23.11: Solution to the Forced Damped Oscillator Equation

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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

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