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During simple pendulum motion at different points it has specific energy.That is how we can precisely predict its velocity as well as its definite position. It seems to violate Heisenberg's uncertainty principle.What contradicts it?

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  • $\begingroup$ Related: physics.stackexchange.com/q/175985/37364. This question is about how quantum mechanics affects an inverted pendulum. $\endgroup$ – mmesser314 Aug 21 '15 at 12:29
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    $\begingroup$ In addition to the nice answers in the question pointed out by @mmesser314, I would point out that your question, basically, boils down to why classical mechanics is not the same as quantum mechanics. But it is, and the extent to which classical mechanics is 'exact' can be gotten at through the techniques in the referenced question/answer. $\endgroup$ – Jon Custer Aug 21 '15 at 13:29
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    $\begingroup$ The Heisenberg uncertainty principle depends on h_bar , a very small number. 6.6x10^-34 Joule seconds. The classical positions and momenta fulfill the inequality without any problem. It is at the quantum level of nano meters and smaller that the effects appear. $\endgroup$ – anna v Aug 21 '15 at 13:44
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    $\begingroup$ Not sure why this is attracting downvotes. It is a common misconception. $\endgroup$ – Floris Aug 21 '15 at 17:08
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You are performing a macroscopic analysis, i.e. working at a scale far above that of the quantum level. Therefore, trying to use quantum theory is not practical, since the scale is ridiculously big in comparison. Its analogous to counting the intermolecular interactions in something as big as a planet.

Instead, we use classical mechanics, like Newton's Laws of Motion. This regards nothing of the mechanics at the quantum level, so it is not exact. However, classical mechanics is certainly accurate enough at the scale of a pendulum, the accuracy depending on the complexity of the model (e.g. including effects due to aerodynamics, internal forces and deformation in the string, etc.).

Basically, the uncertainty principle is not physically violated, because we have not considered its effects in our pendulum model. It would be impractical for the purposes of our analysis. The uncertainties would be highly insignificant, so having a model that says "this velocity at this position" for classical mechanics is no biggie.

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To determine if the pendulum has a specific momentum at each position, verifying your model, you need to do measures.

Let's say that using some kind of laser system you can measure the position of your pendulum with a precision of the size of an atom $\approx1\times10^{-10}~m$. In this case the Heisemberg principle will not be very useful: you still can reach a precision of $1\times10^{-24}~kg~m/s$ in its momentum from which you can extract a very, very good model of its motion.

What if you could measure the position with a precision of $1\times10^{-30}~m$? Well let's think first how to do that. You will need to use photons (or other particles like in an electron microscope) whose wavelength is smaller than $1\times10^{-30}~m$. However such photons would carry a momentum $P=h/\lambda= 6\times 10^{-4}~kg~m/s$ this is approximately the momentum of a wasp flying at full speed! If you start to throw few of them at your pendulum you will alter its motion.

In this context The Heisemberg principle tells that you cannot test your equation of motion to this absurd precision with experimental measures.

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  • $\begingroup$ Maybe the last paragraph would more accurately state "you cannot know the state of the pendulum with sufficient accuracy to violate the uncertainty principle". Otherwise nicely done. $\endgroup$ – Floris Aug 21 '15 at 17:07

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