Damped Oscillations

Question 1. Solve the differential equation

$\displaystyle y'' - 4y' - 5y = e^{2x} \sin(x).$

(Click for Solution)

Solution. Rewriting in $\mathcal D$ -notation,

$(\mathcal D^2 - 4 \mathcal D - 5)y = e^{2x} \sin(x).$

The general solution is given by $y = y_{\mathrm P} + y_{\mathrm C}$ . To obtain the complementary function $y_{\mathrm C}$ , we solve the equation

$(\mathcal D^2 - 4 \mathcal D - 5)y = 0.$

Since the characteristic equation $m^2 - 4m - 5 = 0$ has real and distinct auxiliary roots $m = -1, 5$ , the general solution is given by

$y_{\mathrm C} = C_1 e^{-x} + C_2 e^{5x}.$

To evaluate $y_{\mathrm P}$ , we first replace each $\mathcal D$ with $\mathcal D + 2$ to obtain

$\begin{aligned} (\mathcal D + 2)^2 - 4 (\mathcal D + 2) - 5 &= (\mathcal D^2 + 4\mathcal D + 4) - (4\mathcal D + 8) - 5 = \mathcal D^2 - 9. \end{aligned}$

Hence,

$\begin{aligned} y_{\mathrm P} &= \frac{1}{\mathcal D^2 - 4 \mathcal D - 5}(e^{2x} \sin(x)) \\ &= e^{2x} \cdot \frac{1}{\mathcal D^2 - 9}(\sin(x)) \\ &= e^{2x} \cdot \frac{1}{-1^2 - 9} \cdot \sin(x) \\ &= - \frac 1{10} e^{2x} \sin(x). \end{aligned}$

Consolidating all results,

$y = y_{\mathrm P} + y_{\mathrm C} = - \frac 1{10} e^{2x} \sin(x) + (C_1 e^{-x} + C_2 e^{5x}).$

Question 2. Solve the differential equation

$\displaystyle y'' - 4y' - 12y = e^{2x} \sin(2x).$

(Click for Solution)

Solution. Rewriting in $\mathcal D$ -notation,

$(\mathcal D^2 - 4 \mathcal D - 12)y = e^{2x} \sin(2x).$

The general solution is given by $y = y_{\mathrm P} + y_{\mathrm C}$ . To obtain the complementary function $y_{\mathrm C}$ , we solve the equation

$(\mathcal D^2 - 4 \mathcal D - 12)y = 0.$

Since the characteristic equation $m^2 - 4m - 12 = 0$ has real and distinct auxiliary roots $m = -2, 6$ , the general solution is given by

$y_{\mathrm C} = C_1 e^{-2x} + C_2 e^{6x}.$

To evaluate $y_{\mathrm P}$ , we first replace each $\mathcal D$ with $\mathcal D + 2$ to obtain

$\begin{aligned} (\mathcal D + 2)^2 - 4 (\mathcal D + 2) - 12 &= (\mathcal D^2 + 4\mathcal D + 4) - (4\mathcal D + 8) - 12 = \mathcal D^2 - 16. \end{aligned}$

Hence,

$\begin{aligned} y_{\mathrm P} &= \frac{1}{\mathcal D^2 - 4 \mathcal D - 12}(e^{2x} \sin(2x)) \\ &= e^{2x} \cdot \frac{1}{\mathcal D^2 - 16}(\sin(2x)) \\ &= e^{2x} \cdot \frac{1}{-2^2 - 16} \cdot \sin(2x) \\ &= - \frac 1{20} e^{2x} \sin(2x). \end{aligned}$

Consolidating all results,

$y = y_{\mathrm P} + y_{\mathrm C} = - \frac 1{20} e^{2x} \sin(2x) + (C_1 e^{-2x} + C_2 e^{6x}).$

Question 3. An object with mass $m$ is loaded onto a mass-spring system. It’s displacement is described by the equation

$mx'' = -k(s+x)+mg+(-\gamma x'),$

where $m,k,s,g,\gamma$ are constants and $\gamma > 0$ describes a particular damping force. For a proper derivation, see this post. Given that $mg = ks$ , find an expression for $x$ in terms of $t$ in the case:

$\gamma^2 - 4mk > 0$ ,
$\gamma^2 - 4mk = 0$ ,
$\gamma^2 - 4mk < 0$ .

What is the long-run behaviour of $x$ ? What would happen if there was no damping force (i.e. $\gamma = 0$ )?

(Click for Solution)

Solution. Since $mg = ks$ ,

$mx'' = -kx-\gamma x' \quad \Rightarrow \quad mx'' + \gamma x' + kx = 0.$

Consider the auxiliary equation $mr^2 + \gamma r + k = 0$ , where we used $r$ to reduce ambiguity. By the quadratic equation, the roots $r_{\pm}$ are given by

$\displaystyle r_{\pm} = \frac{-\gamma \pm \sqrt{\gamma^2 - 4mk}}{2m}.$

In the case $\gamma^2 - 4mk > 0$ , the roots $r_{\pm}$ are real and distinct and negative. Thus,

$x(t) = C_1 e^{r_- t} + C_2 e^{r_+ t}.$

This is known as an overdamped system.

In the case $\gamma^2 - 4mk = 0$ , the roots $r_{\pm} = -\gamma/(2m)$ are real and repeated and negative. Thus,

$x(t) = (C_1 t + C_2)e^{rt}.$

This is known as a critically damped system.

In the case $\gamma^2 - 4mk < 0$ , the roots $r_{\pm} = p \pm iq$ are complex conjugates with $p = -\gamma/(2m) < 0$ . Thus,

$x(t) = e^{pt} (C_1 \sin(qt) + C_2 \cos(qt)).$

This is known as an underdamped system.

In all instances, the negative term $-r/(2m)$ induces all exponential terms to $\to 0$ , so that $x(t) \to 0$ as $t \to \infty$ . In the case $\gamma = 0$ , we only have the case $-4mk < 0$ , so that