4.9: Damped Resonance

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We now solve the dimensionless equations given by (4.8.5), (4.8.6) and (4.8.7), written here as $$\label{eq:1}\overset{..}{x}+a\overset{.}{x}+x=\cos\beta t,$$ where the physical constraints of our three applications requires that $\alpha > 0$. The homogeneous equation has characteristic equation $$r^2+\alpha r+1=0,\nonumber$$ so that the solutions are $$r_{\pm}=-\frac{\alpha}{2}\pm\frac{1}{2}\sqrt{\alpha^2-4}.\nonumber$$

When $\alpha^2 − 4 < 0$, the motion of the unforced oscillator is said to be underdamped; when $\alpha^2 − 4 > 0$, overdamped; and when $\alpha^2 − 4 = 0$, critically damped. For all three types of damping, the roots of the characteristic equation satisfy $\text{Re}(r_±) < 0$. Therefore, both linearly independent homogeneous solutions decay exponentially to zero, and the long-time asymptotic solution of $\eqref{eq:1}$ reduces to the non-decaying particular solution. Since the initial conditions are satisfied by the free constants multiplying the decaying homogeneous solutions, the long-time asymptotic solution is independent of the initial conditions.

If we are only interested in the long-time asymptotic solution of $\eqref{eq:1}$, we can proceed directly to the determination of a particular solution. As before, we consider the complex ode $$\overset{..}{z}+\alpha\overset{.}{z}+z=e^{i\beta t},\nonumber$$ with $x_p=\text{Re}(z_p)$. With the ansatz $z_p=Ae^{i\beta t}$, we have $$-\beta^2A+i\alpha\beta A+A=1,\nonumber$$ or $$\begin{align}A&=\frac{1}{(1-\beta^2)+i\alpha\beta}\nonumber \\ &=\left(\frac{1}{(1-\beta^2)^2+\alpha^2\beta^2}\right)\left((1-\beta^2)-i\alpha\beta\right).\label{eq:2}\end{align}$$

To determine $x_p$, we utilize the polar form of a complex number. The complex number $z = x + iy$ can be written in polar form as $z = re^{i\phi}$, where $r =\sqrt{x^2+y^2}$ and $\tan\phi = y/x$. We therefore write $$(1-\beta^2)-i\alpha\beta =re^{i\phi},\nonumber$$ with $$r=\sqrt{(1-\beta^2)^2+\alpha^2\beta^2},\quad\tan\phi=-\frac{\alpha\beta}{1-\beta^2}.\nonumber$$

Using the polar form, $A$ in $\eqref{eq:2}$ becomes $$A=\left(\frac{1}{\sqrt{(1-\beta^2)^2+\alpha^2\beta^2}}\right)e^{i\phi},\nonumber$$ and $x_p=\text{Re}(Ae^{i\beta t})$ becomes $$\begin{align}x_p&=\left(\frac{1}{\sqrt{(1-\beta^2)^2+\alpha^2\beta^2}}\right)\text{ Re}\left\{e^{i(\beta t+\phi)}\right\}\nonumber \\ &=\left(\frac{1}{\sqrt{(1-\beta^2)^2+\alpha^2\beta^2}}\right)\cos (\beta t+\phi).\label{eq:3}\end{align}$$

We conclude with a couple of observations about $\eqref{eq:3}$. First, if the forcing frequency $\omega$ is equal to the natural frequency $\omega_0$ of the undamped oscillator, then $\beta = 1$ and $A = 1/i\alpha$, and $x_p = (1/\alpha ) \sin t$. The oscillator position is observed to be $π/2$ out of phase with the external force, or in other words, the velocity of the oscillator, not the position, is in phase with the force. Second, the value of $\beta$ that maximizes the amplitude of oscillation is the value of $\beta$ that minimizes the denominator of $\eqref{eq:3}$. To determine $\beta_m$ we thus minimize the function $g(\beta^2)$ with respect to $\beta^2$, where $$g(\beta^2)=(1-\beta^2)^2+\alpha^2\beta^2.\nonumber$$

Taking the derivative of $g$ with respect to $\beta^2$ and setting this to zero to determine $\beta_m$ yields $$-2(1-\beta_m^2)+\alpha^2=0,\nonumber$$ or $$\beta_m=\sqrt{1-\frac{\alpha^2}{2}}\approx 1-\frac{\alpha^2}{4},\nonumber$$ the last approximation valid if $\alpha << 1$ and the dimensionless damping coefficient is small. We can interpret this result by saying that small damping slightly lowers the “resonance” frequency of the undamped oscillator.