Forced oscillations: solution of Differential Equation at resonance

[jellicorse]

Homework Statement

Following a worked example in my book, I have been trying to get a solution for the equation

\frac{d^2u}{dt^2} + \frac{k}{m}u = Fcos\omega t

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

Homework Equations

The Attempt at a Solution

To get the full solution:

Complementary Function:

Acos\omega_0t+Bsin\omega_0t, where \omega_0=\sqrt{\frac{k}{m}}To get the Particular Integral:
Assume u= Ctcos\omega t +Dtsin\omega t

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

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

And \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

\frac{d^2u}{dt^"}=(2D\omega -c \omega^2t)cos\omega t +(-2C\omega -D\omega^2t)sin \omega tBack substituting these into the original differential equation:

(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:

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

and (\frac{kDt}{m} -2 C \omega - D\omega^2 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 \frac{F}{2m\omega_0^2} (and I assume since we are talking about a situation in which \omega_0=\omega, D= \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?

 

[vela]

From the form of the particular solution, you're apparently assuming that $\omega=\omega_0 = \sqrt{k/m}$, so the two equations reduce to
\begin{align*}
2\omega D &= \frac Fm \\
-2\omega C &= 0.
\end{align*} The solution you're trying to obtain is not correct: $\frac{F}{m\omega_0^2}t\sin\omega_0 t$ does not have units of length.

 

[jellicorse]

Thanks a lot Vela. I should have noticed that when equating coefficients.

I don't quite understand this though:

The solution you're trying to obtain is not correct: $\frac{F}{m\omega_0^2}t\sin\omega_0 t$ does not have units of length.

I can't see anything to do with units of length in the equation...

 

[vela]

The denominator is $m\omega^2 = k$ where $k$ has units of force/length, so $\frac{F}{m\omega^2}$ will have units of length. That would be fine if the coefficient multiplied only $\sin \omega t$, but it doesn't. It multiplies $t \sin\omega t$. The factor of $t$ introduces a unit of time.

 

[jellicorse]

Oh, I see... That makes sense, thanks!