d mx kx dt ( t) Note first that imposed variation in the mass term is easily dealt with, by simply redefining the time dt mt ( ).
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1 . Parametric Resonance Michael Fowler Introduction (Following Landau para 7 A one-dimensional simple harmonic oscillator, a mass on a spring, d mx kx ( = has two parameters, m and k. For some systems, the parameters can be changed externally (an example being the length of a pendulum if at the top end the string goes over a pulley. We are interested here in the system s response to some externally imposed periodic variation of its parameters, and in particular we ll be looking at resonant response, meaning large response to a small imposed variation. Note first that imposed variation in the mass term is easily dealt with, by simply redefining the time variable to dτ = / m( t, meaning τ = Then mt. d dx 1 d 1 dx 1 d x m = m = m dτ m dτ m dτ, d x and the equation of motion becomes m( τ kx. dτ = This means we can always transform the equation so all the parametric variation is in the spring constant, so we ll just analyze the equation d x ω ( t x=. Furthermore, since we re looking for resonance phenomena, we will only consider a small parametric variation at a single frequency, that is, we ll take ω ( t = ω ( h Ω t 1 cos, where h 1, and h is positive (a trivial requirement just setting the time origin. (Note: We preferω where Landau uses γ, which is often used for a resonance wih these days.
2 We have now a driven oscillator: d x ω x = ωxh cos Ω t. How does this differ from our previous analysis of a driven oscillator? In a very important way! The amplitude x is a factor in the driving force. For one thing, this means that if the oscillator is initially at rest, it stays that way, in contrast to an ordinary externally driven oscillator. But if the amplitude increases, so does the driving force. This can lead to an exponential increase in amplitude, unlike the linear increase we found earlier with an external driver. (Of course, in a real system, friction and nonlinear potential terms will limit the growth. What frequencies will prove important in driving the oscillator to large amplitude? It responds best, of course, to its natural frequency ω. But if it is in fact already oscillating at that frequency, then the driving force, including the factor of x, is proportional to cosω tcos Ω t = cos Ω ω t cos Ω ω t, with no component at the natural frequencyω for a general Ω. The simplest way to get resonance is to take Ω= ω. Can we understand this physically? Yes. Imagine a mass oscillating backwards and forwards on a spring, and the spring force increases just after those points where the mass is furthest away from equilibrium, so it gets an extra tug inwards twice a cycle. This will feed in energy. (You can drive a swing this way. In contrast, if you drive at the natural frequency, giving little push inwards just after it begins to swing inwards from one side, then you ll be giving it a little push outwards just after it begins to swing back from the other side. Of course, if you push only from one side, like swinging a swing, this works but it isn t a single frequency force, the next harmonic is doing most of the work. Resonance near Double the Natural Frequency From the above argument, the place to look for resonance is close to Ω= ω. Landau takes x ω 1 hcos ω ε t x= and, bearing in mind that we re looking for oscillations close to the natural frequency, puts in with at, bt slowly varying. ( ω ε ( ω ε x= atcos t btsin t,
3 3 It s important to realize that this is an approximate approach. It neglects nonresonant frequencies which must be present in small amounts, for example and the 3( ω 1 ε 1 ( ω ε ( ω ε = ( ω ε ( ω ε cos t cos t cos 3 t cos t term is thrown away. And, since the assumption is that at, too, leaving just This must equal bt are slowly varying, their second derivatives are dropped ω ( 1 ( 1 sin ω ε ω ωε cos ω ε ω ( ω ( 1 ε ω ω ε ( ω 1 ε x = a t t a t t bt cos t bt sin t. ω 1 hcos ω ε t atcos ω ε t btsin ω ε t. 1 Keeping only the resonant terms, we take cos( ω ε cos( ω ε cos( ω ε 1 sin ( ω ε t cos( ω ε t = sin ( ω ε t, so this expression becomes t t = t and ( ( cos sin cos sin ω 1 hcos ω ε t atcos ω ε t btsinω ε t = ω a t ω ε t b t ω ε t ha t ω ε t hb t ω ε t The equation becomes: ω ( ω ( 1 ε ω ω ε ( ω 1 ε x = a t sin t a t cos t bt cos t bt sin t ω ( ω ( 1 ε ω ω ε ( ω 1 ε cos sin cos sin == ω a t ω ε t b t ω ε t ha t ω ε t hb t ω ε t The zeroth-order terms cancel between the two sides, leaving at ω sin ω ε t atω εcos ω ε t bt ω cos ω ε t btω εsin ω ε t ( ( ( ( ha ( t cos t hb( t sin t = ω ω ε ω ε Collecting the terms in sin ( ω ε t, cos( ω ε t : ( a bε hω bω sin ω ε t bt aε hω aω cos ω ε t=.
4 4 The sine and cosine can t cancel each other, so the two coefficients must both be identically zero. This gives two first order differential equations for the functions at, increasing functions, proportional to st st a t ae, b( t be The amplitude growth rate is therefore bt, and we look for exponentially = =, which will be solutions provided ( ε ω ( ε ω sa h b =, h a sb =. s hω ε = 4. Parametric resonance will take place if s is real, that is, if < < hω ε hω, a band of wih hω about ω. Example: Pendulum Driven at near Double the Natural Frequency A simple pendulum of length, mass m is attached to a point which oscillates vertically Measuring y downwards, the pendulum position is x= sin φ, y = acos Ω t cos φ. y = acos Ω t. The Lagrangian 1 ( cosφ ( L = m x y mg = m cos φ φ m aωsin Ω t φsinφ mg cosφ 1 d = m φ ma Ωsin Ω t cosφ a Ωsin Ω t mg cosφ The purely time-dependent term will not affect the equations of motion, so we drop it, and since the equations are not affected by adding a total derivative to the Lagrangian, we can integrate the second term by parts (meaning we re dropping a term d ( ma Ωsin Ω t cosφ to get L = m ma Ω Ω t mg 1 φ cos cosφ cos φ. (We ve also dropped the term mga cos Ω t from the potential energy term it has no φ orφ dependence, so will not affect the equations of motion.
5 5 The equation for small oscillations is Comparing this with we see that 4 a/ 1 3 hω = a g/. φ ω 1 4 a/ cos ω ε t φ =, ω = g /. x ω 1 hcos ω ε t x= h, so the parametric resonance range around ω = g / is of wih
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