The damped and forced oscillator

Consider an block of mass \(m\) over the surface of a table and attached to a wall by a spring. If we let the scalar \(x\) measure the displacement of this object from equilibrium position, then Newton’s second law establishes that \[m\ddot{x}=F,\] where \(F\) stands for the sum of the forces acting on the block. In this case we have that \(F=F_{s}+F_{f},\) where \(F_{s}=-kx\) is the force due to the spring (Hooke’s law) and \(F_{f}=-\gamma\dot{x}\) is the frictional force on the object due to its contact with the table. We conclude that \[m\ddot{x}+\gamma\dot{x}+kx=0.\] We can solve this equation by introducing the complex variable \(z=x+iy\) and considering the corresponding differential equation on \(z\), i.e \[m\ddot{z}+\gamma\dot{z}+kz=0.\] If for \(a,\omega\in\mathbb{C}\) we propose the curve \(z=a\exp(\omega t)\) as a solution, we obtain the characteristic equation \[m\omega^{2}+\gamma\omega+k=0,\] which has solutions \(\omega_{1}\) and \(\omega_{2}\) given by \[\begin{aligned} \omega_{1} & =\frac{-\gamma-\sqrt{\gamma^{2}-4mk}}{2m},\\ \omega_{2} & =\frac{-\gamma+\sqrt{\gamma^{2}-4mk}}{2m}.\end{aligned}\] By the principle of superposition the general solution to our differential equation has the form \[z=a_{1}\exp(\omega_{1}t)+a_{2}\exp(\omega_{2}t).\] The coefficients \(a_{0}\) and \(a_{1}\) can be computed from the initial conditions \[z(0)=z_{0},\dot{z}(0)=w_{0},\] which lead to the formula \[\left(\begin{array}{cc} 1 & 1\\ \omega_{1} & \omega_{2} \end{array}\right)\left(\begin{array}{c} a_{1}\\ a_{2} \end{array}\right)=\left(\begin{array}{c} z_{0}\\ w_{0} \end{array}\right).\] We conclude that \[a_{1}=\frac{z_{0}\omega_{2}-w_{0}}{\omega_{2}-\omega_{1}},a_{2}=\frac{w_{0}-z_{0}\omega_{1}}{\omega_{2}-\omega_{1}}.\]

We can model a periodic force of frequency \(\omega\in\mathbb{R}\) acting on the block by using a term of the form \(f\exp(i\omega t)\), where \(f\in\mathbb{C}\). This leads to the non-homogeneous equation

\[m\ddot{z}+\gamma\dot{z}+kz=f\exp(i\omega t).\] If we let \[\omega_{0}=\sqrt{\frac{k}{m}}\] we can write the above equation as \[\ddot{z}+\gamma/m+\omega_{0}^{2}=(f/m)\exp(i\omega t).\] Letting \(z=a\exp(i\omega t)\) gives us the algebraic equation \[a(-\omega^{2}+i(\gamma/m)\omega+\omega_{0}^{2})=f/m.\] Solving for \(a\), we obtain \[a=\frac{f}{m(\omega_{0}^{2}-\omega^{2})+i\gamma}.\] Observe \[|a|^{2}=\frac{|f|^{2}}{m^{2}(\omega_{0}^{2}-\omega){{}^2}+\gamma^{2}}.\] reaches its maximum value when \(\omega=\omega_{0}\). Furthermore, \[\arg(a)=\arg(f)-\arg(m(\omega_{0}^{2}-\omega)+i\gamma).\] The general solution to our equation is then given by \[z=a_{1}\exp(\omega_{1}t)+a_{2}\exp(\omega_{2}t)+a\exp(i\omega t).\] The initial conditions \[z(0)=z_{0},\dot{z}(0)=w_{0}\] lead to the equation \[\left(\begin{array}{cc} 1 & 1\\ \omega_{1} & \omega_{2} \end{array}\right)\left(\begin{array}{c} a_{1}\\ a_{2} \end{array}\right)=\left(\begin{array}{c} z_{0}-a\\ w_{0}-i\omega a \end{array}\right)\] and hence \[a_{1}=\frac{(z_{0}-a)\omega_{2}-(w_{0}-i\omega a)}{\omega_{2}-\omega_{1}},a_{2}=\frac{(w_{0}-i\omega a)-(z_{0}-a)\omega_{1}}{\omega_{2}-\omega_{1}}.\] Finally, we can obtain \(x\) as the real part of \(z\) and the initial condition \(x(0)=x_{0}\) and \(\dot{x}(0)=v_{0}\) can by satisfied by letting \(\text{re}(z_{0})=x_{0}\) and \(\text{re}(w_{0})=v_{0}\); we can choose \(\text{im}(z_{0})\) and \(\text{im}(w_{0})\) however we like.

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