# I X variable in damping force equation for damped oscillation?

#### Protea Grandiceps

Summary
What is the meaning of the x variable in the equation for damping force in respect of non-driven, damped oscillations?
Hi,

for ease of reference this posting is segmented into :

1. Background
2. Focus
3. Question

1. Background:

Regarding (one, linear, second-order, homogeneous, ordinary, differential) equation describing the force in a non-driven, damped oscillation:

F = m.a = -k.x - b.v

F = m.(d^2.x/dt^2) = -k.x - b(dx/dt)

Rearranging:

m.(d^2.x/dt^2) + k.x + b(dx/dt) = 0

...

(d^2.x/dt^2) + b/m(dx/dt) + k/m.x = 0

2. Focus:

But x in k/m.x and dx/dt are in respect of x1= f(t) for simple harmonic oscillation while x in d^2.x/dt^2 and d^2.x/dt^2 is in respect of equation x2 = f(t) for damped oscillation.

3. Question:

So there are in fact variables x1 and x2 but they are both x in the equation (and treated as such).

How can this be correct?

--

Apologies for unicode. Thanks for bearing with me.

Regards

PG

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#### Dale

Mentor
Welcome to PF!

I really appreciate the care and thought that went in to your question. You made it organized and clear. I will respond to the content shortly, but just had to give you “kudos” for the presentation!

#### Protea Grandiceps

Thanks, Dale. I really should edit this though: "...while x in d^2.x/dt^2 and d^2.x/dt^2..." should read "...while x in d^2.x/dt^2...".

#### Dale

Mentor
So there are in fact variables x1 and x2 but they are both x in the equation (and treated as such).

How can this be correct?
There are no distinct x1 and x2 here distinguishing between damped and simple harmonic oscillation. In this equation x is simply position. If the mass is 10 cm away from the equilibrium position then x=0.1 regardless of whether there is any damping or not.

The distinction between damped and simple oscillation is not in x but instead is entirely contained in the parameter b. For b=0 the system is simple harmonic motion, and for b>0 it is damped harmonic oscillation.

#### berkeman

Mentor
Welcome to the PF. Great first thread!
Apologies for unicode.
Have a look at the LaTeX Primer to help you with posting math equations here at the PF:

#### Protea Grandiceps

Hi Dale,

thank you for your considered answer which answers my question to a T. I appreciate your swift response earlier.

I agree with you that the variable x represents distance from equilibrium regardless of whether oscillation is simple harmonic, damped harmonic or driven harmonic for that matter.

Unfortunately, I had intended to ask a different question altogether and was a bit careless in my choice of terms - I'm sorry for this.

Specifically, I should have focused my question on the use of the equation defining x as in x=f(t) in order to determine the equation for a damped harmonic oscillation, rather than on the variable x.

x is defined differently in damped and simple harmonic oscillations. We know this right away since the damping coefficient reduces force (except at all peaks and troughs of the equation), therefore reduces velocity and therefore increases the period (decreases frequency). So the first and second order derivatives in

(d^2.x/dt^2) + b/m(dx/dt) + k/m.x = 0

will be in respective of different equations for x as in x = f(t).

d^2.x/dt^2 is the second order derivative of x = f(t) in a damped harmonic oscillation.

dx/dt is the first order derivative of x = f(t) in a simple harmonic oscillation.

While x in k/m.x refers to x = f(t) in a simple harmonic oscillation.

So far this isn't a problem.

But we use the equation for damping force as the basis to arrive at the equation for a damped harmonic oscillation. Toward this end, there is a step where we replace both x= f(t) for the simple and damped harmonic oscillations with e^(lambda*t), including for the purposes of determining the first and second order derivatives, thereby treating both x= f(t) for the simple and damped harmonic oscillations as if they were the same equation - which they aren't. I don't understand how it is possible to do this - I've a hunch, though, that I've overlooked some property of e^(lambda*t) which renders differences in lambda for the purpose of determining derivatives of e^(lambda*t) negligible.

If it would be of help, I'd be glad to detail the complete derivation of (this particular) equation describing damped, non-driven harmonic oscillation for the sake of context.

Best regards,

PG

#### Protea Grandiceps

Welcome to the PF. Great first thread!

Thanks for your welcome, Berkeman!

Have a look at the LaTeX Primer to help you with posting math equations here at the PF:

Great advice - I'll head right over asap.

#### Protea Grandiceps

There are no distinct x1 and x2 here distinguishing between damped and simple harmonic oscillation. In this equation x is simply position. If the mass is 10 cm away from the equilibrium position then x=0.1 regardless of whether there is any damping or not.

The distinction between damped and simple oscillation is not in x but instead is entirely contained in the parameter b. For b=0 the system is simple harmonic motion, and for b>0 it is damped harmonic oscillation.
Hi Dale,

thanks, I replied to your post without referencing it earlier so I'm just doing referencing it now, post- reply, in case you're only notified about referenced replies.

Best regards,

PG

#### Dale

Mentor
x is defined differently in damped and simple harmonic oscillations. We know this right away since the damping coefficient reduces force (except at all peaks and troughs of the equation), therefore reduces velocity and therefore increases the period (decreases frequency).
No, x is defined the same: in both cases x is the distance from the equilibrium position. The reduced force does not change the meaning of x, nor does the decreased velocity, nor the increased period. The displacement still has the same meaning and definition for both damped harmonic motion and simple harmonic motion (SHM).

will be in respective of different equations for x as in x = f(t).
I am not sure if you are intending f(t) as just a generic function of time or if you intend it as the force as a function of time. I assume from context that you intend it as just an ordinary function of time (specifically the function that solves the differential equation), and all of my subsequent comments follow that assumption.

d^2.x/dt^2 is the second order derivative of x = f(t) in a damped harmonic oscillation.

dx/dt is the first order derivative of x = f(t) in a simple harmonic oscillation.

While x in k/m.x refers to x = f(t) in a simple harmonic oscillation.
No, in all three terms x is the position in a damped harmonic oscillation. There are not separate x for separate terms, and it is not relevant if a given term is similar to a term in SHM (or any other system).

Those terms do appear in the SHM equations and in those SHM equations the f(t) which solves the differential equation will be different. So the meaning of x is the same in both cases but the f(t) that solves the equation differs in the two cases. However, in the case of damped oscillation the same f(t) is used for all terms and for all terms it is the f(t) that solves for x in all of the damped oscillation terms together.

Toward this end, there is a step where we replace both x= f(t) for the simple and damped harmonic oscillations with e^(lambda*t), including for the purposes of determining the first and second order derivatives, thereby treating both x= f(t) for the simple and damped harmonic oscillations as if they were the same equation - which they aren't.
They are the same equation, there is only one x and it is the same x for all the terms in the damped oscillator equation. The substitution is correct, and trying to substitute a different expression for different terms would be mathematically wrong.

#### Protea Grandiceps

Hi Dale,

it's heartening to see that you've been giving my modest interest a bit of your no doubt highly-prized attention - thank you. If I may I'd just like to go over my understanding of select aspects of simple harmonic motion (SHM) and damped harmonic motion (DHM) and then attempt to make an argument, so I'll be drawing on your forbearance once again. For the sake of attractiveness (which at this point might well be an issue, I totally get that) I'll format my text once again with the sub-headings:

1. Background
2. Focus
3. Question

1. Background:

Let's assume a SHM oscillating system described by $$x_{SHM}=A~cos (\omega.t + \theta)$$
Further, let's assume a model of the spring force in the SHM oscillating system, being $$k.x=m.a=F_{SHM}$$
At any point at a distance $x$ from equilibrium ($x=0$) a restorative force $F_{SHM}$ acts on the system that can be expressed either in the form $m.a$ or $k.x$, where $k$ is expressed in units of mass per units of time squared. $k$ is defined as a linear coefficient since it is assumed that the relationship between $x$ and $F_{SHM}$ is usually linear.

Therefore, in order to "perfectly counter" the "restorative force" ignoring velocity $F_{SHM}$, i.e. the force required to lock the system into position at a distance $x$ ignoring velocity (except where velocity =0), a force of $F_{COUNTERSHM} = -F_{SHM}$ is required, i.e. $F_{SHM}$ equal in absolute value, but opposite in vector to $F_{SHM}$, whereby $+$ and $-$ are defined by convention. Since this "perfect countering force" $F_{COUNTERSHM}$ is not spring-driven and depends on $F_{SHM}$, we express it as $m.a$.

Thus we arrive at $$-k.x=m.a=-F_{SHM}=F_{COUNTERSHM}$$
But in practice we express the above equation as $$-k.x=m.a=F_{SHM}$$

Which can rewritten as $$-k.x=m.\frac {d^2x} {dt^2}$$

Now we introduce a damped harmonic motion (DHM) oscillating system, for which the equation is $$x_{DHM}=A~e^{-(\beta/2m)(t)}(cos\sqrt{\omega^2-(\beta/2m)^2}*t+\theta)$$ (Below, under (2.) Focus I'd like to question this derivation).

In DHM, the damping force $F_{D}$ can be described as $$F_{D}=-b.v$$

The linear damping coefficient $b$ is observed in low velocity motion systems damped by fluid. We determine the spring force $F_{DHM}$ (which is the result of damping) in terms of the effect of damping on the spring force $F_{SHM}$. Again, we also describe the "perfect countering force" $F_{COUNTERDHM}$, ignoring velocity, as opposite in vector to $F_{DHM}$.

We describe the damped spring force: $$-k.x-b.v=m.a=-F_{DHM}=F_{COUNTERDHM}$$
Again in practice, we express the above equation as $$-k.x-b.v=m.a=F_{DHM}$$

2. Focus:

Departing from $$-k.x=m.a=F_{SHM}$$ and $$-k.x-b.v=m.a=F_{DHM}$$ as derived above, we can see right away that $$for~a~given~value~for~x~and~b.v \neq 0,~the~value~for~a~in~F_{SHM}\neq the~value~for~a~in~F_{DHM}$$ (Not sure if my notation is correct here.)

Which is not surprising intuitively, since we expect $x_{SHM}$ to differ from $x_{DHM}$ and more specifically, expect the higher period caused by damping, given the same initial $x$ at $t=0$, to result in lower acceleration $a$ relative to $a$ in SHM.

However, in the derivation of $$x_{DHM}=A~e^{-(\beta/2m)(t)}(cos\sqrt{\omega^2-(\beta/2m)^2}*t+\theta)$$ we rewrite $$-k.x-b.v=m.a=F_{DHM}$$ as $$\frac {d^2x}{dt^2} + \left (\frac b m \right) \frac{dx}{dt} + \left (\frac k m \right)x=0$$ and substitute $$x=e^{\lambda(t)}$$$$\frac{dx}{dt}=\lambda(e^{\lambda(t)})$$$$\frac {d^2x}{dt^2}=(\lambda^2)(e^{\lambda(t)})$$

3. Question:

By proceeding with this derivation, it is incorrectly ignored that, as stated above, $$for~a~given~value~for~x~and~b.v \neq 0,~the~value~for~a~in~F_{SHM}\neq the~value~for~a~in~F_{DHM}$$

and falsely assumed what the above rules out, namely that $$\frac {d^2x}{dt^2}_{SHM} = \frac {d^2x}{dt^2}_{DHM}$$ which isn't the case.

in fact by calculation, $$\frac {d^2x}{dt^2}_{SHM} = A~\left( \frac k m \right)\cos (\sqrt {\left ( \frac k m \right)}*t)$$

and

$$\frac {d^2x}{dt^2}_{DHM} = A~(\frac {b} {2m})^2*e^(-(\frac {b} {2m})x)(x \cos\sqrt{\omega^2-(\frac {b} {2m})^2} +t)-2A(\frac {b} {2m})*e^(-(\frac {b} {2m})x)*cos(\sqrt{\omega^2-(\frac {b} {2m})^2}$$

How can it then be right to substitute $\frac {d^2x}{dt^2}=(\lambda^2)(e^{\lambda(t)})$
into $F_{DHM}$ as indicated above?

I've obviously missed something but can't see it.

Thanks for your consideration - if you haven't thrown your hands up over my dilettantism yet.

Best regards,

PG

#### Dale

Mentor
falsely assumed what the above rules out, namely that $$\frac {d^2x}{dt^2}_{SHM} = \frac {d^2x}{dt^2}_{DHM}$$ which isn't the case.
That is certainly not an assumption that is made, and I am not sure at all why you would believe this. Can you explain? (If you look at the equations you can clearly see that they are not the same so that assumption is definitely not made, so why do you believe it was)

Therefore, in order to "perfectly counter" the "restorative force" ignoring velocity FSHMFSHMF_{SHM}, i.e. the force required to lock the system into position at a distance xxx ignoring velocity (except where velocity =0), a force of FCOUNTERSHM=−FSHMFCOUNTERSHM=−FSHMF_{COUNTERSHM} = -F_{SHM} is required,
This counter force is very confusing and completely unnecessary. Perhaps it is the root of your confusion. I would recommend completely abandoning it and any conclusions associated with it.

To derive the equations of motion, all that is needed is a free body diagram. In the case of SHM there is one force (the spring) and in the case of DHM there are two (the spring and the drag). By Newton’s 2nd law the net force is equal to the mass times the acceleration. That is all you need for this derivation.

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#### Protea Grandiceps

Hi,

while I won't be addressing my posts to any particular individual (as much as I may appreciate any particular individual's interest) in order to make it clear that I'm directing them at anybody who is interested in them, I would like to give a quick shout-out to Dale in thanks for the effort that he's been putting into this thread. Apologies for the 3-day delay in my response.

This counter force is very confusing and completely unnecessary. Perhaps it is the root of your confusion. I would recommend completely abandoning it and any conclusions associated with it.

To derive the equations of motion, all that is needed is a free body diagram. In the case of SHM there is one force (the spring) and in the case of DHM there are two (the spring and the drag). By Newton’s 2nd law the net force is equal to the mass times the acceleration. That is all you need for this derivation.

Yes, I see this now. I made the mistake of thinking $k<0$. But $k>0$.

Protea Grandiceps said:

falsely assumed what the above rules out, namely thatd2xdt2SHM=d2xdt2DHMd2xdt2SHM=d2xdt2DHMwhich isn't the case.
That is certainly not an assumption that is made, and I am not sure at all why you would believe this. Can you explain? (If you look at the equations you can clearly see that they are not the same so that assumption is definitely not made, so why do you believe it was)
I'm still grappling with this.

Beginning with
$$-k.x=F$$ and $$-k.x-b.v=F$$
Just to stress aspects of these equations that are important here I want to rewrite them - even though this is redundant - as:
$$-k.x_{SHM}=F_{SHM}$$ and $$-k.x_{SHM}-b.v_{SHM}=F_{DHM}$$
It follows that if $b.v\neq0$, which is observed in DHM except at the peaks and troughs of the oscillating motion (assuming sinusoidal characteristics), $$F_{SHM}\neq F_{DHM}$$
Implicit in our discussion of the spring force equations is also our agreement that we are refining our modelling of a single given oscillating system by progressing from where we are describing it as SHM to where we are describing it as DHM. Therefore, $m$ is the same in $$F_{SHM}=m.a~~and~~F_{DHM}=m.a$$
Now if $F_{SHM} \neq F_{DHM}$, as we have established as indicated further above, and $m$ is the same in $F_{SHM}=m.a$ and $F_{DHM}=m.a$, then it follows that$$a_{F_{SHM}} \neq a_{F_{DHM}}$$
I'd like to write this as
$$-k.x_{SHM}-.v_{SHM}=m.a_{DHM}=F_{DHM}$$
But $$a=\frac {d^2x}{dt^2}$$
Specifically here $$a=\frac {d^2x_{SHM}}{dt^2}$$
and $$a=\frac {d^2x_{DHM}}{dt^2}$$
and as shown above $$a_{F_{SHM}} \neq a_{F_{DHM}}$$
then $$\frac {d^2x_{SHM}}{dt^2}\neq \frac {d^2x_{DHM}}{dt^2}$$
Now we recall that the equation describing the motion of an oscillating system in DHM is derived from the equation for the spring force $F_{DHM}$. To this end we write
$$-k.x_{SHM}-b.v_{SHM}=m.a_{DHM}=F_{DHM}$$as
$$\frac {d^2x_{DHM}}{dt^2}+(\frac{b}{m})\frac{dx_{SHM}}{dt}+(\frac{k}{m})x_{SHM}=0$$
And now comes the part with which I struggle to come to terms: There is a method by which second-order equations of the type $x'' + ax' + bx = 0$ are solved by making substitutions of $x=e^{\lambda t}$ for $x$ - I understand that. But the above equation is not of the type $x'' + ax' + by = 0$ since $d^2x$ refers to the DHM term and $\frac {dx}{dt}$ and $x$ refer to the SHM terms and therefore I wonder how it can be solved using this method. The above equation is really of the type $y'' + ax' + bx = 0$, I'd say. There obviously is a way to solve it that I've overlooked.

So ignoring my question for a moment, the above equation is solved by making the following substitutions:
$$x_{SUBST}=e^{\lambda t}~~~for~~~x_{SHM}$$
$$x_{SUBST}'=\frac{dx_{SUBST}}{dt}= \lambda {e^{ \lambda t}}~~~for~~~\frac{dx_{SHM}}{dt}$$
$$x_{SUBST}''=\frac{d^2x_{SUBST}}{dt^2}=\lambda^{2}{e^{ \lambda t}}~~~for~~~\frac{d^2x_{DHM}}{dt^2}$$
The equation describing the motion of an oscillating system in DHM $x_{DHM}$ is then, following further steps that I find fairly straightforward, determined to be
$$x_{DHM}=A~e^{(-b/2m)t}(\cos\sqrt{\omega^2-(\frac{b}{2m})^2}+\theta)$$
Personally, I find this form more commonly cited:
$$x_{DHM}=A~e^{-\gamma t}(\cos\omega' t + \theta)$$
where $\omega=\sqrt{\frac {k} {m}}~~~$,$~~~\omega'=\sqrt{\omega^2-\gamma^2}~~~$,$~~~\gamma=\frac {b} {2m}$

Thanks for listening as always and have a great weekend.

PG

#### Dale

Mentor
Beginning with
$$-k.x=F$$ and $$-k.x-b.v=F$$
Just to stress aspects of these equations that are important here I want to rewrite them - even though this is redundant - as:
$$-k.x_{SHM}=F_{SHM}$$ and $$-k.x_{SHM}-b.v_{SHM}=F_{DHM}$$
This part is not redundant, it is wrong.

To avoid using an excessive amount of symbols I will consistently use $\dot x$ instead of $v$ and $\ddot x$ instead of $a$, and I will leave the net force as $m\ddot x$ instead of writing $F$. Where you did so above, that is understood, but I prefer more consistent notation.

So here $x_{SHM}$ is the solution to the equation $-k x = m \ddot x$ and $x_{DHM}$ is the solution to the equation $-k x - b \dot x = m \ddot x$. So

$$-k x_{SHM} = m \ddot x_{SHM}$$ $$-k x_{DHM} - b \dot x_{DHM}=m \ddot x_{DHM}$$ $$m \ddot x_{SHM} \ne m \ddot x_{DHM}$$ $$-kx_{SHM}-b\dot x_{SHM}\ne m \ddot x_{DHM}$$
Indeed, although the term $b \dot x_{SHM}$ can be written down it doesn’t make any sense physically. The parameter b does not apply for the SHM system, as you specifically excluded $b \dot x =0$

Implicit in our discussion of the spring force equations is also our agreement that we are refining our modelling of a single given oscillating system by progressing from where we are describing it as SHM to where we are describing it as DHM.
If you want to do that then you simply solve the DHM equation and set b=0 and see what solution you obtain. Since you know that the DHM differential equation reduces to the SHM equation when b=0 then you should get that the SHM solution is equal to the DHM solution for b=0.

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#### Protea Grandiceps

Terrific, the penny finally dropped. That's what you already intended in an earlier response, that the reduced force in DHM vs SHM doesn't change the meaning of $x$, $x$ changes the same.

I saw your earlier message last night in which you quoted me without comment. You were drawing attention to my incorrect equation $$-k.x_{SHM}-b.v_{SHM}=F_{DHM}$$ It dawned on me what I hadn't seen, but I couldn't respond immediately and left it for the next morning. When I signed in this morning I saw that you had made some careful annotations, writing:$$-k.x_{DHM}-b.\dot x_{DHM}=m.\ddot x_{DHM}$$
and knew that I had finally been put on the right track.

Good notation it seems is not just a matter of form, but has an effect on the way you think. Straight out of Wittgenstein.

Thank you for your golden advice over the course of several posts.

Best regards,

PG

#### Dale

Mentor
I saw your earlier message last night in which you quoted me without comment.
Yes, that was a mistake on my part. I accidentally clicked the Post Reply button far too early. I hope that didn’t cause any confusion

#### vanhees71

Gold Member
I'm not sure what the question really is, but I guess it's about solving the equation, right?

Let me put it in a more convenient form
$$\ddot{x}+2 \gamma \dot{x}+\omega_0^2 x=0.$$
Here $\gamma=b/(2m)$ and $\omega_0^2=k/m$.

There are many ways to solve it. One is to use complex numbers and make the ansatz $x(t)=A \exp(\lambda t)$. In the following I'll use another way to be able to do everything within the realm of real functions.

The first step is to make the ansatz
$$x(t)=\exp(-\gamma t) y(t),$$
where $y$ is a new unknown function, we want to calculate. First put the ansatz into the equation. For that purpose we need
$$\dot{x}(t)=\exp(-\gamma t) [-\gamma y(t)+\dot{y}(t)],$$
$$\ddot{x}(t)=\exp(-\gamma t)[\gamma^2 y(t) -2 \gamma \dot{y}(t) + \ddot{y}(t)].$$
The merit of this ansatz becames clear now, when plugging it into the equation. Cancelling the overall factor $\exp(-\gamma t)$ you get
$$\ddot{y}+(\omega_0^2-\gamma^2) y=0.$$
As you see, the merit is that now there's no more term with $\dot{y}$.

Now the equation looks similar to the equation for the undamped oscillator, but we have to carefully discuss the three cases

(1) $\omega_0^2-\gamma^2=\omega^2>0$
(2) $\omega_0^2-\gamma^2=0$
(3) $\omega_0^2-\gamma^2=-\tilde{\gamma}^2<0$

Start with case (1). Then you have
$$\ddot{y}+\omega^2 y=0,$$
and this is the equation for a harmonic oscillator with the angular frequency $\omega$. The general solution is
$$y(t)=A \cos(\omega t) + B \sin(\omega t).$$
The corresponding general solution for the original equation is now given by the ansatz which introduced $y(t)$ above:
$$x(t)=[A \cos(\omega t) + B \sin(\omega t)]\exp(-\gamma t),$$
i.e., you have still oscillations but with an amplitude damped by the exponential factor; $A$ and $B$ are integration constants which can be used to fulfill the inital conditions, $x(0)=x_0$ and $\dot{x](0)=v_0$.

Case (2) is even simpler. Now the equation reads
$$\ddot{y}=0$$
with the general solution
$$y(t)=A t + B.$$
For $x$ we find
$$x(t)=(A t + B) \exp(-\gamma t).$$
It's not oscillating anymore but going to 0.

Case (3) gives the equation
$$\ddot{y}-\Gamma^2 y=0.$$
Here obviously the general solution is
$$y(t)=A \exp(\Gamma t) + B \exp(-\Gamma t)$$
and thus we find
$$x(t)=A \exp[(\Gamma-\gamma)t]+ B \exp[-(\Gamma+\gamma)t].$$
Since $\Gamma=\sqrt{\gamma^2-\omega^2}<\gamma$, both exponential functions are damped, and nothing oscillates.

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