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Over Damped and Critically Damped Oscillator The equation for a damped harmonic oscillator is

&&x + ! &x +"

0

2x = 0

The solution may be obtained by assuming an exponential solution of the form x(t) = Aept so that

x(t) = Ae

ptso &x = pAe

pt, and &&x = p

2Ae

pt

Substituting the trial solution into the original equation it becomes:

p2 + ! p +"

0

2( )Aept= 0 In order to satisfy this equation either A = 0 or the quantity in parenthesis is zero; if we assume the former then x(t) =0, and while a valid solution is not all that interesting. The other alternative is to find the values of p that satisfies the quadratic equation, which are:

p =!"

2±

" 2

4!#

0

2= !a ±m

The constants a and m have been introduced to simplify later expressions and their definitions should be obvious. There are three cases to consider: real distinct roots, degenerate roots, and complex roots. In the case of over damping the roots are real and distinct, in the case of critical damping the roots are degenerate, and in the case of under-damped motion the roots are complex. Consider first the over-damped case in which the roots are real and distinct.

x(t) = e!at Aemt + Be!mt( ) with the velocity given as:

v(t) = !ae!at Aemt + Be!mt( ) +me!at Aemt ! Be!mt( ) The coefficients A and B are determined by specification of the initial conditions x(0) and v(0) which give rise to two equations that are solved simultaneously for A and B. If x(0) is x0 and v(0) is v0 then the equations are:

x0= A+ B and v

0= !a A+ B( ) +m A ! B( )

Solving for A and B and using the expression obtained the expression for x(t) is

x(t) =e!at

2mmx

0+ ax

0+ v

0( )emt + mx0! ax

0! v

0( )e!mt( )

By inspection x(0) and v(0) return the appropriate initial conditions. The graph in figure 1 shows the displacement of the oscillator on the vertical axis vs. time on the horizontal axis. The initial position and velocity were set to x(0) = 1 and v(0) = +3 m/sec (black curve) and x(0) = 1 and v(0) = -3 m/sec (blue curve).


Over Damped a = .6, m=.3

The critically damped case occurs when the roots of the quadratic or characteristic equation are equal, which implies that m is zero. Examination of the solution shows that for m = 0 the form of x(t) is indeterminate as the exponentials inside the brackets go to 1 yielding a solution of the form 0/0.

x(t) =e!at

2mmx

0+ ax

0+ v

0( )emt + mx0! ax

0! v

0( )e!mt( )

x(t) may be found by finding the limit of the above expression as m tends to zero by expanding the exponentials and letting m go to zero or more simply by using L’Hopital’s rule, which may be done in this case by differentiating of the numerator and denominator with respect to m and then letting m tend to zero. Doing so given the result for x(t) as

x(t) = x0e!at+ (ax

0+ v

0)te

!at

The new solution is of the form x(t) = Ae-at +Bte-at, and we have thus found the critically damped solution. It should be noted there are many ways to find this solution, for example using the method of variation of parameters, however this is perhaps the simplest. The graph above is duplicated, however now the critically damped solutions are included as well as the over damped solutions. As the graph suggests the critically damped solutions approach the equilibrium position more rapidly than their over damped counterparts.


Over damped and critically damped m = 0 for critically damped solutions.

Over Damped Solution – Alternate Form

The over damped solution stated above was

x(t) =e!at

2mmx

0+ ax

0+ v

0( )emt + mx0! ax

0! v

0( )e!mt( )

and may be written as

x(t) =e!at

mmx

0

emt + e!mt

2

"

#$

%

&'+ ax

0+ v

0( )emt ! e!mt

2

"

#$

%

&'

"

#$

%

&' =

e!at

mmx

0coshmt + ax

0+ v

0( )sinhmt( )

This alternate formulation is included as it will be used in exercises that follow. The importance of damping is illustrated in many applications; for example pneumatic door closers should be designed so that the return mechanism will not impart excessive force causing the door to overshoot equilibrium and thus bang into the door jam causing objectionable noise and doing potential damage to the door. Pneumatic closers are generally equipped with a damping adjustment so that the closer can be critically damped. Another important example is in the design of an automotive suspension. The suspension should be designed in such a way that the car will return to its equilibrium as quickly as possible with little or no overshoot. The usual design is for slight overshoot as this will give a smoother ride, however from a handling point of view the closer the suspension to critical damping the better.


The Under Damped Case

Before tackling the problem with damping the solution for the undamped oscillator is obtained. The purpose of doing so is to show the general techniques used to solve the problem with damping and to illustrate the role of complex numbers in the treatment of this and many other problems. The equation without damping is

&&x +!

0

2x = 0 with p

2 +!0

2( )Aept= 0" p = ±i!0

Thus the solution may be written as

x(t) = Aei!0 t + Be

"i!0 t

Using the same technique as above the values of A and B may be obtained by satisfying the initial conditions namely the initial position x(0) and the initial velocity v(0) which are:

x0= x(0) = A+ B and v

0= v(0) = i!

0A " B( )

There appears to be a problem with these equations since both the initial conditions are real quantities, however the equations at first glance suggest the initial velocity is imaginary. In order for there to exist meaningful solutions it is necessary to assume A and B are complex numbers, namely

A =! + i" and B = # + i$ so that

x0= A+ B = ! + "( ) + i #+$( )%# = &$

v0= i'

0! & "( ) + i #&$( )( ) = i'0

! & "( ) +'0$ &#( )%! = "

The fact that the initial position and velocity are both real necessitates picking A and B with the stipulations required by the above equations. After examination it should be clear that α = β requires that α = β = x0/2, likewise ε = - δ requires ε = v0/2ω0 so that A = x0/2 - i v0/2ω0 and B = x0/2 + i v0/2ω0 and thus the solution that satisfies the initial conditions is

x(t) =x0

2!v0

2"0

#

$%

&

'(e

i"0 t +x0

2+v0

2"0

#

$%

&

'(e

!i"0 t = x0

ei"0 t + e

!i"0 t

2

#

$%

&

'(! i

v0

"0

ei"0 t ! e!i"0 t

2

#

$%

&

'(

But the sum of exponentials divided by 2 is just cos(ω0t) and the sum of the difference divided by 2 is i*sin(ω0t), so the solution may be written as:

x(t) = x0cos!

0t +

v0

!0

sin!0t

In order to check to see that this is the proper solution, note that the solution returns the appropriate initial conditions. Finally, before leaving this example, it should be noted that x(t) may be written as either a sine or cosine function with an amplitude and phase both of which depend on the initial


conditions of the problem. To illustrate assume x(t) may be written as Acos(ω0t - δ) then it follows that:

x0cos!

0t +

v0

!0

sin!0t = Acos(!

0t "#) = Acos!

0t cos# + Asin!

0t sin# $

x0= Acos# and

v0

!0

= Asin#

From the last equation it follows that A and φ are given as:

A = x0

2+

v0

!0

"

#$

%

&'

2

and ( = tan)1v0

x0!0

The same technique may be used to show that any choice of sine or cosine with plus or minus a phase angle can be used. The amplitude will always be the same, but the phase angle will change sign for the various possibilities. Try all possible combinations out as an exercise – it illustrates the many ways the solution to this problem may be expressed.

The Oscillator with Damping

The solution to the damped oscillator is found by following the same program used to solve the un-damped oscillator. The algebra is just a touch more tedious but the methodology is identical: solve the auxiliary equation to find its roots, write the equations for the two unknown constants A and B that satisfy the initial conditions, allow A and B to be complex, find their solutions, and then use them to find the expression for x(t). The auxiliary equation, as it is often called, in p is:

p2 + ! p +"

0

2( )Aept= 0 and p is given as where the magnitude of the natural frequency is assumed to be greater than γ2/4:

p =!"

2±

" 2

4!#

0

2=!"

2± i #

0

2 !" 2

4=!"

2± i#

1

The expression for p has been modified for this case in which ω0 is greater than γ and thus the expression for p is complex. In terms of the values for p the solution of the oscillator is

x(t) = e!"t /2 Aei#1t + Be!i#1t( )

Notice that ω1 and ω0 are different numbers; ω1 is the angular frequency of the damped oscillator whereas ω0 is the frequency of the undamped oscillator called the natural frequency of the oscillator. Since we already know the values of A and B are going to have to be complex for real solutions as in the previous case the initial conditions for the system may be written immediately as

A =! + i" and B = # + i$

x0= A+ B = ! + #( ) + i "+$( )%" = &$


Which is exactly the same as the previous case, the expression for the velocity, however is more complex as v(t) is given as:

v(t) = &x(t) =!"

2e!"t /2

Aei#1t + Be!i#1t( ) + i#e!"t /2 Ae

i#1t ! Be!i#1t( )

v0=!"

2# + i$( ) + % + i&( )( ) + i'1 # + i$( ) ! % + i&( )( )

v0=!"

2# + $( ) +%1(& !')! i

"

2' +&( ) + i%1 # ! $( )

From this equation in order for the imaginary part to be zero α must equal β and as was the case before ε = -δ. But if α = β then α = x0/2 = β using the expression for x0. In a similar fashion using the expression for v0 the relationship between α, β, δ, and γ it follows that

v0+!x

0

2= 2"#

1or " =

v0

2#1

+!x

0

4#1

= $%

Putting it all back together the solution for x(t) may be written as

x(t) = e!"t /2 Aei#1t + Be!i#1t( ) = e!"t /2

x0

2! i

v0

2#1

+"x

0

4#

$

%&

'

()

$

%&&

'

())e

i#1t +x0

2+ i

v0

2#1

+"x

0

4#1

$

%&

'

()

$

%&&

'

())e

!i#1t*

+,,

-

.//

Now using the Euler identity – recall the expressions for sine and cosine in terms of the exponentials and by inspection x(t) may be written as: (Maybe careful inspection – but inspection nonetheless.)

x(t) = e!"t /2

x0cos#

1t +

v0

#1

+"x

0

2#1

$

%&

'

()sin#1t

*

+,

-

./

As a check it is easily verified that this expression returns the appropriate initial conditions. As an exercise show this to be the case. The graph on the next page shows a lightly damped oscillator with initial position x = 2.4 meters and initial velocity 10 m/Sec. The green dashed line is the decay envelope and the red line is the actual displacement of the oscillator as a function of time. How would the graph differ is the initial velocity were -10 m/Sec?


Damped Oscillator & Decay Envelope

Summary – What have we accomplished?

The solution to the damped oscillator has been found for the cases of over damping, critical damping and under damping. The solutions presented are general solutions in the sense that each solution includes arbitrary initial conditions so that the solutions presented may be used for any problem by simply substituting the appropriate initial conditions. The method used to find each solution was the same. Often one encounters a mix of approaches that can be confusing initially and thus an effort has been made to adopt a consistent approach for all three cases. The expression for the overdamped case expressed as a decaying exponential times the hyperbolic functions (page 3) is not generally discussed but will be used in several exercises and is particularly useful when considering the transition from one case to the next, for example when the roots transition between real distinct and degenerate the expression may be used to easily determine the change in the character of the solution.


The Damped Driven Oscillator

The systems that we have looked at to this point have been free in the sense that no outside agent is acting upon them, that is the oscillator is given an initial nudge that imparts an initial velocity at some position at a time we call zero and then the solution represents the time evolution of the oscillator. Next consider an oscillator that is driven by some outside agent with a force F(t). What do we expect to happen? Suppose the system is acted upon by a force that varies as cos(ωdt) and may be written as F(t)=F0cosωdt, where ωd is called the forcing or driving frequency. Now we have three ω’s, the natural frequency ω0, the damped frequency ω1, and the driving frequency ωd. The equation that describes this situation is

&&x + ! &x +"

0

2x = F

0cos"

dt

The differential equation that describes this system is a linear second order equation and as such if you have a general solution to the homogenous equation and a particular solution to the inhomogeneous equation then the sum of these two is also a solution, about which more will be said later. However, at present we shall focus on only the steady state solution and will try to make an intelligent guess as to what a steady state solution might look like. Suppose the oscillator is initially at rest at the equilibrium position when the forcing function is applied. For this situation the homogenous solution is x(t) = 0 and the resulting particular solution will be the steady state solution. Even if the homogeneous solution were not zero after a period of time passes the homogeneous solution will contribute nothing to the motion since it decays exponentially. Thus we conclude that if a steady state is reached the solution to the homogeneous equation will disappear since it is exponentially damped and all that will remain is the steady state solution caused by the driving force, referred to as the forcing function. The driving force, however has a frequency ωd and thus we might well guess that the steady state solution will have the same frequency of oscillation. The name “forcing function” is used to suggest this latter point. In what other ways can the steady state solution differ from the driving force? It would appear the most general answer is that the steady state solution can also differ in amplitude and phase, but will have the same frequency as the forcing function, therefore we shall assume a solution of the form:

xp (t) = Aei !d t"#( )

If the postulated solution, xp(t) is a solution, it will satisfy the inhomogeneous equation above with the appropriate choice of A and δ. Upon taking the derivatives of xp(t) and substituting them into the equation above one obtains:

!"d

2 + i#"d+"

0

2( )Aei "d t!$( ) =F0

mei"d t

Where the driving force is the real part of the exponential on the right. The term eiωt cancels out and the expression above may be written as:

!"d

2 + i#"d+"

0

2( )A =F0

mei$

Separating the real and imaginary parts of the above expression gives:

!0

2 "!d

2( )A =F0

mcos# and $!

dA =

F0

msin#


Squaring, adding and simplifying gives A and δ as

A(!d) =

F0

m

1

!0

2 "!d

2( )2

+ #!d( )2

and $ = tan"1#!

d

!0

2 "!d

2( )

This is then the steady state solution to the driven oscillator. We are now left the task of interpreting this rather complicated looking solution. A, the amplitude, has been expressed a function of the driving frequency to emphasize that the amplitude and phase both depend on its value. As was alluded to all of the values that determine the amplitude and phase are fixed and thus cannot be used to satisfy the initial conditions. The amplitude of oscillation may be expressed in a second form by doing a bit of algebra and introducing a quantity Q defined as Q=ω0/γ. This second form is frequently encountered and very useful.

A(!d) =

F0

m

1

!0

2 "!d

2( )2

+ #!d( )2

=F0

m

1

!0!

d

!0

!d

"!

d

!0

$

%&

'

()

2

"# 2

!0

2

!0

2 = k / m so

A(!d ) =F0

m

1

!0

2 "!d

2( )2

+ #!d( )2

=F0

k

!0

!d

!0

!d

"!d

!0

$

%&

'

()

2

+1

Q2

The algebra is a bit obtuse and thus it is recommended you make certain you see how the result was obtained. There are three important cases; The magnitude of A in the low frequency limit, the value of A when the driving frequency matches the natural frequency, and the high frequency limit. For ωd = 0 the amplitude is given as:

A(0) =F0

m!0

2=F0

k

For a very low frequency, such as ωd near zero, the oscillator behaves as if it is being “dragged” by the forcing function and when the forcing function reaches its maximum or minimum value the oscillator will be F0/k = ±A(0) meters from the equilibrium position. Indeed if the forcing function did not vary with time, but were just equal to F0, the oscillator would be displaced to A(0) and remain at that position. The phase angle, which is the phase between the displacement and the forcing function is zero when ωd = 0, and this means the displacement and applied force are in phase with one another. The amplitude of the oscillator at ω0 = ωd is seen to be A(ω0 = ωd) = Q(F/k)

A(!0=!d ) =

F0

k

"

#$

%

&'Q


As the driving frequency increases well past the resonant condition the amplitude decreases rapidly.

Resonance – the driving frequency is varied and the natural frequency is fixed.

When the driving frequency and the natural frequency are equal the amplitude gets very large and the phenomenon is referred to as resonance. In the case we are discussing it is assumed that the driving frequency is being varied, and in this case the relative maximum of the peak in the graph above is not at ωd = ω0, but slightly below it. If on the other hand ωd is held constant and the natural frequency is allowed to vary the resonant peak occurs at ω0 = ωd. In either case the amplitude is maximum when

the denominator is minimum, so to find the maximum find the derivative of !0

2 "!d

2( )2

+ #!d( )2

,

set it equal to zero and solve for the appropriate ω. If this is done in the case in which the driving frequency is varied the maximum occurs at a value just a bit below ω0 as seen by examination of the following equation. (The value computed is for the values used in the example.)

d

d!d

!0

2 "!d

2( )2

+ #!d( )2

= 0$!d =!01"

1

2Q2= .968

The case that is being described is very common and occurs often; a simple mass spring system driven by a sinusoidal force will exhibit this kind of resonance and while the maximum does not occur at exactly ωd = ω0 the difference is very small and frequently not experimentally measurable for a high Q system. If the differentiation is carried out with respect to ω0 the natural frequency of the oscillator then:

d

d!0

!0

2 "!d

2( )2

+ #!d( )2

= 0$!d=!

0


This type of situation occurs when the oscillator is “tuned”, such as in the case of a tunable LC circuit used in the first RF stage of a radio. The oscillator’s natural frequency is varied by means of a variable capacitor and the current in the circuit reaches a maximum when the natural frequency matches that of a signal delivered by the antenna. A mechanical oscillator could likewise be tuned by changing the mass or for that matter the spring constant. In the first case, that is the case in which the driving frequency is varied it is interesting to note that the resonant peak occurs below either the natural frequency or the damped frequency of the oscillator. Using the numbers in the example the various frequencies and their corresponding amplitudes are:

!0=1 with A(!

0) = 2.86

!1=!

01"

1

4Q2= .984 with A(!

1) = 2.89

!max

=!01"

1

2Q2= .968 with A(!

max) = 2.90

Resonant Curve for a Tuned Oscillator

Phase Considerations

We have discussed the amplitude as a function of driving frequency in some detail; however there is also a shift in phase between the oscillator and the driving force that is important to understand and an aspect of the problem that is generally not given the attention it deserves. When the driving


frequency is very low, approaching zero, the oscillator is literally being dragged by the forcing function and the displacement and driving function are in phase as mentioned earlier. The amplitude and phase relationships are reproduced here for convenience.

A(!d) =

F0

m

1

!0

2 "!d

2( )2

+ #!d( )2

and $ = tan"1#!

d

!0

2 "!d

2( )

The phase angle is the phase between the driving force and the displacement. Indeed, as noted above as the driving frequency approaches zero the phase angle likewise tends to zero. As the driving frequency approaches the natural frequency of the oscillator the phase angle approaches π/2, and in the high frequency limit the driving force and the displacement are π radians out of phase with one another or as the expression goes “180 out”. The low frequency limit is easily understood in term of the explanation above and like wise the high frequency limit makes sense; the driver is oscillating so fast the energy can not be absorbed by the oscillator. In the high frequency limit the displacement and driving force are π radians out of phase so that as the oscillator goes positive the driving force is negative and vice versa. The net result is the high frequency driving force tends to oppose the displacement with the subsequent result that the amplitude of oscillation tends to zero. Put another way the driving function’s frequency is so high the oscillator does not have time to respond to the forcing function. Perhaps the most interesting and most difficult case to grasp occurs when the driving force and displacement are π/2 out of phase with one another, which occurs when the driving frequency and natural frequency are identical. When the driving force and displacement are π/2 out of phase the driving force always acts in a way to augment the motion of the oscillator – the driving force is always “pushing” the oscillator in the direction it is going – the driving force never pushes the oscillator in the “wrong” direction. In the graph below the oscillator starts at the origin and the driving force is maximum. As the oscillator approaches its maximum positive displacement the driving force drops to zero and reverses pushing the oscillator back toward the equilibrium poison exerting its maximum value as the oscillator passes through the origin with its maximum velocity. This process is repeated every half cycle and as the driving force is never in opposition to the displacement it makes sense this circumstance will lead to maximum amplitude.

Driving Force vs. Displacement with π/2 phase difference


If the phase is changed from π/2 the force is not maximum when the oscillator passes through points of zero displacement and there are intervals when the driving force opposes the motion and thus decreases the amplitude. In the graph below the driving force and displacement are out pf phase by about π/6 radians and in some regions the driving force clearly is in opposition to the displacement, a situation that does not occur with a 90 degree phase shift. For example when the driving force is at its minimum value the oscillator is still approaching its minimum value, but between the two minima the applied force is in opposition to the displacement driving down the displacement of the oscillator.

Driving Frequency 206 mHz or ωd=1.3, Natural Frequency 159 mHz or ω0=1.

The Total Solution

As was mentioned above the total solution of the damped driven oscillator consists of a so-called general solution, which is the solution to the homogeneous equation and a particular solution that is the solution to the inhomogeneous equation. The general solution has associated with it two arbitrary constants which are used to satisfy the initial conditions of the problem. The steady state solution is referred to as a particular solution and it is the solution to the inhomogeneous equation or the so-called steady state solution. The steady state solution is the solution that remains after the general solution has died away due to exponential damping. Because the system is linear the sum of the general and particular solution constitute the complete solution to the problem. The complete solution is composed of transient terms (those that die) and a steady state term that remains as long as the driving force is in place. The theory of linear differential equations also allows another claim to be made; solutions to linear differential equations are unique. Thus once you have arrived at the complete solution it is the solution. Symbolically:

xcomplete = xgeneral + xparticluar = complete solution which is unique

xgeneral = Bcos(!1t " #) A and # arbitrary constatns

xparticular = F cos(!dt "$) where F = A(!)


Suppose that a damped oscillator is driven and initially the position of the oscillator is x(0) = x0 and v(0) is v0. Find the solution. The solution may be written as:

x(t) = e!"t Acos(#1t)+ Bsin(#

1t)( ) +

F0

k

1

#0

2 !#d

2( )2

+ "#d( )2

cos(#dt !$) with $ = tan!1

"#d

#0

2 !#d

2

However in order to make the solution a bit easier express the general solution as:

x(t) = xg (t)+ xp (t) = Ae!"tcos(#

1t ! $)+ F cos(#dt !%)

Please note that a change in notation has been made. A(ω) has been replace by F(ω) and A and β are now the arbitrary constants used to satisfy the initial conditions. In order to satisfy the initial conditions it follow that:

x0= Acos! + F cos" and v

0= #

$A

2cos! +%

1Asin! +%

dF sin"

From the first equation Acosβ = x0 - Fcosδ, and upon substitution into the second equation it follows that

x0! F cos"= Acos# so that

v0!$dF sin" +

%

2x0! F cos"( ) =$1 x0 ! F cos"( ) tan#

tan# =v0!$dF sin" +

%

2x0! F cos"( )

$1x0! F cos"( )

=%

2$1

+v0!$dF sin"

$1x0! F cos"( )

A =x0! F cos"

cos#Note :$

1=$

01!

1

4Q2

Now x(t) may be written as:

x(t) =x0! F cos"

cos#

$

%&

'

() e

!*t /2cos(+

1t ! #)+ F cos(+

dt !")

where F =F0

k

1

+0

2 !+d

2( )2

+ *+d( )2

Evaluation of x(0) reveals that it reduces to x(0) = x0. The expression is rather messy, however it is clear that at t = 0 the solution reduces to the appropriate value. As an exercise verify that v(0) = v0. The graphs that follow are for lightly damped oscillator (γ=.1, ω0 = 1, and Q = 10) being driven with frequency ωd = 1.1. The upper and lower horizontal lines are the amplitude of the steady state solution and the exponential curve represents the strength of the transient solution. One the transient is dead only the steady state solution remains. If the initial conditions are changed the shape of the oscillation will be altered.


Typical Solution with Transient

This is the same oscillator but with a much higher initial velocity.


Energy of an Oscillator

For a simple harmonic oscillator without damping it is easy to show that the total energy of the system may be expressed as E= ½ kA2, where A is the maximum amplitude. The total energy of the oscillator is just the sum of the kinetic and potential energies and with no damping the energy will remain constant, however with damping the amplitude decays since the dissipative term or damping terms does non recoverable work on the system and thus the oscillator’s amplitude decays. Since the damping force is –bv it follows the work done by friction is:

Wfriction = ! bvdx =x(0)

x(t )

" ! bv2dt < 0

0

t

"

Using the general solution for the damped oscillator an expression for the energy as a function of time can be obtained. The position and velocity may be expressed as:

x(t) = e!"t /2Acos #t ! $( )

v(t) = e!"t /2A!"

2cos #t ! $( ) !# sin #t ! $( )

%

&'

(

)*

But the velocity may be rewritten by factoring ω as

v(t) = e!"t /2#A!"

2#cos #t ! $( ) ! sin #t ! $( )

%

&'

(

)* = e

!"t /2#A!1

2Qcos #t ! $( ) ! sin #t ! $( )

%

&'

(

)*

For a high Q system the cosine term in the velocity may be ignored, as it is much smaller than the remaining term. For example if Q = 10 the cosine term is down by a factor of 20. Thus the total energy may be expressed as:

E =1

2mv

2 +1

2kx

2 = e!"t1

2m#2

A2sin

2 #t ! $( ) +1

2kA

2cos

2 #t ! $( )%

&'

(

)*

However mω2 is very nearly equal to k since the difference between ω and ω0 is so small for a high Q system. So that if the energy at t = 0 is E0 it follows that E(t) = E0e--γt.

E = E0e!"t

Thus a remarkably simple result is obtained for the energy decay as a function of time. The energy decays at twice the rate as the amplitude. Armed with this expression for the energy as a function of time it is now easy to interpret the meaning of Q, the quality factor.

Quality Factor The quality factor of an oscillator is related to the energy lost per oscillation, δE, and the total energy stored in the oscillator. It was shown above that the oscillator’s energy decreases over time due to the work done by friction and further it was shown that E(t) is an exponential decay, so the change in energy is:


E = E0e!"t

sodE

dt= !"E

0e!"t= !"E

Q =#0

"$#

"because # =#

01!

1

4Q2

The approximation is very good and gets better as the Q of the system gets larger; a Q of only 5 changes ω by only one part in 200 – nearly as pure as Ivory soap. Rearranging the equation for dE/dt one can write

!E "dE

dt#t = $E#t

But the time required for the oscillator pass through one radian is δt =1/ω so

Q =E

!E=Energy stored in Oscillator

Energy lost per radian

So far the quantity Q has been used in several instances but now a definition has been developed that gives it physical meaning. Recall that Q is used to find the difference between the natural frequency and the damped frequency, and Q times A(0)=F0/k is equal to the amplitude of a driven oscillator when the driving frequency equals the natural frequency. And while we have not yet shown it Q is related to a quantity called the half maximum of the resonant peak, which is a measure of the sharpness of the resonance with respect to frequency. Q is a very useful quantity and once known gives one a great deal of insight into the nature of the system.

Power The damped driven oscillator must be furnished with energy in order to maintain a constant steady state oscillation. The instantaneous power of the oscillator may be computed as P = Fv, which for the damped driven oscillator is given as:

P(t) = Fv =!"

dF0

2

m "0

2 !"d

2( )2

+"d

2# 2cos "

dt( )sin("d

t !$) with $ = tan!1"

d#

"0

2 !"d

2

%

&'

(

)*

This expression is for the steady state condition for which the transient solutions are assumed to have died. Note that unlike the transient parts of the solution the expression for the instantaneous power is independent of the damped frequency. Using the expression for the instantaneous power it is now possible to calculate the average power over a period, or

P(t) = P(t)dt0

T

!

In order to evaluate this integral the double angle formula is used to obtain an expression for the average power.


P(t) =!"

dF0

2

m "0

2 !"d

2( )2

+"d

2# 2cos "

dt( ) sin("d

t)cos$ ! cos "dt( )sin$( )dt

0

T

%

P(t) ="

dF0

2sin$

2m "0

2 !"d

2( )2

+"d

2# 2

The maximum average power occurs when the driving frequency and the natural frequency are equal. In order to express the power at other frequencies the expression for sine δ must be expressed in term of the driving frequency, which is easy to do using the expression for tangent of δ.

Pmax

=!dF0

2sin"

2m !0

2 #!d

2( )2

+!d

2$ 2!0 =!d "=% /2

=F0

2

2m$=F0

2Q

2m!0

Using the expression for tangent of δ one may write sinδ as:

sin! ="d#

"0

2 $"d

2( )2

+"d

2# 2so that P("d ) =

"d

2#F0

2

2m "0

2 $"d

2( )2

+"d

2# 2( )

P("d ) ="d

2#F0

2

2m "0

2 $"d

2( )2

+"d

2# 2( )=

#F0

2

2m"0

2 "0

"d

$"d

"0

%

&'

(

)*

2

+1

Q2

%

&

''

(

)

**

A plot of the power as a function of the driving frequency appears as:

Average Power vs. Driving Frequency, Power in Watts and Frequency in Radians


The power absorption curve is very nearly symmetric about the maximum and it is useful to define what is called the half maximum width of this curve. In order to do so using the expression for the average power as a function of the driving frequency it may be recast as follows.

P(!d ) =!d

2"F0

2

2m !0

2 #!d

2( )2

+!d

2" 2( )=

"F0

2

2m!0

2 !0

!d

#!d

!0

$

%&

'

()

2

+1

Q2

$

%

&&

'

(

))

The expression above is an exact expression for the power expressed as a function of the driving frequency and reduces to the maximum value when the driving frequency and the natural frequency are identical. Assuming that the Q of the system under discussion is high, say greater than 10, then for driving frequencies near the natural frequency the two may be assumed close in value allowing for the following approximate expression:

!0

!d

"!d

!0

#

$%

&

'(

2

+1

Q2=!0

2 "!d

2

!0!d

#

$%

&

'(

2

+1

Q2=

!0+!d( ) !0

"!d( )!0!d

#

$%

&

'(

2

+1

Q2

)2!

0( ) *!0( )!0!d

#

$%

&

'(

2

+1

Q2=2*!

0( )2

!0

2+1

Q2=2*!

0( )2

!0

2++ 2

!0

2

Now substituting this approximate expression into the expression for the average power the expression above becomes:

P(!d) "

#F0

2

2m 4$!0

2 + # 2( )

Now if the driving frequency is set to the natural frequency the expression above reduces to Pmax and if we require:

If 4!"0

2= # 2 =

"0

2

Q2then P("d ) =

F0

2

4#m=Pmax

2

so !" = ±"0

2Q

This expression shows that the higher the Q of the oscillator the more sharply peaked the power absorption curve. Knowledge of Q thus gives one a very good idea of an oscillator’s character. Knowing Q you can compute the difference between the damped frequency and the natural frequency, you can determine the amplitude of the resonant peak, you can determine the peak power absorption in steady state, and you can determine the bandwidth.

critically damped oscillator 02b - oneontaemployees.oneonta.edu/andersae/critically damped... ·...

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