Forced oscillations (resonance) Differential Equation

[jonnburton]

I have been working though a worked example on this topic in my book and have been unable to follow part of the working. I wondered if anyone could point me in the right direction.The equation describing the motion of the system with no damping is $\displaystyle m\frac{d^2u}{dt}+ k u = Fcos\omega t$

The book says that at resonance, i.e. when $\displaystyle \omega_0$ (the natural frequency) = $\displaystyle \omega$ (the forcing frequency), the term $\displaystyle F cos\omega t$ is a solution to the homogenous equation and the solution to the differential equation above is$\displaystyle Acos\omega_0t +Bsin\omega_0t+\frac{F}{2m\omega_o^2}tsin\omega_0t$

I have been trying to see how they got to this solution but keep getting stuck.This is what I have done:

The differential equation can be expressed as:

$\displaystyle \frac{d^2u}{dt}+ \frac{k}{m} u = \frac{F}{m}cos\omega t$

I can see that $\displaystyle F cos\omega t$ is a solution to the homogenous equation $\displaystyle \frac{d^2u}{dt}+ \frac{k}{m} u = 0$


To get the full solution:

Complementary Function:

$\displaystyle Acos\omega_0t+Bsin\omega_0t$, where $\displaystyle \omega_0=\sqrt{\frac{k}{m}}$

To get the Particular Integral:

Assume $\displaystyle u= Ctcos\omega t +Dtsin\omega t$

Then $\displaystyle \frac{du}{dt}= C cos\omega t - Ct\omega sin \omega t + D sin\omega t +Dt \omega cos \omega t$

$\displaystyle \frac{du}{dt}= (Dt\omega +C ) cos\omega t + (D-Ct\omega) sin\omega t$

And $\displaystyle \frac{d^2u}{dt^2} = -C \omega sin \omega t -C\omega sin \omega t - C \omega^2 t cos \omega t + D\omega cos \omega t + D\omega cos \omega t - D \omega^2 t sin \omega t$

$\displaystyle \frac{d^2u}{dt^"}=(2D\omega -c \omega^2t)cos\omega t +(-2C\omega -D\omega^2t)sin \omega t$



Back-substituting these into the original Differential equation:

$\displaystyle (2D\omega -C\omega^2t + \frac{k Ct}{m}) cos\omega t + (\frac{KDt}{m}-2C\omega -D\omega^2t) sin \omega t = \frac{F}{m} cos\omega t$



Equating coefficients:

$\displaystyle (2D\omega -C\omega^2t + \frac{k Ct}{m}) =\frac{F}{m}$

and $\displaystyle (\frac{kDt}{m} -2 C \omega - D\omega t) = 0$

After this, I have tried solving for D and C but it seems to end up in a pretty intractable mess.

I know somehow D should be equal to $\displaystyle \frac{F}{2m\omega_0^2}$ (and I assume since we are talking about a situation in which $\displaystyle \omega_0=\omega$, D= $\displaystyle \frac{F}{2m\omega_0^2} = \frac{F}{2m\omega}$) but can not see how to get there. Can anyone tell me if I have been going in the right direction so far?

 

[CurveGuy]

1. Your above equation (kDt / m − 2Cω − Dωt) = 0 has a typo: the term 'Dωt' should be 'Dω²t', correct?
2. Next, try to eliminate 't' from this two equations leaving one equation dependent on ω . I got ω = (D F / m) / [ 2 (C² + D)] ... make any sense out of this answer?

 

[yoscar04]

You don't need to assume a sin part in your particular solution If you try it, its coefficient will turn out to be 0. Try something of the form u_p=Acos([MATH]\omega[/MATH]t). with [MATH]\omega[/MATH]₀²=k/m. Substitute in your original ODE and find A. You should get something like u_p=Fcos([MATH]\omega[/MATH]t)/(m([MATH]\omega[/MATH]²-[MATH]\omega[/MATH]₀²)).

 

[yoscar04]

I didn't notice before you used tsin[MATH]\omega[/MATH]t and tcos[MATH]\omega[/MATH]t. Why didn't you use pure sin or cos?

 

[HallsofIvy]

The original equation is $\displaystyle m\frac{d^2u}{dt^2}+ ku= F cos(\omega t)$.

The "associated homogeneous equation" is $\displaystyle m\frac{d^2u}{dt^2}+ ku= 0$. That is a "homogeneous linear equation with constant coefficients". Its "characteristic equation" is $\displaystyle mr^2+k= 0$ which has imaginary roots $\displaystyle r= i\sqrt{\frac{k}{m}}$ and $\displaystyle r= -i\sqrt{\frac{k}{m}}$.

That tells us that the general solution to the associated homogeneous equation is $\displaystyle u(t)= C_1\cos\left(\sqrt{\frac{k}{m}}t\right)+ C_2\sin\left(\sqrt{\frac{k}{m}}t\right)$ where $\displaystyle C_1$ and $\displaystyle C_2$ are constants that must be determined from other information such as "initial values" or "boundary values".

Now to get the general solution to the entire equation we only need to add a single solution to the entire equation to that. Using the "method of undetermined coefficients" we a solution of the form $\displaystyle u(t)= A\cos(\omega t)+ B\sin(\omega t)$ where A and B are the "undetermined coefficients" that have to be determined so that this satisfies the equation.

$\displaystyle u(t)= A\cos(\omega t)+ B\sin(\omega t)$
$\displaystyle \frac{du}{dt}= -A\omega \sin(\omega t)+ B\omega \cos(\omega t)$
$\displaystyle \frac{d^2u}{dt^2}= -A\omega^2 \cos(\omega t)- B\omega^2 \sin(\omega t)$.

So $\displaystyle m\frac{d^2u}{dt^2}+ ku= -Am\omega^2 \cos(\omega t)- Bm\omega^2 \sin(\omega t)+ kA\cos(\omega t)+ kB\sin(\omega t)= F \cos(\omega t)$.

$\displaystyle \left(-Am\omega^2+ kA\right) \cos(\omega t)+ \left(-Bm\omega^2+ kB\right) \sin(\omega t)= F \cos(\omega t)$.

Since $\displaystyle \sin(\omega t)$ and $\displaystyle \cos(\omega t)$ are "independent functions", we must have $\displaystyle -Am\omega^2+ kA= (-m\omega^2+ k)A= F$ and $\displaystyle -Bm\omega^2+ kB= (-momega^2+ k)B= 0$. So B= 0 and $\displaystyle A= \frac{F}{k- m\omega^2}$.

The general solution to the entire equation, $\displaystyle \frac{d^2u}{dt^2}+ ku= F \cos(\omega t)$ is $\displaystyle u(t)= C_1 cos\left(\sqrt{\frac{k}{m}}t\right)+ C_2 sin\left(\sqrt{\frac{k}{m}}t\right)+ \frac{F}{k- m\omega^2}\cos(\omega t)$.


Except for one detail!

If it happens that $\displaystyle k= m\omega^2$ then A is undefined! That is the "resonance" case, where the "forcing term", $\displaystyle F cos(\omega t)$ has the same frequency as the "natural frequency", k. So what do we do in that case? Instead of looking for a solution to the entire equation of the form "$\displaystyle A\cos(\omega t)+ B\sin(\omega t)$ we need to look for a solution of the form $\displaystyle At\cos(\omega t)+ Bt\sin(\omega t)$.

Why $\displaystyle t\cos(\omega t)$ and $\displaystyle t\sin(\omega t)$? (That was yoscar04's question.) There are two ways to answer. First, because it works! A lot of mathematics involves "trying" various things until you find one that works, then sticking with it. Second, there is another method for finding specific solutions to non-homogeneous differential equations called "variation of parameters". It does not require "guessing" a particular form for a solution and gives $\displaystyle t\cos(\omega t)$ and $\displaystyle t \sin(\omega t)$ as solutions. But it is a lot more complicated and typically involves doing some difficult integrals so if you can avoid it do so![/tex][/tex]

 

[yoscar04]

The original equation is $\displaystyle m\frac{d^2u}{dt^2}+ ku= F cos(\omega t)$.

The "associated homogeneous equation" is $\displaystyle m\frac{d^2u}{dt^2}+ ku= 0$. That is a "homogeneous linear equation with constant coefficients". Its "characteristic equation" is $\displaystyle mr^2+k= 0$ which has imaginary roots $\displaystyle r= i\sqrt{\frac{k}{m}}$ and $\displaystyle r= -i\sqrt{\frac{k}{m}}$.

That tells us that the general solution to the associated homogeneous equation is $\displaystyle u(t)= C_1\cos\left(\sqrt{\frac{k}{m}}t\right)+ C_2\sin\left(\sqrt{\frac{k}{m}}t\right)$ where $\displaystyle C_1$ and $\displaystyle C_2$ are constants that must be determined from other information such as "initial values" or "boundary values".

Now to get the general solution to the entire equation we only need to add a single solution to the entire equation to that. Using the "method of undetermined coefficients" we a solution of the form $\displaystyle u(t)= A\cos(\omega t)+ B\sin(\omega t)$ where A and B are the "undetermined coefficients" that have to be determined so that this satisfies the equation.

$\displaystyle u(t)= A\cos(\omega t)+ B\sin(\omega t)$
$\displaystyle \frac{du}{dt}= -A\omega \sin(\omega t)+ B\omega \cos(\omega t)$
$\displaystyle \frac{d^2u}{dt^2}= -A\omega^2 \cos(\omega t)- B\omega^2 \sin(\omega t)$.

So $\displaystyle m\frac{d^2u}{dt^2}+ ku= -Am\omega^2 \cos(\omega t)- Bm\omega^2 \sin(\omega t)+ kA\cos(\omega t)+ kB\sin(\omega t)= F \cos(\omega t)$.

$\displaystyle \left(-Am\omega^2+ kA\right) \cos(\omega t)+ \left(-Bm\omega^2+ kB\right) \sin(\omega t)= F \cos(\omega t)$.

Since $\displaystyle \sin(\omega t)$ and $\displaystyle \cos(\omega t)$ are "independent functions", we must have $\displaystyle -Am\omega^2+ kA= (-m\omega^2+ k)A= F$ and $\displaystyle -Bm\omega^2+ kB= (-momega^2+ k)B= 0$. So B= 0 and $\displaystyle A= \frac{F}{k- m\omega^2}$.

The general solution to the entire equation, $\displaystyle \frac{d^2u}{dt^2}+ ku= F \cos(\omega t)$ is $\displaystyle u(t)= C_1 cos\left(\sqrt{\frac{k}{m}}t\right)+ C_2 sin\left(\sqrt{\frac{k}{m}}t\right)+ \frac{F}{k- m\omega^2}\cos(\omega t)$.


Except for one detail!

If it happens that $\displaystyle k= m\omega^2$ then A is undefined! That is the "resonance" case, where the "forcing term", $\displaystyle F cos(\omega t)$ has the same frequency as the "natural frequency", k. So what do we do in that case? Instead of looking for a solution to the entire equation of the form "$\displaystyle A\cos(\omega t)+ B\sin(\omega t)$ we need to look for a solution of the form $\displaystyle At\cos(\omega t)+ Bt\sin(\omega t)$.

Why $\displaystyle t\cos(\omega t)$ and $\displaystyle t\sin(\omega t)$? (That was yoscar04's question.) There are two ways to answer. First, because it works! A lot of mathematics involves "trying" various things until you find one that works, then sticking with it. Second, there is another method for finding specific solutions to non-homogeneous differential equations called "variation of parameters". It does not require "guessing" a particular form for a solution and gives $\displaystyle t\cos(\omega t)$ and $\displaystyle t \sin(\omega t)$ as solutions. But it is a lot more complicated and typically involves doing some difficult integrals so if you can avoid it do so![/tex][/tex]

Thanks for your observation.
I missed the fact he was looking for resonance type solution ([MATH]\omega[/MATH]=[MATH]\omega[/MATH]₀).

 

[yoscar04]

Thanks for your observation.
I missed the fact he was looking for resonance type solution ([MATH]\omega[/MATH]=[MATH]\omega[/MATH]₀).

Actually after reading your reply a second time I do recall that there are cases when solving an ODE with constant coefficients we get solutions of the form tsin([MATH]\omega[/MATH]t). For example considering the operator ([MATH]\frac{d}{dx}[/MATH]-r)^m, if we set u₁=e^rx, u₂(x)=xe^rx,...,u_m(x)=x^m-1e^rx, these m functions are m independent functions annihilated by the operator ([MATH]\frac{d}{dx}[/MATH]-r)^m.

 

[CurveGuy]

Its been a while since my class on ODEs, but looking at the ODE it seems to say that it is not damped. So, your Atcos(ωt)+Btsin(ωt) substitution would be wrong. Those 't's in their give it a ramp affect times a sinusoidal curve.

Have you tried a numerical solution in order to 'see' what's going on? There is a Windows app called FC-Compiler that would solve this with little work. I'm the main force behind this FC compiler ... author died last year.

 

[yoscar04]

Its been a while since my class on ODEs, but looking at the ODE it seems to say that it is not damped. So, your Atcos(ωt)+Btsin(ωt) substitution would be wrong. Those 't's in their give it a ramp affect times a sinusoidal curve.

Have you tried a numerical solution in order to 'see' what's going on? There is a Windows app called FC-Compiler that would solve this with little work. I'm the main force behind this FC compiler ... author died last year.

Try it and you will see that Ftsin([MATH]\omega[/MATH]t)/(2m[MATH]\omega[/MATH]₀) is indeed a particular solution (linearly growing oscillations)

 

[CurveGuy]

The original equation is $\displaystyle m\frac{d^2u}{dt^2}+ ku= F cos(\omega t)$.

The "associated homogeneous equation" is $\displaystyle m\frac{d^2u}{dt^2}+ ku= 0$. That is a "homogeneous linear equation with constant coefficients". Its "characteristic equation" is $\displaystyle mr^2+k= 0$ which has imaginary roots $\displaystyle r= i\sqrt{\frac{k}{m}}$ and $\displaystyle r= -i\sqrt{\frac{k}{m}}$.

Yes, but no. The original ODE $\displaystyle m\frac{d^2u}{dt^2}+ ku= F cos(\omega t)$ is saying that the steady state solution is u=A cos(ωt). So, you then need to find the damped out part of the solution, i.e. $\displaystyle m\frac{d^2u}{dt^2}+ ku= 0$. Thus u = A cos(ωt) + (... ??? ...). where the 2nd part goes to zero with time. How about u = A cos(ωt) + $\displaystyle B{e^{- C t}}$ ( ??? ). This 'B' section would die out with time, correct?

 

[yoscar04]

Yes, but no. The original ODE $\displaystyle m\frac{d^2u}{dt^2}+ ku= F cos(\omega t)$ is saying that the steady state solution is u=A cos(ωt). So, you then need to find the damped out part of the solution, i.e. $\displaystyle m\frac{d^2u}{dt^2}+ ku= 0$. Thus u = A cos(ωt) + (... ??? ...). where the 2nd part goes to zero with time. How about u = A cos(ωt) + $\displaystyle B{e^{- C t}}$ ( ??? ). This 'B' section would die out with time, correct?

There is no damping present in his equation.

 

[Deleted member 4993]

There is no damping present in his equation.

Correct. Damping is generally proportional to $\displaystyle \frac{du}{dt}$ and the equation becomes:

$\displaystyle m\frac{d^2u}{dt^2} + h\frac{du}{dt} + ku = F(t)$

 

[Deleted member 4993]

Correct. Damping is generally proportional to $\displaystyle \frac{du}{dt}$ and the equation becomes:

$\displaystyle m\frac{d^2u}{dt^2} + h\frac{du}{dt} + ku = F(t)$

And of course there will be "no resonance" with damping. The nature of the solution will change - exponentially.