# Driven oscillator solution intuition

For the sinusoidally driven oscillator given by:

$$m\ddot{x} + b \dot{x} + kx = F_0 \cos(\omega t)$$

or

$$\ddot{x} + 2\beta \dot{x} + \omega_0^2x = A \cos(\omega t)$$

The particular solution is:

$$x(t) = \frac A p \cos(\omega t - \delta)$$

where

$$p = \sqrt{ (\omega_0 ^2 - \omega ^2)^2 + 4 \beta ^2 \omega ^2 }$$

and

$$\delta = \arctan(\frac{2 \beta \omega }{\omega_0^2 -\omega^2})$$

Is there some intuition which allows us to arrive at this particular solution by inspection?

What sticks out to me is the that $p$ looks like a distance which depends on $\omega$ and whats more is that the phase shift is an angle in this "$\omega, \omega_0, \beta$ space". Is this idea on the right track?

Yes, the intuition is called the method of phasors. Suppose we guess a solution of the form $x(t) = A_0 e^{i\omega t}$. Then the left-hand side of the equation is $$\left( (i \omega)^2 + 2 \beta (i \omega) + \omega_0^2 \right) A_0 e^{i \omega t}$$ while the driving force is $F_0 e^{i \omega t}$. Therefore, cancelling, we have $$A_0 = \frac{F_0}{(i \omega)^2 + \omega_0^2 + 2 \beta (i \omega)}.$$ To get the picture, we think of $A_0$ and $F_0$ as vectors in the complex plane. Then the denominator gives the ratio of their norms, while the angle of the denominator gives the angle between them, immediately reproducing the expressions you have for $p$ and $\delta$.

• I think it would be better to leave the solution as $x(t)=A_0 e^{i(\omega t+\delta)}$ so that that we can keep $A_0$ real. The left hand side of the equation would then be $A_0 e^\delta$. The right hand side can then also be separated into a real factor and a complex number of unit length. – Johnathan Gross Jul 27 '17 at 19:26
• @JohnathanGross I do not understand why adding a phase $\delta$ is necessary to make $A_0$ real? – Farcher Jul 28 '17 at 10:07
• You have $A_0$ equal to a complex number. By explicitly including a phase, your equation for $A_0$ becomes an equation for $A_0e^{i\phi}$ which is the polar form of a complex number. – Johnathan Gross Jul 28 '17 at 15:17

I write this as an answer because I do not know how to add a diagram in a comment.

I think that @knzhou has produced an excellent answer and you can choose either $A_0$ or $F_0$ as the reference phasor.

Rearranging the equation that links $F_0$ and $A_0$

$F_0 = \left ((i \omega)^2 + \omega_0^2 + 2 \beta (i \omega) \right ) A_0$

and choosing $A_0$ to be real, the phasor diagram looks like this

with $A_0$ real and $F_0$ complex.