Resonance of a damped driven harmonic oscillator

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$m=$1 [kg]

$b=$1 [N s/m]

$k=$1 [N/m]

$F_0=$1 [N]

$\Omega=$1 [rad/s]

$\omega=\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$

= 1 [rad/s] = 1 [Hz]

$\theta=\text{atan}\left(\frac{\Omega b}{k-m\Omega^2}\right)$

= 0 [rad] = 0 [deg]

$|A|=\frac{F_0}{\sqrt{(k-m\Omega^2)^2+\Omega^2b^2}}$

= 1 [m]

$Q=\frac{\sqrt{mk}}{b}$

= 1

Display $F_0e^{i\Omega t}$: 

 Display $|A|e^{i(\Omega t-\theta)}$: 

Display transients $z$: 

 Display $y$: 

The simulation above shows the motion of a damped, driven oscillator. The square blue weight has a mass $m$ and is connected to a spring with a spring constant $k$. There is a damping force $-bv$ where $v$ is the velocity of the weight. A driving force of the form $F_0\cos (\Omega t)$ pushes on the weight. This is indicated by the red arrow. Here $F_0$ is the amplitude of the drive force and $\Omega$ is the drive frequency. The motion of the weight consists of a transient response and a steady-state response. The steady-state solution is $A\cos(\Omega t -\theta)$. The steady-state oscillations of the weight always lag a phase angle $\theta$ behind the drive. The values phase angle and the amplitude $A$ are displayed next to the simulation above. The amplitude of the motion of the weight is maximum when the drive frequency $\Omega$ equals the natural frequency $\omega$. This condition is called a resonance.

A damped driven oscillator is often analyzed using complex numbers. The driving force can be thought of as the real part of circular motion in the complex plane. Select "Display $F_0e^{i\Omega t}$" to see the complex representation as a solid red circle in the simulation. The complex steady-state response to the drive can be seen by selecting "Display $Ae^{i(\Omega t-\theta)}$". The motion of the steady-state response appears as an open blue circle. The angle between the red line and the blue line is $\theta$. Select "Display transients $z$" to see the complex solution that includes the transient response and the steady-state solution. This appears as a solid blue circle. When the solid blue circle is centered in the open blue circle, the transients have decayed away. The complex solution $z = x +iy$ has a real part and an imaginary part. The imaginary part is the solution for a drive force $F_0\sin (\Omega t)$. By selecting "Display $y$" this solutions is displayed as a light-blue square. Every time a slider is moved, a random initial condition is chosen so the transient response will be different every time.

The transient response consists of solutions for zero drive force. Slide $F_0$ to zero to observe the transient response. The differential equation that describes the motion of the of an undriven damped oscillator is,

\[\begin{equation} \label{eq:e1} m\frac{d^2x}{dt^2}+b\frac{dx}{dt} + kx = 0, \end{equation}\]

When solving this problem, it is common to consider the complex differential equation,

\[\begin{equation} \label{eq:e2} m\frac{d^2z}{dt^2}+b\frac{dz}{dt} + kz = 0, \end{equation}\]

where $z$ is a complex number $z(t)=x(t)+iy(t)$. Here $x(t)$ and $y(t)$ are both real functions and $\sqrt{-1}=\pm i$. Substituting $z$ into Eq. \eqref{eq:e2} yields,

\[\begin{equation} \label{eq:e3} m\frac{d^2x}{dt^2}+b\frac{dx}{dt} + kx +im\frac{d^2y}{dt^2}+ib\frac{dy}{dt} + iky= 0. \end{equation}\]

Since a real number cannot be equal to an imaginary number, if a solution $z$ is found to Eq. \eqref{eq:e2}, both the real, $x$, and imaginary, $y$, parts will be solutions of Eq. \eqref{eq:e1}.

We seek a solution to Eq. \eqref{eq:e2} of the form $z=e^{\lambda t}$. Substituting this into Eq. \eqref{eq:e2} yields,

\[\begin{equation} m\lambda^2+b\lambda + k = 0. \end{equation}\]

The solutions for $\lambda$ are,

\[\begin{equation} \lambda_{\pm} = \frac{-b\pm\sqrt{b^2-4km}}{2m}. \end{equation}\]

If $b^2-4km > 0$ the two solutions for $\lambda$ are real and the two solutions to Eq. \eqref{eq:e1} are,

\[\begin{equation} x_1(t) = C_1 \exp \left(\frac{-b+\sqrt{b^2-4km}}{2m}t\right)\hspace{1cm}\text{and}\hspace{1cm} x_2(t) = C_2 \exp \left(\frac{-b-\sqrt{b^2-4km}}{2m}t\right). \end{equation}\]

The constants $C_1$ and $C_2$ are determined by the initial conditions. These are called overdamped solutions. When a damped mass-spring system with these parameters is pulled away from its equilibrium position and then released, the return to the equilibrium position is described by an exponential decay and there are no oscillations. Overdamped solutions will occur if the quality factor $Q =\frac{\sqrt{mk}}{b} < \frac{1}{2}$.

If $b^2-4km=0$ there is only one solution for $\lambda=-\frac{b}{2m}$ and the two solutions to Eq. \eqref{eq:e1} are,

\[\begin{equation} x_1(t) = C_1 \exp \left(\frac{-b}{2m}t\right)\hspace{1cm}\text{and}\hspace{1cm} x_2(t) = C_2t \exp \left(\frac{-b}{2m}t\right). \end{equation}\]

The constants $C_1$ and $C_2$ are determined by the initial conditions. These are called critically damped solutions. When a damped mass-spring system with these parameters is pulled away from its equilibrium position and then released, it returns to equilibrium position as rapidly as possible without oscillations. Critically damped solutions will occur if the quality factor $Q = \frac{\sqrt{mk}}{b} = \frac{1}{2}$.

If $b^2-4km < 0$ the two solutions for $\lambda$ are complex and the two solutions for $z$ to Eq. \eqref{eq:e2} have the form,

\[\begin{equation} z_1(t) \propto \exp \left(\frac{-b}{2m}t+i\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}t\right)\hspace{1cm}\text{and}\hspace{1cm} z_2(t) \propto \exp \left(\frac{-b}{2m}t-i\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}t\right). \end{equation}\]

The real solutions to Eq. \eqref{eq:e1} are the real and imaginary parts of $z_1$ and $z_2$,

\[\begin{equation} x_1(t) = C_1 \exp \left(\frac{-b}{2m}t\right)\cos (\omega t)\hspace{1cm}\text{and}\hspace{1cm} x_2(t) = C_2 \exp \left(\frac{-b}{2m}t\right)\sin (\omega t). \end{equation}\]

The constants $C_1$ and $C_2$ would be determined by the initial conditions. These are underdamped solutions which oscillate with an angular frequency of $\omega = \sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$. Underdamped solutions will occur if the quality factor $Q = \frac{\sqrt{mk}}{b} > \frac{1}{2}$.

When a driving term is added to the system the differential equation becomes,

\[\begin{equation} m\frac{d^2x}{dt^2}+b\frac{dx}{dt} + kx = F_0\cos(\Omega t). \end{equation}\]

The harmonic motion of the drive can be thought of as the real part of circular motion in the complex plane. The complex differential equation that is used to analyze the damped driven mass-spring system is,

\[\begin{equation} \label{eq:e10} m\frac{d^2z}{dt^2}+b\frac{dz}{dt} + kz = F_0e^{i\Omega t}. \end{equation}\]

This is equivalent to the equations for the real and imaginary parts,

\[\begin{equation} m\frac{d^2x}{dt^2}+b\frac{dx}{dt} + kx = F_0\cos(\Omega t)\hspace{1cm}\text{and}\hspace{1cm}m\frac{d^2y}{dt^2}+b\frac{dy}{dt} + ky = F_0\sin(\Omega t). \end{equation}\]

The driving function $F_0 e^{i\Omega t}$ describes a point that moves in the complex plane in a circle of radius $F_0$ at a constant angular velocity of $\Omega$ (the solid red point in the simulation). The real part of this motion is the original driving force, $F_0\cos (\Omega t)$. In addition to the two solutions we found for the undriven case, this equation also has a solution that oscillates at the drive frequency and has the form $z = Ae^{i\Omega t}$. Substituting this solution into Eq. \eqref{eq:e10} results in the following equation,

\[\begin{equation} \left( -m\Omega^2+i\Omega b+k\right)A = F_0. \end{equation}\]

The factor in parenthesis can be written in polar form,

\[\begin{equation} \left( -m\Omega^2+i\Omega b+k\right) = \rho e^{i\theta}, \end{equation}\]

where

\[\begin{equation} \rho = \sqrt{(k-m\Omega^2)^2+\Omega^2b^2}\hspace{1cm}\text{and}\hspace{1cm} \tan\theta = \frac{\Omega b}{k-m\Omega^2}. \end{equation}\]

Solving for $A$ yields,

\[\begin{equation} A=\frac{F_0}{\rho}e^{-i\theta}. \end{equation}\]

The solution to the complex differential equation is $z=\frac{F_0}{\rho}e^{i(\Omega t -\theta)}$. This solution moves in a circle in the complex plane (the open blue point in the simulation). The steady-state motion that is observed is the real part of the complex solution, $x_1=\frac{F_0}{\rho}\cos (\Omega t -\theta)$. The response lags behind the drive by a phase $\theta$.

The form of the total solution including the transient terms depends on the value of $b^2-4km$,

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