23.11: Solution to the Forced Damped Oscillator Equation

We shall now use complex numbers to solve the differential equation

$$F_{0} \cos (\omega t)=m \frac{d^{2} x}{d t^{2}}+b \frac{d x}{d t}+k x \nonumber$$

We begin by assuming a solution of the form

$$x(t)=x_{0} \cos (\omega t+\phi) \nonumber$$

where the amplitude $x_{0}$ and the phase constant $\phi$ need to be determined. We begin by defining the complex function

$$z(t)=x_{0} e^{i(\omega t+\phi)} \nonumber$$

Our desired solution can be found by taking the real projection

$$x(t)=\operatorname{Re}(z(t))=x_{0} \cos (\omega t+\phi) \nonumber$$

Our differential equation can now be written as

$$F_{0} e^{i \omega t}=m \frac{d^{2} z}{d t^{2}}+b \frac{d z}{d t}+k z \nonumber$$

We take the first and second derivatives of Equation (23.D.3),

$$\frac{d z}{d t}(t)=i \omega x_{0} e^{i(\omega t+\phi)}=i \omega z \nonumber$$

$$\frac{d^{2} z}{d t^{2}}(t)=-\omega^{2} x_{0} e^{i(\omega t+\phi)}=-\omega^{2} z \nonumber$$

We substitute Equations (23.D.3), (23.D.6), and (23.D.7) into Equation (23.D.5) yielding

$$F_{0} e^{i \omega t}=\left(-\omega^{2} m+b i \omega+k\right) z=\left(-\omega^{2} m+b i \omega+k\right) x_{0} e^{i(\omega t+\phi)} \nonumber$$

We divide Equation (23.D.8) through by $e^{i \omega t}$ and collect terms using yielding

$$x_{0} e^{i \phi}=\frac{F_{0} / m}{\left(\left(\omega_{0}^{2}-\omega^{2}\right)+i(b / m) \omega\right)} \nonumber$$

where we have used $\omega_{0}^{2}=k / m$. Introduce the complex number

$$z_{1}=\left(\omega_{0}^{2}-\omega^{2}\right)+i(b / m) \omega \nonumber$$

Then Equation (23.D.9) can be written as

$$x_{0} e^{i \phi}=\frac{F_{0}}{m y} \nonumber$$

Multiply the numerator and denominator of Equation (23.D.11) by the complex conjugate $\bar{z}_{1}=\left(\omega_{0}^{2}-\omega^{2}\right)-i(b / m) \omega$ yeilding

$$x_{0} e^{i \phi}=\frac{F_{0} \bar{z}_{1}}{m z_{1} \bar{z}_{1}}=\frac{F_{0}}{m} \frac{\left(\left(\omega_{0}^{2}-\omega^{2}\right)-i(b / m) \omega\right)}{\left(\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+(b / m)^{2} \omega^{2}\right)} \equiv u+i v \nonumber$$

where

$$u=\frac{F_{0}}{m} \frac{\left(\omega_{0}^{2}-\omega^{2}\right)}{\left(\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+(b / m)^{2} \omega^{2}\right)} \nonumber$$

$$v=-\frac{F_{0}}{m} \frac{(b / m) \omega}{\left(\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+(b / m)^{2} \omega^{2}\right)} \nonumber$$

Therefore the modulus $x_{0}$ is given by

$$x_{0}=\left(u^{2}+v^{2}\right)^{1 / 2}=\frac{F_{0} / m}{\left(\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+(b / m)^{2} \omega^{2}\right)} \nonumber$$

and the phase is given by

$$\phi=\tan ^{-1}(v / u)=\frac{-(b / m) \omega}{\left(\omega_{0}^{2}-\omega^{2}\right)} \nonumber$$