A question came once into my mind
There is a pendulum having length of string 1metre and was initially at rest. Now the point of suspension suddenly starts to move in uniform horizontal circular motion of radius 10 cm and angular velocity 1 second inverse. The bob of the pendulum will initially show some erratic motion but after some time it itself will also begin to move in a uniform horizontal circular motion with same angular velocity. Your task is to find the radius of this circular motion of bob.
The question appeared simple at first sight.
This is how I thought the motion will be. It looks like a part of a conical pendulum.It is similar to that seen while riding in
merry-go-round.(centrifugal causes the horse to move slightly away from the axis). Note that at any instant the bob and point of suspension has same angular position. The surface of revolution of the string is that of a frustum. Now I can easily apply Newton's laws of motion and solve it. I did got an answer but I don't remember it and it's irrelevant, if you are convinced that this is a possible solution.
I asked this question to my friend and he solved and gave a different answer. I thought he did mistake but I was surprised to see his approach. This is how he thought the motion should occur
Here the bob has a slightly smaller radius. The surface of revolution of string is in the shape of 2 inverted cones joined at vertex. At any instant the angular position of bob is opposite to the point of suspension. Here also we got a second (also valid) answer to the question.
You can try to perform this experiment practically by attaching some heavy weight to the end of a rope and then revolving it by hand. Someone can make a good multiple option correct question from this giving both the answers in the option but most of the people will only be able to think of 1 answer.
Anyways,my query is regarding laplace determinism which states that if all the conditions and parameters of a system are known at an instant then the values of these parameters can be accurately predicted at any other instant. There seems to be a violation of that here because initial conditions were known and we are getting 2 possible answers. How is this possible?
Best Answer
Petr's answer is making some good points, but there is actually a much simpler, philosophy-less answer that largely resolves the issue (with a caveat that I'll get into later). Essentially, what happens is this: if the exciting motion is slower than the (linearized) natural frequency of the pendulum, then it will settle into the merry-go-round configuration. If it's faster, it'll go into the double-cone configuration.
(That's assuming there is some kind of friction in the system, which there always is. If there was no friction, then it would never settle into either of these configurations, but instead keep forever wobbling irregularly!)
A simple qualitative reason why it behaves like this is an energy argument: at slow frequency, the merry-go-round has the pendulum hanging almost straight down, i.e. the mass is low down (low potential energy), and the kinetic energy is anyways low at slow speed. Whereas at high frequency, the merry-go-round has the bob moving on a longer path, thus with faster speed and higher kinetic energy.
Meanwhile, the double-cone always requires the pendulum to cross over the center, which requires some potential energy even at very low frequency. But at high frequency it can make the bob move at an arbitrarily small radius around the center, and thus minimise kinetic energy.
The system will tend towards the configuration with lower total energy, because any air friction etc. losses will consistently remove energy. Although the exciting circular motion can both add and remove energy, it has no inherent preference, and the state where it only replentishes energy lost to friction around the lowest-energy steady state is most stable.
All of this rather hand-wavey argument can be easily made more rigorous in the special case where the pendulum is much longer than the scale of the initial motion, because then the pendulum behaves as a simple harmonic oscillator, and the solution is that of the classic driven oscillator, known by different names in different applications. The electronic analogue is the LC filter, which has a well-known behaviour of flipping the phase as you cross the resonance frequency. This transition is exactly analogue to the pendulum ending up in the two different states, depending on the frequency with which you excite it.
Now, the caveat: there is an important difference with the pendulum, namely that it's (for non-infinitesimal displacements) not linear, and the nonlinearity can actually stabilise the merry-go-round even at high frequency. The extreme case is where it's centrifugeing the bob outwards at almost horizontal angle. This will happen in particular if you start at low frequency and make it ever faster, provided friction is sufficiently small (as it is with a heavy bob in air, but not with e.g. a feather or in water). I believe the nonlinearity can also stabilize the double-cone at frequencies below resonance, but not sure about this. These effects are what the other answers mean when they say it depends on the initial condition. Indeed, in the edge case at almost exactly the resonance frequency I would expect that tiny changes in the initial state can cause the pendulum to end up in the less energetically favourable configuration. But still, the movement is deterministic, and with reasonably accurate start conditions you can reliably achieve long-term predictable movement, unlike with the textbook chaotic systems like the Lorenz attractor.