Forced Oscillator with unfamiliar forcing function and constants

[oddjobmj]

Homework Statement

A force F_ext(t)=F₀[1-e^(-a*t)] acts, for time t>0, on an oscillator which is at rest at x=0 at time 0. The mass is m; the spring constant is k; and the damping force id -b dx/dt. The parameters satisfy these relations:

b=mq and k=4mq² where q is a constant with units of inverse time.

Find the motion. Determine x(t); and hand in a qualitatively correct graph of x(t).

(B) Determine the final position.

Homework Equations

Typical form of a damped oscillator:

$\ddot{x}$+2β$\dot{x}$+w²x=0

The Attempt at a Solution

This question is from an online, self-study course and we have no lecture for forced oscillators. I am left to figure this out with my book and the internet. I would greatly appreciate a few pokes in the right direction. All the examples I am finding on forced oscillators are of the more common form of: $\ddot{x}$+2β$\dot{x}$+w²x=Acos(wt)

Here is what I am working with:

$\ddot{x}$+2β$\dot{x}$+w²x=F[1-e^(-at)]

I understand that I need to find a solution to the homogeneous equation and also a particular solution and their sum will be my x(t). My solution to the homogeneous equation is simple if I can determine the nature of the damping (which I have problems with below). The tricky part then becomes the particular solution.

I also need to determine the nature of the damping. Is this overdamped, underdamped, or critically damped? How do I relate b=mq to b²=4mk which is the form I am more accustomed to if it were critically damped, for example? I believe that if someone helps me figure out the nature of the damping I will be able to piece together a particular solution.

(B) I see that the driving force quickly approaches F₀ and I could take the limit as t->∞ to find the final position. I need x(t) first, though.

Thank you for your time and help!

 

[vanhees71]

First you should write down the equation of motion with the quantities given in the question and then compare it to your standard form in order to sort out the constants of your equation properly.

For the particular solution, do you know the Green's function of the harmonic oscillator? Then you simply have to fold the external force with it.

 

[vela]

You've been given the forces, so start by applying F=ma to get a differential equation for x(t) in terms of b, k, and m. When you solve the homogeneous equation, you should be able to deduce which case you have for the given values of b and k.

 

[rude man]

Homework Statement

A force F_ext(t)=F₀[1-e^(-a*t)] acts, for time t>0, on an oscillator which is at rest at x=0 at time 0. The mass is m; the spring constant is k; and the damping force id -b dx/dt. The parameters satisfy these relations:

b=mq and k=4mq² where q is a constant with units of inverse time.

Find the motion. Determine x(t); and hand in a qualitatively correct graph of x(t).

(B) Determine the final position.

Homework Equations

Typical form of a damped oscillator:

$\ddot{x}$+2β$\dot{x}$+w²x=0

The Attempt at a Solution

This question is from an online, self-study course and we have no lecture for forced oscillators. I am left to figure this out with my book and the internet. I would greatly appreciate a few pokes in the right direction. All the examples I am finding on forced oscillators are of the more common form of: $\ddot{x}$+2β$\dot{x}$+w²x=Acos(wt)

Here is what I am working with:

$\ddot{x}$+2β$\dot{x}$+w²x=F[1-e^(-at)]

That equation is dimensionally inconsistent. The equation is ƩF = mx''. Fix that first.

Then rewrite 2β = 2ζw. What is the relation between ζ and the conditions of underdamping, critical amping and overdamping?

BTW you should use w0, not w in your starting equation. w is the actual frequency whereas w0 is the natural (undamped) frequency.

You should realize that it is unnecessary to know a priori if the system is underdamped, overdamped etc. But you need to be comfortable with imaginary numbers if the system is underdamped.
Remember that exp(ix) = cos(x) + i sin(x).

 

[oddjobmj]

Thank you both for your help.

m$\ddot{x}$+b$\dot{x}$+kx=0

Dividing by m in the homogeneous form and replacing b/m=2B and k/m=w₀²

Then plugging in the known relations b=mq and k=4mq² I get:

2B=q and w₀²=4q²

B=q/2

w₀=2q

If that is the case then W₀ is clearly larger than B telling me this is underdamped.

Is that what you mean vanheese71?

-----

Alternatively, I can look for a general solution to the homogeneous equation:

m$\ddot{x}$+b$\dot{x}$+kx=0

x(t)=C₁e^{$\frac{t(sqrt(b^2-4km)+b)}{2m}$}+C₂e^{$\frac{t(sqrt(b^2-4km)-b)}{2m}$}

I can plug in b=mq and k=4mq² but it is messy and I don't quite see where to go from there.

 

[vela]

The square root will turn out to be imaginary, which gives rise to oscillating solutions, i.e., it's underdamped.

For the particular solution, try using the method of undetermined coefficients.

 

[rude man]

It's messy but it can be done.

However, the best way is for you to find the solution to the homogeneous equation in the underdamped case. Many websites can give you this.

As for the forcing function, you probably haven't had Green's function, and I assume also not the Laplace transform, so you need to guess the form of x(t) for the forcing function case. To help you with this, note that F contains a constant plus an exponential, and that you're likely to have sin and cos in there also since it's underdamped.

HOWEVER: don't even try to come up with the exact x(t)! It's VERY messy! (It would take me 1/2 hr to write it down for you!) All you're asked is to produce a qualitatively correct curve.

So, realizing that F = 0 for t = 0 you know that initially it's just the homogeneous response. And for the steady-state response, which means many time constants of the homogeneous solution, F = constant = F0. So you can assume the steady-state equation to be just mx'' + bx' + kx = F0.

For part B: if you use 'common sense' and think of the physics of the situation as in a mass-spring situation so that damping is zero, x(infinity) should be obvious.(hint: x'' = x' = 0 at t = infinity.)

Put all the pieces together and you should be able to come up with a plausibly accurate graph of x(t) vs. t.

 

[oddjobmj]

That equation is dimensionally inconsistent. The equation is ƩF = mx''. Fix that first.

mx''=-bx'-kx+F[1-e^(-at)]

mx''+bx'+kx=F[1-e^(-at)]

x''+2Bx'+w₀²=[1-e^(-at)]/m, 2B=b/m, w₀²=k/m

Is that what you meant by fixing the equation?

Then rewrite 2β = 2ζw. What is the relation between ζ and the conditions of underdamping, critical amping and overdamping?

I'm not sure where zeta comes in. I have yet to come across that symbol in my study of oscillators that I can recall.

BTW you should use w0, not w in your starting equation. w is the actual frequency whereas w0 is the natural (undamped) frequency.

You're absolutely right. Sorry I missed it!

You should realize that it is unnecessary to know a priori if the system is underdamped, overdamped etc. But you need to be comfortable with imaginary numbers if the system is underdamped.
Remember that exp(ix) = cos(x) + i sin(x).

You're right. I was subconsciously trying to skip right to the typical form for a particular type of damping. I did work out a general solution to the homogeneous equation in my last post (I missed your post when replying) which I think allowed me to relate b, m, k, and q with and find out that it is underdamped. If my conclusion is correct I hope that I arrived at the correct conclusion in a sensible way because I definitely want/need to understand it instead of happening across the correct solution.

EDIT: Just noticed two new responses. Checking them out now!

 

[oddjobmj]

The square root will turn out to be imaginary, which gives rise to oscillating solutions, i.e., it's underdamped.

For the particular solution, try using the method of undetermined coefficients.

Ahh, fantastic. Both methods have brought me to the same conclusion. I feel that the first route was more natural for me. Does what I did seem to hold water because that's the work I will show if so?

However, the best way is for you to find the solution to the homogeneous equation in the underdamped case. Many websites can give you this.

Yes, that is sort of why I was trying to jump ahead earlier and guess the nature of the damping. I'm pretty comfortable with finding the solutions to the homogeneous portion.

As for the forcing function, you probably haven't had Green's function, and I assume also not the Laplace transform, so you need to guess the form of x(t) for the forcing function case. To help you with this, note that F contains a constant plus an exponential, and that you're likely to have sin and cos in there also since it's underdamped.

I did happen across a solution that used Green's theorem which is, you are correct, unfamiliar to me. I should learn that method because I will definitely use it eventually but as noted below it may not be immediately necessary. Thank you for pointing that out.

So, realizing that F = 0 for t = 0 you know that initially it's just the homogeneous response. And for the steady-state response, which means many time constants of the homogeneous solution, F = constant = F0. So you can assume the steady-state equation to be just mx'' + bx' + kx = F0.

I'm not entirely sure what you mean by many time constants of the homogeneous solution but I can see that the 1-e^(-at) portion quickly approaches 1 as t increases making the forcing function F₀ after short t. So, I agree, the steady state equation is what you have shown.

How, though, do I produce a graph that considers that short period of dynamic force? Sure, the initial response is simply the homogeneous solution and after short t it is what you have shown but between t=0 and t=short?

In addition to producing the graph I have to show x(t). Can I represent x(t) as a homogeneous solution plus the steady state function solution? I thought I had to use a particular solution that included the 1-e^(-at)

 

[vela]

For the particular solution, assume it has the form A+Be^-at. Plug it into the differential equation and then solve for A and B.

 

[oddjobmj]

For the particular solution, assume it has the form A+Be^-at. Plug it into the differential equation and then solve for A and B.

Thank you.

I was told my original differential equation was dimensionally inconsistent. Is the equation below correct?

x''+2*Bx'+w₀²x=$\frac{F_0(1-e^{-at})}{m}$

Of course, at t=0 there is no driving force so it would be homogeneous.

 

[rude man]

mx''=-bx'-kx+F[1-e^(-at)]

mx''+bx'+kx=F[1-e^(-at)]

x''+2Bx'+w₀²=[1-e^(-at)]/m, 2B=b/m, w₀²=k/m

Is that what you meant by fixing the equation?

Yes.

I'm not sure where zeta comes in. I have yet to come across that symbol in my study of oscillators that I can recall.

It's called the damping coefficient. There is a very simple relationship between it and the condition of damping.

You're absolutely right. Sorry I missed it!

You're right. I was subconsciously trying to skip right to the typical form for a particular type of damping. I did work out a general solution to the homogeneous equation in my last post (I missed your post when replying) which I think allowed me to relate b, m, k, and q with and find out that it is underdamped. If my conclusion is correct I hope that I arrived at the correct conclusion in a sensible way because I definitely want/need to understand it instead of happening across the correct solution.

EDIT: Just noticed two new responses. Checking them out now!

OK, will keep tabs.

 

[oddjobmj]

I'm trying to figure out which way to go. You have suggested, if I am interpreting your response correctly, that finding a particular solution to the transient function is unnecessary. Whereas Vela has suggested an educated guess on what the solution would look like, plugging it in, and solving for the particular.

I have responded since your most recent post and I believe that more recent post better expresses my current understanding and weaknesses.

 

[rude man]

I'm trying to figure out which way to go. You have suggested, if I am interpreting your response correctly, that finding a particular solution to the transient function is unnecessary. Whereas Vela has suggested an educated guess on what the solution would look like, plugging it in, and solving for the particular.

I have responded since your most recent post and I believe that more recent post better expresses my current understanding and weaknesses.

The homogeneous solution is x(t) = 0 for all t since the initial conditions are both zero x(0) = x'(0) = 0.

So divide the forcing function into two sections: F1 = F0 and F2 = -F0 exp(-at), t > 0. Then the complete x(t) is the sum of the effects of F1 and F2.

There is no separate 'transient' solution. There is only one solution, caused by F1 + F2. The steady-state solution is x(infinity), when x = x' = 0. So x(infinity) = a constant ≠ 0.

Choosing trial x(t) = A + B exp(-at) will not work. (I know 'cause I worked the problem!). There will be a simple ~ exp(-at) term but then also a second term ~ exp(-dt) sin(g)t where d and g are very involved expressions. It should not be surprising to find sin terms for an underdamped system.

Anyway - pick whatever floats your boat!

 

[oddjobmj]

The homogeneous solution is x(t) = 0 for all t since the initial conditions are both zero x(0) = x'(0) = 0.

That should have been obvious from the get go. I'm sort of embarrassed to not have realized that earlier. Thank you. Looking back at the thread, though, I realize we may have all be missing that at first. :tongue:

F1 = F0 and F2 = -F0 exp(-at)

If that is not a typo (did you mean F₀[1-e^(-at)]? otherwise, where did that come from?) then at t=∞, F2=0 and the sum of F1 and F2 is simply F₀. This is a bit confusing, really, because the way I'm interpreting your comment regarding the homogeneous solution is that, effectively, there is no oscillation. We're dealing with a driving force and a resisting force. I don't know how else to interpret this besides simple motion equations:

F=ma => a=F/m => dv/dt=F/m => dv=F/m dt

Plug in F and solve.

Or do you mean proceed with x''+2Bx'+w₀²=Forces and just don't expect a contribution from the homogeneous solution. If that is the case then I still am not sure how to solve this differential equation. Sorry for the misunderstanding... I've never come across a problem like this before. Perhaps I am over thinking it. An entire day of crunching physics problems tends to make me a little loopy.

(B) Given the above I imagine this problem as a spring attached to wall and someone grabbing the spring, giving it a pull, and holding it at its final stretched position. The response to the 'driving' force is the spring constant which is dependent on x. Once the force of kx=F₀ the system will stop moving. I just need to solve for x at that point which I guess will have to be in terms of k and F₀ seeing as how I'm not given any values. Does this sound right?

 

[oddjobmj]

Eh, just realized when I lied down for bed where those two pieces F1 and F2 came from. Multiply through... I'll have to re-think the problem tomorrow but my confusion still stands.

 

[vela]

The homogeneous solution is x(t) = 0 for all t since the initial conditions are both zero x(0) = x'(0) = 0.

That should have been obvious from the get go. I'm sort of embarrassed to not have realized that earlier. Thank you. Looking back at the thread, though, I realize we may have all be missing that at first. :tongue:

No one missed that point because it's not correct. The initial conditions need to be satisfied by the complete solution. You can't apply it to only the homogeneous solution to solve for the constants.

As I said earlier, just set the particular solution to $x_p(t) = A + Be^{-at}$, plug it into the differential equation, and solve for A and B. Once you have the complete solution, then apply the initial conditions x(0)=x'(0)=0 to solve for the constants c₁ and c₂.

 

[rude man]

I have to say that this problem is hard to address without invoking the complete expression for x(t) which is very elaborate, involving an exponential and an exponential multiplied by a sine term. This is due to the fact that unfortunately the system is underdamped.

Vela's approach will not work unless his "a" can be complex. This is obvious because no products of exponentials and sines can otherwise ensue. But maybe he did have a complex "a" in mind. I'm not saying that will work either, but it might.

I solved this by the Laplace transform which makes it a piece of cake IF you have a good table of transforms or are willing to shell out $$ to Wolfram Alpha.

 

[vela]

The particular solution doesn't contain any oscillating terms. I'm not sure why you keep insisting it does. Those terms are part of the homogeneous solution.

 

[vanhees71]

It's not so difficult. Either you use the Green's function (which leads to a somewhat elaborate integration) or you use the already suggested "ansatz of the form of the right-hand side", i.e.,
$$x(t)=A+B \exp(-a t)$$
and determine values for $A$ and $B$. That's perhaps even simpler than the method with the Green's function, although you have to add the general solution of the homogeneous equation and determine the coefficients to fulfill the initial condition.

 

[rude man]

The particular solution doesn't contain any oscillating terms. I'm not sure why you keep insisting it does. Those terms are part of the homogeneous solution.

The homogeneous solution is x(t) = 0 for all t since the two initial conditions are both zero.

But x(t) with F as the forcing function does include a sinusoid - a decaying sinusoid, due to the fact that this is an underdamped system. I don't know why you persist in stating that it does not.

FYI the term is
(a²F₀/m) exp(-ζω₀t) sin[ω₀√(1 - ζ²)t - ψ]

divided by
√(1 - ζ²) (1 - 2ζTω₀ + T²ω₀²)^1/2.

where T = 1/a, ζ is the damping coefficient (I think 1/4 here, not sure) and ω₀ is the natural frequency. (My Laplace tables are engineering-oriented so please excuse the substitution T = 1/a).

Your and vanhees' "guess" of x = A + B exp(-at) covers a part of the particular solution only. The complete particular solution is my term plus (presumably) your term.

I won't elaborate further, for example giving the full expression for the phase angle ψ, nor will I give the rest of the solution icluding a constant plus a decaying exponential term which I believe is the result of the x = A + B exp(-at) guess. My fingers would probably go numb.

 

[Office_Shredder]

rude man, "the particular solution" is not the solution of the problem, it's a solution to the differential equation (NOT including the initial condition). Once you have that, you can add the homogeneous solution and then find the coefficients that make the initial conditions work.

 

[Mark44]

rude man, what you are saying is incorrect. The particular solution does not have a sinusoidal component. The only oscillating component of the general solution comes from the complementary function - the solution to the homogeneous problem.

As has already been stated twice, the best choice for a particular solution is y_p = A + Be^-at. After the A and B coefficients have been found, then combine the particular solution with the complementary function to find the solution of the initial value problem.

 

[rude man]

rude man, "the particular solution" is not the solution of the problem, it's a solution to the differential equation (NOT including the initial condition). Once you have that, you can add the homogeneous solution and then find the coefficients that make the initial conditions work.

I realizr that.

OK, semantics aside, I have described the total solution to x(t) and it is apparently still under dispute, in particular, whether or not a sinusoid is included.

When using Laplace there is no distinction made between 'homogeneous', 'particular', etc. You get the total solution including initial conditions by the expedient of finding the transform in a set of tables. Unfortunately, most physicists are not familiar with this transform, judging by my experience, so please don't take this as some kind of insult to physicists, whom I on the whole consider a lot smarter than the average EE.

 

[rude man]

rude man, what you are saying is incorrect. The particular solution does not have a sinusoidal component. The only oscillating component of the general solution comes from the complementary function - the solution to the homogeneous problem.

As has already been stated twice, the best choice for a particular solution is y_p = A + Be^-at. After the A and B coefficients have been found, then combine the particular solution with the complementary function to find the solution of the initial value problem.

Semantics, semantics.

I would really like to see the derivation of the complete x(t) using x = A + B exp(-at) as the guess.

It can only happen if the characteristic equation's roots are complex. I concede that point. I also concede that my knowledge of what is 'particular', what is 'complementary', etc. is wanting. As I explained, this is because with the Laplace approach such distinctions are simply not made.

I still would like to see the derivation though. If the total x(t) does not wind up in the form I presented then I know it's wrong.

 

[Office_Shredder]

Semantics, semantics.

I would really like to see the derivation of the complete x(t) using x = A + B exp(-at) as the guess.

You solve for A and B then add the homogeneous solution to the differential equation, which is NOT 0 as you claimed before. It (the homogeneous plus the particular solution) has two arbitrary coefficients, which you plug in the initial conditions to solve for. Something that is taught in any introductory differential equations course.

 

[oddjobmj]

Wow, I greatly appreciate all the help here. Thank you all!

I'm going to use C₁ and C₂ because my differential equation already has β and that would be unnecessarily confusing.

If we assume x=C₁+C₂e^(-at) then:

x'=-aC₂e^(-at)
and
x''=a²C₂e^(-at)

I can plug those into my differential equation:

x''+2βx'+w₀²x=$\frac{F_0(1-e^{-at})}{m}$

If I do that and solve for C₁ and C₂ I get:

C₁=$\frac{-F_0e^{-at}(e^{at}-1)}{am(a-2β)}$

and

c₂=$\frac{-C_1w^2me^{at}-F_0e^{at}+F_0}{m(a^2-2aβ+w^2)}$

If I plug either one of those into the original x=C₁+C₂e^(-at) I find that both constants are working out to 0 when x=t=0. Which makes sense because the force is not acting at t=0. Should I proceed before solving for the constants? The advice in the thread is diverging even besides the homogeneous solution not being zero bit.

x(t)=C₁+C₂e^(-at)+[Homogeneous Solution with C₁ and C₂]

THEN plug in my initial conditions to solve for the constants?

Anyway, the typical form of the homogeneous solution for an underdamped oscillator is:

x(t)=e^-βt(C₁cos(w₁t)+C₂sin(w₁t)), where w₁=$\sqrt{w_0^2-β^2}$

 

[rude man]

I am leaving this to others to finish. See post

I can only say again that the complete solution for x(t) is VERY involved and I'm not sure how it can be simplified enough for you to come up with a qualitatively correct graph of x(t).

I would be surprised and chastened to see anyone else come up with a solution for you.

 

[oddjobmj]

I appreciate your help, rude man. I have no reason to disregard your advice and have tried to move forward with it despite missing some information. I have mathematica so extended computing time on alpha is not an issue. If I need to reduce the equation it can do that. I haven't actually been able to produce an equation for x(t) complicated or not because my constants keep working out to zero.

Even if you humor them and help me through their suggestion so I can see why their solution will not work I would appreciate that. Or perhaps it does work but it is rather messy and I can do the simplifying, if possible, to figure out how to graph it.

 

[rude man]

I appreciate your help, rude man. I have no reason to disregard your advice and have tried to move forward with it despite missing some information. I have mathematica so extended computing time on alpha is not an issue. If I need to reduce the equation it can do that. I haven't actually been able to produce an equation for x(t) complicated or not because my constants keep working out to zero.

Even if you humor them and help me through their suggestion so I can see why their solution will not work I would appreciate that. Or perhaps it does work but it is rather messy and I can do the simplifying, if possible, to figure out how to graph it.

Thing is, I know the solution and I don't know that they do. And the only reason I know it is that I use the Laplace transform. I have often bewailed the fact that this is not typically taught in a course on differential equations. I know the reason: the Laplace is very complex to describe and justify, is useless without tables, but - if you want a solution to a linear diff. eq. (including partial diff. eq.!) with constant coefficients then it makes the task exceedingly simple.

All hinges on whether their "guess" is good or not. All in all I think you should follow their advice since I doubt you have had the Laplace. If you or anyone else ever comes up with a total x(t) I can verify whether it's right or not. That's the one thing I do know, is that I know the answer!
So, doing it their way: your characteristic equation is
r^2 + 2βr + w₀² = (r + r₁)(r + r₂) = 0

x = C1 exp(r₁t) + C2 exp(r₂t).
r₁ and r₂ will be complex conjugates so you have to know how to handle this to wind up with a real function of time eventually.

Then, process the forcing function: their 'guess": x = A + B exp(-at).
Take 1st & 2nd derivatives, susbstitute into the original homogeneous equation and solve for A and B. Do this by equating coefficients of 1 and exp(-at). (2 eq., 2 unknowns).

Solve for C1 and C2 by taking the total x(t) with C1 and C2 and the solved-for A and B, and solve for C1 and C2 with the given initial conditions (x = x' = 0). Again, that's 2 eq. & 2 unknowns. That would give you the total x(t).

Neither C1 nor C2 will be zero. One reason is, as I said, r₁ and r₂ have to be complex conjugates of each other & that is not possible unless C1 and C2 are finite.

(Finally: demand that they teach you the Laplace transform method! If you plan on an engineering career, that is essential. If you go the physics route it seems not to be very popular.).

Happy headaches!

 

[ehild]

Wow, I greatly appreciate all the help here. Thank you all!

I'm going to use C₁ and C₂ because my differential equation already has β and that would be unnecessarily confusing.

If we assume x=C₁+C₂e^(-at) then:

x'=-aC₂e^(-at)
and
x''=a²C₂e^(-at)

I can plug those into my differential equation:

x''+2βx'+w₀²x=$\frac{F_0(1-e^{-at})}{m}$

If I do that and solve for C₁ and C₂ I get:

C₁=$\frac{-F_0e^{-at}(e^{at}-1)}{am(a-2β)}$

and

C₂=$\frac{-C_1w^2me^{at}-F_0e^{at}+F_0}{m(a^2-2aβ+w^2)}$

C1 and C2 are constant. You considered them functions of t which is wrong.

If you substitute x, x', and x'' into the differential equation, you have constant terms and terms with the common factor e^-at on both sides. The equation has to be valid for all t, so the coefficients of the exponent must be the same and so are the constant terms.

With the given condition, your differential equation is x"+qx'+4q²x=(F_o/m) (1-e^-at) .

The particular solution is supposed to be $x=C_1+C_2e^{-at}$.

Substituting x, x', x'' :

a²C₂e^-at-qaC₂e^-at+4q²(C₁+C₂e^-at)=F_o/m (1-e^-at)

The constant terms have to be the same on both sides:

4q²C₁=F_o/m-->$C_1=\frac{F_o}{4q^2m}$.

Collect the exponential terms:
e^-atC₂(a²-qa+4q²) =-F_o/me^-at, so

$C_2=\frac{-F_o}{m(a^2-qa+4q^2)}$

Add the particular solution to the general solution of the homogeneous equation $y_h=e^{-0.5qt}(A\sin(ωt)+B\cos(ωt))$ and find A, B from the initial conditions.

ehild

 

[rude man]

Oddjobmj, ehild took a good step in rewriting the equation in therms of q. It makes life easier.

In case you're scratching your head over how he got his y_h equation, that derives from what I showed you in re: the characteristic roots r₁ and r₂.

These are complex so in order to go from the usual A exp(r₁t) + B exp(r₂t) you just use Euler's rule & if all went well you will get his y_h. Note that he reversed A and B with C1 and C2 from what we were using.

 

[oddjobmj]

These are complex so in order to go from the usual A exp(r1t) + B exp(r2t) you just use Euler's rule & if all went well you will get his yh. Note that he reversed A and B with C1 and C2 from what we were using.

Ah yes, this simplification was covered in the lectures for damped oscillations.

The constant terms have to be the same on both sides:

Ah, wow. That's a nifty trick I didn't consider, thank you!

And... this is one hell of an equation. I've been trying to reduce this for two hours now. No luck.

I didn't use q until later.

My particular solution to x(t):

$\frac{F_0}{mw_0^2}$-$\frac{F_0e^{-at}}{m(a^2-2βa+w_0^2)}$

My homogeneous solution to x(t):

e^-βt[C₁cos(w₁t)+C₂sin(w₁t)]

Full solution:

x(t)=$\frac{F_0}{mw_0^2}$-$\frac{F_0e^{-at}}{m(a^2-2βa+w_0^2)}$+e^-βt[C₁cos(w₁t)+C₂sin(w₁t)]

Setting x=t=0 and solving for the constants:

C₁=$\frac{F_0}{m(a^2-2βa+w_0^2)}$-$\frac{F_0}{mw_0^2}$

C₂=$\frac{-aF_0(β(a-2β)+w_0^2)}{mw_0^2w_1(a(a-2β)+w_0^2)}$

You can imagine what the full solution for x(t) looks like. I can simplify some variables with the initial relationships:

b=mq

k=4mq²

β=$\frac{b}{2m}$=$\frac{q}{2}$

w₀²=$\frac{k}{m}$=4q²

w₁=$\sqrt{w_0^2-β^2}$=$\sqrt{k/m-b^2/4m^2}$=$\sqrt{4q^2-q^2/4}$=$\frac{sqrt(15)q}{2}$

So now I have x(t) in terms of F₀, m, a, q, and t. Any attempt to plug this into something like mathematica ends poorly. I'll give it another go with a more formal syntax but any suggestions would be welcome.

Thanks again!

EDIT: For fun, here is my one-line notation:

[F/(m*4*q^2)]-[F*e^(-a*t)/(m*((a^2)-2*(q/2)*a+(4*q^2)))]+e^((-q/2)*t)*[((F/(m*(a^2-2*(q/2)*a+4*q^2)))-(F/(m*4*q^2)))*cos(t*q*sqrt(15)/2)+((-a*F*((q/2)*(a-2*(q/2))+4*q^2))/(m*(4*q^2)*(q*sqrt(15)/2)*(a*(a-2*(q/2))+4*q^2)))*sin(t*q*sqrt(15)/2)]

Here is alpha's response:

"Using closest interpretation: cos"

Yeah... that's what I meant...

Mathematica came up with this as the most reduced form it could find:

http://i.imgur.com/waPS3Ei.png

I'll have to go back through and double-check the work to make sure I didn't make a silly mistake. (i.e. to find the mistake I made)

 

[rude man]

Ah yes, this simplification was covered in the lectures for damped oscillations.

Ah, wow. That's a nifty trick I didn't consider, thank you!

And... this is one hell of an equation. I've been trying to reduce this for two hours now. No luck.

I didn't use q until later.

My particular solution to x(t):

$\frac{F_0}{mw_0^2}$-$\frac{F_0e^{-at}}{m(a^2-2βa+w_0^2)}$

My homogeneous solution to x(t):

e^-βt[C₁cos(w₁t)+C₂sin(w₁t)]

Full solution:

x(t)=$\frac{F_0}{mw_0^2}$-$\frac{F_0e^{-at}}{m(a^2-2βa+w_0^2)}$+e^-βt[C₁cos(w₁t)+C₂sin(w₁t)]

Setting x=t=0 and solving for the constants:

C₁=$\frac{F_0}{m(a^2-2βa+w_0^2)}$-$\frac{F_0}{mw_0^2}$

C₂=$\frac{-aF_0(β(a-2β)+w_0^2)}{mw_0^2w_1(a(a-2β)+w_0^2)}$

You can imagine what the full solution for x(t) looks like. I can simplify some variables with the initial relationships:

b=mq

k=4mq²

β=$\frac{b}{2m}$=$\frac{q}{2}$

w₀²=$\frac{k}{m}$=4q²

w₁=$\sqrt{w_0^2-β^2}$=$\sqrt{k/m-b^2/4m^2}$=$\sqrt{4q^2-q^2/4}$=$\frac{sqrt(15)q}{2}$

So now I have x(t) in terms of F₀, m, a, q, and t. Any attempt to plug this into something like mathematica ends poorly. I'll give it another go with a more formal syntax but any suggestions would be welcome.

Thanks again!

EDIT: For fun, here is my one-line notation:

[F/(m*4*q^2)]-[F*e^(-a*t)/(m*((a^2)-2*(q/2)*a+(4*q^2)))]+e^((-q/2)*t)*[((F/(m*(a^2-2*(q/2)*a+4*q^2)))-(F/(m*4*q^2)))*cos(t*q*sqrt(15)/2)+((-a*F*((q/2)*(a-2*(q/2))+4*q^2))/(m*(4*q^2)*(q*sqrt(15)/2)*(a*(a-2*(q/2))+4*q^2)))*sin(t*q*sqrt(15)/2)]

Here is alpha's response:

"Using closest interpretation: cos"

Yeah... that's what I meant...

Mathematica came up with this as the most reduced form it could find:

http://i.imgur.com/waPS3Ei.png

I'll have to go back through and double-check the work to make sure I didn't make a silly mistake. (i.e. to find the mistake I made)

You're doing admirably, but I strongly recommend you use ehild's equation in q. I wish I would have thought of it myself. It takes care of several unneeded parameters. You wind up with just q, F and a.

In fact, his post #31 summarizes things very neatly. I suggest you follow his steps and then we can see how all winds up.

 

[oddjobmj]

Thanks again.

Same deal, actually. The constants have the same form and I'd be willing to bet the two solutions I found are equivalent.

I'm not sure how you got rid of m, t, and w, though. To get rid of w I have to substitute back in the imaginary portion of the root. If you were able to get rid of t and m I will walk back through the problem and type it out if necessary.

It is still a huge mess after the simplification. I don't think that is an issue, though. The graph doesn't have to be quantitatively correct. It just has to qualitatively represent the behavior this arrangement would exhibit. The plot of the final function looks like I would expect it to if I strategically pick values for the variables.