`` NEWTON: Simple Harmonic Oscillatory

 Simple Harmonic Oscillatory ``` Name: Aaron Status: student Grade: 9-12 Location: NC Country: USA Date: Summer 2013 ``` Question: How do I tell how many cycles a simple harmonic oscillator can go through in total? What I mean is: imagine I have an oscillator that moves based on the function x(t)= cos (2t). I know that the period (which is pi in this case) is the time it would take for one cycle and I know that the frequency (which is 1/pi) is the amount of cycles per second, but I do not know the maximum number of cycles this oscillator can take or the maximum amount of time that this oscillator can move. The domain of this function is all real numbers after all. This means that I can just keep plugging in higher and higher numbers for t and watch the oscillator keep moving in between x(t)=1 and x(t)= -1. The derivative of this function (a.k.a the velocity of the system in this case) is -2 sin (2t). There are no values of t where both the original function and its derivative equal 0 at the same time. This means that mathematically, this system will never come to rest at its natural state! However, we all know that the system has to come to rest at some point. This is why I'm asking how I would know when the oscillations would stop. My only guess is that it has something to do with potential energy (1/2 k x^2). Replies: Hi Aaron, Thanks for the question. The solution for simple harmonic motion, x(t) = cost(2t) will oscillate forever. This model does not have any energy loss in the sense that mechanical energy is removed from the system. You can prove (and I assign this as a homework question to my students) that the total mechanical energy in a system undergoing simple harmonic motion is constant, meaning that it will oscillate forever. If you want to include some damping, say due to air drag, you will obtain a solution of the form x(t) = A*exp(-bt)cos(wt) which is a decaying exponential and a cosine wave. This situation is called damped simple harmonic motion, and the oscillations will eventually stop (for all practical purposes). I hope this helps. Thanks JeffGrell A true simple harmonic oscillator experiences ONLY a Hooke's law force. There is no dissipative force such as friction. This is a theoretical fiction; no such thing exists in reality. Instead, "non-conservative" forces reduce the total energy of the system, eventually stopping it. But these forces are not included in the simple system modeled by the sinusoidal function. The simplest way to include a dissipative force into the system is to add a retarding force that is negatively proportional to the velocity of the oscillator, F = -kx - bv. The -kx term is the Hooke's law spring; the -bv is a force opposing the velocity. (Viscous drag acts like that in certain conditions). When such a term is included, if the drag coefficient b is not too large, the amplitude of the oscillation gradually decreases with time. The position equation then becomes a sinusoid multiplied by a decaying exponential. And you are right, the energy plays a part in this decay. The -kx is a conservative force; it converts kinetic energy into potential energy and vice versa. But the viscous drag force -bv only steals energy from the system; what goes away never returns. Richard E. Barrans Jr., Ph.D., M.Ed. Aaron, No, the answer has to do with loss of energy. You have described a “perfect” oscillator with no loss. Consider a bouncing ball. A perfect ball in a vacuum would bounce forever. Every bounce transfers the kinetic energy of the ball hitting the ground to exactly the energy it takes to compress the ball. The ball would bounce back in exactly the same speed but opposite direction so it would rise until all the kinetic energy goes into potential energy. The process would repeat forever, as you have noticed. We all know that balls do not compress perfectly. If we stretch a rubber band, it will become slightly warmer. This effect is just some energy going into heat. We don not have perfect systems. Even if our ball were perfect, we would have air drag on the ball which would eventually steal all the energy of the ball’s motion. What your equations are missing is a dampening factor, usually expressed as an exponential function (exp(-kt)) multiplying the amplitude of the oscillations. The dampening constant k is determined by such forces as friction or other loss. t is time, and as it grows, the exponential gets smaller and the ball stops bouncing. If you go back to the original equations from which you derived the oscillations and put in the energy loss, you will see this term pop out. It is a more difficult problem, so often we start with the idealized harmonic oscillator. Kyle J. Bunch, PhD, PE The solution x(t)=cos(2t) is for an oscillator that is not damped and will continue to oscillate forever. You want the solution for a damped oscillator, which you can find from Wikipedia, for example. Basically, the answer might look something like this: x(t) = exp(-at)cos(bt) where a and b depend on the restoring force and the damping force. Tim Mooney Click here to return to the Physics Archives

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