Forced Oscillator with unfamiliar forcing function and constants

[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.

 

[rude man]

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.

Sorry, I meant A and B (his C1 and C2) contain only F₀, q, m and a. Certainly, you still wind up with m, t and w as part of the complete x(t).

Note that his w is not the natural frequency w₀, it's w₀√(1-ζ²)
where ζ is the damping coefficient = β/w₀.

Anyway, I have no reason to think your sol'n isn't right. As I warned you long ago, this thing is a mess!

BTW I would combine the sin and cos terms into a sin(wt + ψ) term. Might make the graphing easier.

 

[ehild]

The x(t) function is the sum of a constant term, an exponentially decreasing term and an oscillating term with exponentially decreasing amplitude. And both x(t) and its derivative are zero at t=0. You can tell at once how the function behaves as the time tends to infinity. Can you sketch it ?

ehild

 

[oddjobmj]

Here is what I came up with. The image contains both my x(t) function and a picture of the plot. If that is not how it is supposed to look I'll walk back through the steps again.

http://imgur.com/R5ImlBm

Edit:

As t-> infinity x approaches that first piece:

F/4mq^2

 

[Mark44]

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 can't speak for all college instructors, but in the differential equations courses I taught in a community college over many years, one of the techniques I presented was using the Laplace transform to solve DEs.

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.

This is a silly remark. With a forcing function as simple as 1 - e^-at it's not rocket science to come up with a particular solution. Once you have the solution to the homogeneous DE and a particular solution, you combine the two solutions, and use the initial conditions to find the solution to the given differential equation.

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.

As can the OP. Once you have arrived at a solution, it's always a good idea to check two things:
1. Do x(t) and x'(t) satisfy the initial conditions?
2. Does x(t) and its derivatives satisfy the differential equation?
If you can answer "yes" to these two questions, then all is good.

 

[vela]

Here is what I came up with. The image contains both my x(t) function and a picture of the plot. If that is not how it is supposed to look I'll walk back through the steps again.

http://imgur.com/R5ImlBm

Edit:

As t-> infinity x approaches that first piece:

F/4mq^2

It looks pretty except for one thing. Remember that $\omega$ isn't independent of $q$. It's equal to $\omega = \frac{\sqrt{15}q}{2}$.

What sometimes helps with the algebra is expressing things in terms of a dimensionless combination of constants. For example, you found that the particular solution was given by
$$x_p(t) = \frac{F_0}{4mq^2} - \frac{F_0}{m(a^2-aq+4q^2)}e^{-at}.$$ $a$ and $q$ both have units of 1/time. Notice how $4q^2$ shows up in both terms. If you were to factor $4q^2$ out of the quadratic and define $\alpha=\frac{a}{2q}$, you'd end up with
$$x_p(t) = \frac{F_0}{4mq^2} - \frac{F_0}{4mq^2(\alpha^2-\frac{1}{2}\alpha+1)}e^{-at} = \frac{F_0}{4mq^2}\left[1-\frac{1}{\alpha^2-\frac{1}{2}\alpha+1}e^{-at}\right].$$ The factor out front is a constant, and you can pretty easily check that it has dimensions of length, which is what you expect for x(t). It pretty much just comes along for the ride in the rest of your calculations since the time dependence is all contained in the stuff in the square brackets. You can write the complete solution as
$$x(t) = \frac{F_0}{4mq^2}\left[1-\frac{1}{\alpha^2-\frac{1}{2}\alpha+1}e^{-at}+c_1 \cos \frac{\sqrt{15}q}{2}t + c_2 \cos \frac{\sqrt{15}q}{2}t\right].$$ When solving for the constants by applying the initial conditions, you don't have to deal with $a$ and $q$ separately for the most part, but only with $\alpha$, which makes it much less likely you'll make a mistake. (Plus it's less writing.)

 

[oddjobmj]

Thank you, Vela.

The equation seems to exhibit the characteristics that I would expect from the description of the system. The graph matches my expectations as well other than x(0)!=0 or at least it seemed that way on the plot.

It seems you are suggesting that I have more to do. Are you suggesting I run through it again or just pointing out an easier way to go about it if I end up having to work through it again?

I will plug the actual value back in for w which I did forgot to do before screen capping that plot.

 

[vela]

I can't speak for all college instructors, but in the differential equations courses I taught in a community college over many years, one of the techniques I presented was using the Laplace transform to solve DEs.

I agree it's a pretty standard topic for an introductory differential equations course as far as I can tell.

I don't think using Laplace transforms really buys you anything for this problem. You're just trading one set of somewhat tedious algebra for another. It's pretty straightforward to solve either way (though I know the OP probably doesn't think so ).

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.

Real engineers don't use tables; they evaluate the Bromwich integral.

 

[vela]

Thank you, Vela.

The equation seems to exhibit the characteristics that I would expect from the description of the system. The graph matches my expectations as well other than x(0)!=0 or at least it seemed that way on the plot.

If your solution doesn't satisfy x(0)=0, then you have a problem. In Mathematica, use the option Plotrange->All to see the entire plot.

It seems you are suggesting that I have more to do. Are you suggesting I run through it again or just pointing out an easier way to go about it if I end up having to work through it again?

Other than fixing $\omega$, I don't think you have anything more to do. That said, if you're inclined to, I would encourage you to try going over the problem again to see if you can figure out how to do the algebra more efficiently. It never hurts to develop your algebra skills so that you can do future problems more quickly and with fewer mistakes.

 

[oddjobmj]

I was wrong. The plot does show x(0)=0 but with the values for the variables I plugged in it just didn't display the plot with a high enough resolution to see that.

Other than fixing ω, I don't think you have anything more to do. That said, if you're inclined to, I would encourage you to try going over the problem again to see if you can figure out how to do the algebra more efficiently. It never hurts to develop your algebra skills so that you can do future problems more quickly and with fewer mistakes.

I've been through this problem probably 5-6 times now. I will do that, though, because I have to condense the work to fit on to one page. I've been nabbed a significant number of points for not deriving what I thought to be elementary relations more than once already so... I have a lot to fit in.

Thanks again for your help! The discussion has been fruitful well beyond the scope of this problem.

 

[rude man]

I agree it's a pretty standard topic for an introductory differential equations course as far as I can tell.

I don't think using Laplace transforms really buys you anything for this problem. You're just trading one set of somewhat tedious algebra for another. It's pretty straightforward to solve either way (though I know the OP probably doesn't think so ).


Real engineers don't use tables; they evaluate the Bromwich integral.

Funny!

But - what 'tedious algebra'? It's all printed out for you. I hereby offer to race you any time with a similar problem. I solved this equation in the time it took to find the appropriate transform. If the equation had not been so nicely simplifiable (coefficients in q only) plus a fancier forcing function (how about F = δ(t = T)? Your "guess" here?) I doubt you could have competed with the Laplace method.

 

[oddjobmj]

This is getting interesting :D

 

[Mark44]

Funny!

But - what 'tedious algebra'? It's all printed out for you. I hereby offer to race you any time with a similar problem. I solved this equation in the time it took to find the appropriate transform. If the equation had not been so nicely simplifiable (coefficients in q only) plus a fancier forcing function (how about F = δ(t = T)? Your "guess" here?) I doubt you could have competed with the Laplace method.

As you should have learned by now, most of us posting in this thread are well acquainted with using Laplace transforms to solve differential equations. The "tedious algebra" that vela referred to can be seen in examples such as finding the inverse Laplace transform of, say, (2s - 3)/(s²(s² + 2s + 5}). I made this one up, but I assure you that I have worked through similar problems in the past. I think I can guaranteed that you're not going to find the inverse Laplace transform of this function without doing a fair amount of "tedious algebra."

I have nothing against using Laplace transforms to solve DEs. This technique is just one of several tools that someone should have in his toolbox. For some DEs, solution using Laplace transforms is the obvious choice. For others, solution using the method of undetermined coefficients might be the better choice.

From what you (rude man) wrote earlier in this thread, I infer that solution using Laplace transforms is the only technique that you have. As the only saying goes, "If your only tool is a hammer, everything looks like a nail."

 

[rude man]

As you should have learned by now, most of us posting in this thread are well acquainted with using Laplace transforms to solve differential equations. The "tedious algebra" that vela referred to can be seen in examples such as finding the inverse Laplace transform of, say, (2s - 3)/(s²(s² + 2s + 5)). I made this one up, but I assure you that I have worked through similar problems in the past. I think I can guaranteed that you're not going to find the inverse Laplace transform of this function without doing a fair amount of "tedious algebra."

I have nothing against using Laplace transforms to solve DEs. This technique is just one of several tools that someone should have in his toolbox. For some DEs, solution using Laplace transforms is the obvious choice. For others, solution using the method of undetermined coefficients might be the better choice.

From what you (rude man) wrote earlier in this thread, I infer that solution using Laplace transforms is the only technique that you have. As the only saying goes, "If your only tool is a hammer, everything looks like a nail."

Oh brother. What can I say to that jejune cliche that won't get me reprimanded.

I took a full semester of diff. eq. But once you know Laplace, providing it's apposite to the problem, then you just forget all that other crap. In my 40 yrs. that has worked 100% for me, thank you very much. (fortunately, most systems involve constant coefficients).

It would seem that you are not at all familiar with the expediency associated with the Laplace transform, including how easily initial conditions are included. No "guessing". No particular integrals vs. complementary solutions vs. homogeneous vs. inhomogeneous, all that claptrap.

BTW Your made-up transfer function represents a non-minimum-phase, underdamped system, but no matter, it's found as #01.101 in my tables except that an extra s in the denominator means you have to do a supplemental integration in the time domain after you get f(t) corresponding to 01.101. A whole lot easier than your classical approach.

 

[rude man]

This is getting interesting :D

Well, with the world series over (yea Red Sox!) we have to provide some alternative spectator sport

As far as you're concerned, though, you probably won't be exposed to the Laplace transform unless you go into engineering. I have worked with dozens of physicists, many of them PhD's, predominantly MIT and Caltech graduates but also other schools. NONE of them knew the Laplace. They do however know the Fourier transform which in the right hands can do the same thing (and more) than the one-sided Laplace. (There is also a two-sided Laplace for stochastic systems but this is rarely encountered.) The big plus for the Fourier is it can handle signals over all time whereas the one-sided Laplace is suitable only for t > 0 problems.

If you're majoring in math you will probably run into it but your field is based on mostly theory whereas the Laplace is the quick-and-easy cookbook way.

If you're an undergraduate in physics though, you will probably have to live with the classical claptrap. So let our Worthies show you the way, which they're very good at, apparently.

 

[Mark44]

It would seem that you are not at all familiar with the expediency associated with the Laplace transform, including how easily initial conditions are included.

Why would you make that assumption? As it turns out, I am very familiar with how to use the Laplace transform on an initial value problem.

No "guessing". No particular integrals vs. complementary solutions vs. homogeneous vs. inhomogeneous, all that claptrap.

The "guess" part is really a misnomer. For a large set of forcing functions, finding the particular solution is not guesswork.

BTW Your made-up transfer function represents a non-minimum-phase, underdamped system, but no matter, it's found as #01.101 in my tables except that an extra s in the denominator means you have to do a supplemental integration in the time domain after you get f(t) corresponding to 01.101. A whole lot easier than your classical approach.

Still sounds like some "tedious algebra" to me.

To repeat myself, I view using Laplace transforms as a useful technique for solving differential equations. I find it useful to have a variety of techniques to draw from, and have no need to disparage the techniques that I don't know with deprecating remarks about "guessing" or "claptrap" or the like.

 

[oddjobmj]

If you're an undergraduate in physics though, you will probably have to live with the classical claptrap. So let our Worthies show you the way, which they're very good at, apparently.

Yep, physics undergrad. I'm working towards industry though so my education won't be as mathematically rigorous as many of my peers although I do have a great many hoops to jump through.

 

[rude man]

Yep, physics undergrad. I'm working towards industry though so my education won't be as mathematically rigorous as many of my peers although I do have a great many hoops to jump through.

You'll be fine. You've shown more interest and perseverance than 90% of our OP's.

And you'll learn to take advantage of all the software out there, like Wolfram Alpha, which makes solving problems a jiff. For the moment you just need to 'go with the flow' i.e. pass those courses, most of which material you won't ever need!
.