Under, Over and Critical Damping

7272019 Under Over and Critical Damping

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UnderOverandCriticalDamping1 ResponsetoDamping

As we saw the unforced damped harmonic oscillator has equation

mx+bx+kx=0 (1)

with mgt0 bge0 and k gt0 It has characteristic equation

ms2 +bs+k =0with characteristic roots

minusbplusmnradic b2 minus4mk (2)

2mThere are three cases depending on the sign of the expression under the

square root

i) b2lt4mk (this will be underdamping bis small relative to mand k )

ii) b2gt4mk (this will be overdamping bis large relative to mand k )

iii) b2 = 4mk (this will be criticaldamping b is just between over andunderdamping

Mathematically the easiest case is overdamping because the roots are

real However most people think of the oscillatory behavior of a dampedoscillator Since this is connected to underdamping we start with that case

Case(i)Underdamping(non-real complex roots)If b2

lt 4mk then the term under the square root is negative and thecharacteristic roots are not real In order for b2

lt4mk the damping constantbmust be relatively small

First we use the roots (2) to solve equation (1)Let ωd = b2 minus4mk 2m Then we have|

minus|b

characteristic roots2mplusmniωd leading to

complex exponential solutions e(minusb2m+iωd)t e(minusb2mminusiωd)tThe basic real solutions are eminusbt2mcos(ωdt) and eminusbt2msin(ωdt)The general real solution is found by taking linear combinations of the two basic solutions that is

x(t) =c1eminusbt2mcos(ωdt) +c2eminusbt2msin(ωdt)



Under Over and Critical Damping OCW 1803SC

or

1048573

x(t) =eminusbt2m(c1 cos(ωdt) +c2 sin(ωdt))= Aeminusbt2mcos(ωdtminusφ) (3)

Letrsquos analyze this physically When b = 0 the response is a sinusoidDamping is a frictional force so it generates heat and dissipates energyWhen the damping constant bis small we would expect the system to stilloscillate but with decreasing amplitude as its energy is converted to heatOver time it should come to rest at equilibrium This is exactly what wesee in (3) The factor cos(ωdtminusφ)shows the oscillation The exponentialfactor eminusbt2m has a negative exponent and therefore gives the decayingamplitude As t infin the exponential goes asymptotically to 0 so x(t)rarralso goes asympotically to its equilibrium position x

=0

We call ωd the dampedangular(orcircular)frequencyof the systemThis is sometimes called a pseudo-frequencyof x(t) We need to be carefulto call it a pseudo-frequency because x(t)is not periodic and only periodicfunctions have a frequency Nonetheless x(t)does oscillate crossing x=0twice each pseudo-period

Example1 Show that the system x+1x+3x=0 is underdamped find itsdamped angular frequency and graph the solution with initial conditions

x(0) =1 x(0) =0

Solution Characteristic equation s2 +s+3 =0Characteristic roots

minus12

plusmniradic 112

Basic real solutions eminust2 cos(radic 11 t2) eminust2 sin(radic 11 t2)General solution

x(t) =eminust2(c1 cos(radic

11 t2) +c2 sin(radic

11 t2))= Aeminust2 cos(radic

11 t2minusφ)

Since the roots have nonzero imaginary part the system is underdampedThe damped angular frequency is ωd =radic 112The initial conditions are satisfied when c1 =1 and c2 =1

radic 11 So

x(t) = eminust2 cos(radic

11 t2) +radic 111

sin(radic

11 t2)radic 12 eminust2= radic 11

cos(radic 11 t2minusφ)

where φ=tanminus1(1radic

11)

2




Figure 1 The damped oscillation for example 1

Case(ii)Overdamping(distinct real roots)If b2

gt 4mk then the term under the square root is positive and the characteristic roots are real and distinct In order for b2

gt 4mk the damping

constant bmust be relatively large

One extremely important thing to notice is that in this case therootsarebothnegative You can see this by looking at the formula (2) Theterm under the square root is positive by assumption so the roots are realSince b2 minus4mk lt b2 the square root is less than band therefore the rootminusb+radic b2 minus4mk lt0 The other root is clearly negative

Now we use the roots to solve equation (1) in this case

minusb+radic b2 minus4mk minusbminusradic b2 minus4mk Characteristic roots r1 = r2 =

2m 2mExponential solutions er1t er2tGeneral solution

x(t) =c1er1t+c2er2tLetrsquos analyze this physically When the damping is large the frictional

force is so great that the system canrsquot oscillate It might sound odd butan unforced overdamped harmonic oscillator does not oscillate Since bothexponents are negative every solution in this case goes asymptotically tothe equilibrium x=0

At the top of many doors is a spring to make them shut automaticallyThe spring is damped to control the rate at which the door closes If thedamper is strong enough so that the spring is overdamped then the door just settles back to the equilibrium position (ie the closed position) without oscillating ndashwhich is usually what is wanted in this case

Example2 Show that the system x+4x+3x = 0 is overdamped and

graph the solution with initial conditions x(0) =1 x(0) =0 Which rootcontrols how fast the solution returns to equilibrium

3




Solution Characteristic equation s2 +4s+3 =0

Characteristic roots (this factors) minus1minus3Exponential solutions eminust eminus3tGeneral solution

x(t) =c1eminust+c2eminus3tBecause the roots are real and different the system is overdampedThe intial conditions are satisfied when c1 = 32 c2 =minus12 So x(t) = 3eminust2minuseminus3t2

0 1 2 3 4 5

0

0

0

2

0

4

0

6

0

8

1

0

t

x

Figure 2 The overdamped graph for example 2

Because eminustgoes to 0 more slowlythan eminus3t2 it controls the rate at whichxgoes to 0 (Remember it is the term that goes to zero slowest term thatcontrols the rate)

Case(iii)

Critical

Damping

(repeated real roots)If b2 =4mk then the term under the square root is 0 and the characteristic

polynomial has repeated roots minusb2m minusb2m

Now we use the roots to solve equation (1) in this case We have onlyone exponential solution so we need to multiply it by tto get the secondsolutionBasic solutions eminusbt2m teminusbt2mGeneral solution

x t eminusbt2m( ) = (c1 +c2t)

As in the overdamped case this does not oscillate It is worth notingthat for a fixed mand k choosing bto be the critical damping value givesthe fastest return of the system to its equilibrium position In engineeringdesign this is often a desirable property This can be seen by consideringthe roots but we will not go through the algebra that shows this (See figure(4))

4




Example3 Show that the system x+4x+4x=0 is critically damped and

graph the solution with initial conditions x(0) =1 x(0) =0Solution Characteristic equation s2 +4s+4 =0Characteristic roots (this factors) minus2minus2Exponential solutions (only one) eminus2tGeneral solution

x(t 2t) =eminus (c1 +c2t)

Because the roots are repeated the system is critically dampedThe intial conditions are satisfied when c1 =1 c2 =2 Sox t eminus2t( ) = (1 +2t)

0 1 2 3 4 5

0 0

0

2

0

4

0 6

0

8

1 0

x

t

Figure 3 The critically damped graph for example 3

Notice that qualitatively the graphs for the overdamped and criticallydamped cases are similar

The following figure shows plots for solutions to x+bx+x = 0 with

initial conditions x(0) = 1 x(0) = 0 The three plots are b = 1 under-damped b=2 critically damped (dashed line) b=3 overdamped Noticethat the critically damped curve has the fastest decay

0 2 4 6 8

- 0

2

0

2

0

6

1

0

t

x 1

Figure 4 Plots of solutions to x+bx+x=0

5



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1803SC Differential Equations

Fall 2011

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or

1048573

x(t) =eminusbt2m(c1 cos(ωdt) +c2 sin(ωdt))= Aeminusbt2mcos(ωdtminusφ) (3)

Letrsquos analyze this physically When b = 0 the response is a sinusoidDamping is a frictional force so it generates heat and dissipates energyWhen the damping constant bis small we would expect the system to stilloscillate but with decreasing amplitude as its energy is converted to heatOver time it should come to rest at equilibrium This is exactly what wesee in (3) The factor cos(ωdtminusφ)shows the oscillation The exponentialfactor eminusbt2m has a negative exponent and therefore gives the decayingamplitude As t infin the exponential goes asymptotically to 0 so x(t)rarralso goes asympotically to its equilibrium position x

=0

We call ωd the dampedangular(orcircular)frequencyof the systemThis is sometimes called a pseudo-frequencyof x(t) We need to be carefulto call it a pseudo-frequency because x(t)is not periodic and only periodicfunctions have a frequency Nonetheless x(t)does oscillate crossing x=0twice each pseudo-period

Example1 Show that the system x+1x+3x=0 is underdamped find itsdamped angular frequency and graph the solution with initial conditions

x(0) =1 x(0) =0

Solution Characteristic equation s2 +s+3 =0Characteristic roots

minus12

plusmniradic 112

Basic real solutions eminust2 cos(radic 11 t2) eminust2 sin(radic 11 t2)General solution

x(t) =eminust2(c1 cos(radic

11 t2) +c2 sin(radic

11 t2))= Aeminust2 cos(radic

11 t2minusφ)

Since the roots have nonzero imaginary part the system is underdampedThe damped angular frequency is ωd =radic 112The initial conditions are satisfied when c1 =1 and c2 =1

radic 11 So

x(t) = eminust2 cos(radic

11 t2) +radic 111

sin(radic

11 t2)radic 12 eminust2= radic 11

cos(radic 11 t2minusφ)

where φ=tanminus1(1radic

11)

2







gt 4mk the damping











3







0 1 2 3 4 5

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Case(iii)

Critical

Damping






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gt 4mk the damping











3







0 1 2 3 4 5

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Case(iii)

Critical

Damping






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0 1 2 3 4 5

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Case(iii)

Critical

Damping






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0 1 2 3 4 5

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0 6

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1 0

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0 2 4 6 8

- 0

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0 1 2 3 4 5

0 0

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1 0

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0 2 4 6 8

- 0

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Fall 2011






Fall 2011


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Under, Over and Critical Damping