I'm having some trouble while trying to understand one task.. The task is as follows:

$$\ddot{x}(t) + \dot{x}(t) + 2x(t) = \sin(\omega t)$$ where $x(0) = 7, t\geq 0$

The solution is in the following form:

$$x(t) = f(t) + A\sin(\omega t + \varphi)$$

And the task is: find $\omega$ so that $A$ is max.

My understanding of this is that $f(t)$ is the solution of the homogeneous differential equation and the rest is the special solution of the nonhomogeneous equation. Still that does not give me any clue about how to evaluate the relationship between $A$ and $\omega$. Any clues?


2 Answers 2


Since $f$ is the (generic) homogeneous solution, you can choose to set it equal to zero; then you just have the special solution $x(t) = A\mathrm{sin}(\omega t+\phi)$. You should be able to find $\ddot{x}(t)$ and $\dot{x}(t)$, then plug them all into your core differential equation and see what it means for both sides to be equal - if you've done it right, this will give you the relation between $\omega$ and $A$ that you're looking for.

  • $\begingroup$ Thank you. I'm certainly able to do that, but is it right to just "choose the solution to be equal to 0" ? What are the consequences? $\endgroup$
    – kubal5003
    Feb 22, 2011 at 20:36
  • $\begingroup$ kubal: if you plug in the generic form for $x(t)$ into the equation and then use the fact that $f(t)$ solves the homogeneous version of the equation, you'll find that it cancels right out and you're left with exactly the same terms that you would've had if you had chosen $f(t)$ to be $0$ to begin with. $\endgroup$ Feb 22, 2011 at 21:37
  • $\begingroup$ I guess I'll have to do that in order to prove that the solution is not just a special case. This might appear on tommorows exam (I hope not) $\endgroup$
    – kubal5003
    Feb 22, 2011 at 22:12

The correct value of $\omega$ is not $\frac{\sqrt{7}}{2} \approx 1.32$. It's $\sqrt{\frac{3}{2}} \approx 1.22$. We are looking for a near resonance effect, but you can't actually have resonance when the harmonic oscillator is damped. This makes the analysis is a little different.

A reference is the section on sinusoidal forcing in the Wikipedia page on the harmonic oscillator. From the formulas there you can see that the relationship between $A$ and $\omega$ is given by

$$A = \frac{1}{\sqrt{\omega^2 + (2 - \omega^2)^2}}.$$

Solving $\frac{dA}{d \omega} = 0$ yields $\omega = \sqrt{\frac{3}{2}}$.

Update: Here's the derivation.

The trick is to generalize, and solve the differential equation $x'' + x' + 2x = e^{i \omega t}$. The resulting solution will have a real part and an imaginary part. Since $e^{i \omega t} = \cos \omega t + i \sin \omega t$, you actually want the imaginary part of the solution.

As the driving force is an exponential, we know that the particular solution must be of the form $x_p(t) = c e^{i \omega t}$. Subbing that into the differential equation produces the auxiliary equation $-c \omega^2 + i c\omega + 2c = 1$. Solving that for $c$ yields $$c = \frac{1}{a + i b} = \frac{a - i b}{a^2 + b^2},$$ where $a = 2 - \omega^2$ and $b = \omega$. Thus the particular solution to the complex differential equation is $$x_p(t) = \frac{a - i b}{a^2 + b^2} (\cos \omega t + i \sin \omega t),$$ of which the imaginary part is $$-\frac{b}{a^2 + b^2} \cos \omega t + \frac{a}{a^2+b^2} \sin \omega t.$$ Since $A$ is just the magnitude of this solution (you're doing a rotation to the vertical axis when converting to $A \sin (\omega t + \phi)$), we get $$A = \frac{1}{\sqrt{a^2+b^2}} = \frac{1}{\sqrt{\omega^2 + (2 - \omega^2)^2}}.$$

  • $\begingroup$ You could also get this by assuming a solution of the form $c_1 \cos \omega t + c_2 \sin \omega t$. Once you find $c_1$ and $c_2$ in terms of $\omega$, you'll have $A = \sqrt{c_1^2 + c_2^2}$. I like the solution with the complex exponential, though; it seems more intuitive to me. $\endgroup$ Feb 23, 2011 at 3:37

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