transfer is always the time integral of force over the time intervall of impact.

But thatis theory only.

Newtons decription of force by time derivative of momentum transfer is a concept, that did not survive the times of Euler, when it was replaced by the far more fruitful concept of $$m \frac{d^2x}{dt^2}=\frac{dp}{dt} =- \frac{d \text{Energy}}{dx}$$.

The mathematical clarification came at the cost of about 200 years of discussion in theoretical physics, what is "energy" and what is the difference between "living force", "momentum", and "impulse".