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Home , Damping, Harmonic oscillator

The oscillator with

Definition: Summary: ¥ body of m attached to with spring The equation of is constant k is released from position x0 (measured 2 from equilibrium position) with v0; d x( t ) + β dx(t ) + ω 2 = 2 2 0 x(t ) 0 ¥ resistance due to F = − bv , b = non- dt dt , res negative constant (possibly zero) where

x(t) b k ¥ β = and ω = . 2m 0 m There are three distinct kinds of motion:

Prerequisites: ¥ β<ω0 : underdamped

¥ fundamentals of Newtonian ¥ β=ω0 : critically damped

¥ the simple without resis- ¥ β>ω0 : overdamped tance Go to derivation.

Why study it? ¥ a very simple dynamical system with an exact solution in closed form; ¥ occurs frequently in everyday applications Go to Javaª applet In an earlier section, we studied the simple har- to allow the door to close in the least amount of monic oscillator. Our example was an ideal without slamming. Here is a movie illus- spring with no damping - no friction trating these three cases. between the body and the track. Now let’s see Now let’s see how well our intuition holds up. what happens when we relax this restriction, and Adding up the on the body and using include a resistive force of the form Newton’s second law, we find F = Ðbv , res ma (t ) = − kx(t ) − b v( t ) , where b is a positive constant and v is the where the first term on the right-hand side is the velocity. (We considered such a force in another restoring force due to the spring, and the second context in a previous section.) is the damping force due to friction. Rearranging, First, we try to picture what will happen. The we find the resistive force will act to slow the motion down, 2 d x( t ) + b dx(t ) + ω 2 = and the of the will 2 0 x(t ) 0 , continually decrease. We might expect this dt m dt decrease to be exponential, based on our earlier ω = / where 0 k m is the natural freqency of the experience with this form of resistive force. oscillator. We might also expect there to be several cases, It takes some inspiration to solve this equation. depending on the amount of damping. Imagine a The standard trick is to try a solution of the form screen door which is released from its open position. If the damper is worn out and doesn’t exp(αt) , provide enough damping, the door will slam shut. because this function just reproduces itself when (If the door frame were not there, the door would differentiated. The value of the constant α is oscillate back and forth a few before determined by plugging this form back into the stopping.) differential equation. We find If the damper is really stiff, the door might take α 2 + b α + ω 2 α = too long to shut, letting in all kinds of bugs. ( 0 ) exp( t) 0 , m Somewhere in between these two cases will be a case where the damping is just the right amount which can only be true for all t if the first factor is zero. Using the quadratic equation, we find ä β + ë å x 0 v 0 ì 2 1 / 2 = − β å ω + ω ì ä ë x(t ) exp á t é å x 0 cos 1 t sin 1 t ì α = b ± å b ω 2 ì å ω ì Ð å Ð 0 ì . ã 1 í . 2 m ã 4 m 2 í We will make the following shorthand notation: A second case occurs when b β ω (“critically damped”) β ≡ . 2) = 0 2 m In this case, there is only one value of α: There will be three cases, depending on the size α = Ðω . of β (the amount of damping). 0 The solution is the limit of the underdamped case 1) β<ω0 (“underdamped”) as ω1 goes to zero: This is the case of small damping. The argument of the square root in α is negative, so α = β + β + x(t ) exp á Ð t é á x 0 á x 0 v 0 é t é can have either of the two complex values α = β ± ω 2 β 2 Ð i 0 Ð , The linear dependence on t is characteristic of the case in which the two possible values of α are where i2=Ð1. (If you are weak on complex equal. numbers, now would be the time to review.) As a β ω shorthand, we will write 3) > 0 (“overdamped”) In this case, there are again two values of α, this ω = ω 2 Ð β 2 time both real: 1 0 . α = Ðβ ± β 2 Ð ω 2 . One of the basic results of is 0 that exp(iω1) = cos ω1 + i sin ω1. Hence, our As a shorthand, we will write solutions are linear combinations of the functions ω = β 2 ω 2 β ω β ω 2 Ð 0 exp á Ð t é cos 1 t and exp á Ð t é sin 1 t . . The solution is easily found to be The solution is size of β:

åä βx + v ìë x(t ) = exp á − βt é å x cosh ω t + 0 0 sinh ω t ì å 0 2 ω 2 ì ã 2 í

where cosh and sinh are hyperbolic functions. t The following plot shows typical curves for each

of the three cases, all with the same x0 and with β v0=0:

Damping: Here is a movie illustrating the three kinds of damping.

critical

over under 2. Damped, driven oscillator

under You may recall our earlier treatment of the driv-

en harmonic oscillator with no damping. We t found that if the driving is equal to the natural frequency of the system, then the amplitude becomes arbitrarily large as time goes on. We see that the motion is a decaying oscillation This is unphysical, however. In practice,there is in the underdamped case; the amplitude decays always some damping present. As the velocity according to the envelope exp(Ðβt). We see that becomes larger, this damping leads to greater and equilibrium is approached fastest for the greater loss from the system, as we critically damped case, hence its name. discussed earlier. As time goes on, the rate of energy lost due to damping balances the energy Here is a three-dimensional plot showing how the gained due to the external driving force, and a three cases go into one another depending on the steady-state oscillation is achieved. Let’s see how this is reflected in the 2 βω mathematics. Newton’s law now reads tan δ = ω 2 − ω 2 2 0 . d x( t ) + dx(t ) + = ω m 2 b k x(t ) F 0 cos ( t ) . dt dt ω 2 = ω 2 − β 2 (The quantity 1 0 was defined earlier.) The solution is quite messy, but otherwise These are very important equations, and it is straightforward. It can be arranged into the sum worth spending some time studying their of two pieces, one of which is proportional to properties (rather than deriving them). exp(Ðβt) and the other of which is not. The former goes away as time becomes large, and is The amplitude of the steady-state solution is therefore called a transient. We will ignore this F A( ω ) = 0 . piece. ω2 − ω 2 2 + β 2 ω 2 m ( R ) 4 1 The other piece, the steady-state solution, is the β one we are interested in. It is easy to show by For the , let’s regard the damping as ω direct substitution into the differential equation fixed and the driving frequency as variable. ω that it is given by The denominator of A( ) is smallest when the driving frequency is equal to the resonant ω F frequency. So A( ) itself is a maximum for that x (t ) = 0 cos(ω t − δ) value of ω. This is called . (In the last s ω 2 − ω 2 2 + β 2 ω 2 m ( ) 4 1 , ω = ω R section, we found resonance at 0 when no damping is present. This agrees with our present where the resonant frequency is given by formula for the resonant frequency, when β=0.) Looking at our formula for the resonant ω = ω 2 − 2β 2 R 0 frequency, we see that the resonance effect only ω > β occurs when 0 2 , and we restrict ourselves and the satisfies to this case. If we plot the amplitude of the steady-state solution versus ω, we get a curve with a peak at the resonant frequency. The next diagram shows Now let’s plot δ, the phase by which the steady- several such curves, each for a different value of state solution lags behind the driving force. the damping constant β. δ

π A(ω) 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

decreasing π/2 damping β

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ω ω ω ω 0 0 The height and width of the peak are controlled (The curves have the same color as their partners mainly by the value of β. The height of the peak in the previous plot.) is We see that there is always a delay between the F 0 action of the driving force and the response of the ω β , system. When the system is being forced at its 2m 1 natural frequency ω0, the phase lag is π/2. You and its width is proportional to might be familiar with this if you have ever played on a swingset. You can increase your 2ω β . 1 amplitude the fastest by leaning back right at the The smaller the damping β, the higher and bottom of your swing, one-quarter cycle behind narrower the peak gets, and the closer the your maximum amplitude. resonant frequency gets to the natural frequency ω If the driving frequency is much smaller than the 0. natural frequency, the driving force and the response are in phase. This makes sense, because the system can adjust itself to match the slow driving force. On the other hand, if the driving frequency is much higher than the natural frequency, the driving force and the response are out of phase (phase lag π).