### Author Topic: Problem of the week 4a  (Read 6307 times)

#### Victor Ivrii ##### Problem of the week 4a
« on: January 31, 2013, 05:35:09 PM »
Consider equation describing harmonic oscillations
\begin{equation}
y'' + \omega^2 y=0
\label{eq-1}
\end{equation}
\begin{equation}
y'' + \alpha y'+ \omega^2 y=0,\qquad \alpha>0.
\label{eq-2}
\end{equation}

a) Describe solutions

Now add external force
\begin{equation}
y'' + \alpha y'+ \omega^2 y=\cos(\beta t)
\label{eq-3}
\end{equation}
and find solution in the form
\begin{equation}
y=A(\beta ) \cos(\beta t)+B(\beta )\sin(\beta t).
\label{eq-4}
\end{equation}

b) Find resonance frequency $\beta$ such that amplitude $C(\beta)=\sqrt{A(\beta)^2+B(\beta)^2}$ reaches maximum.

#### Brian Bi ##### Re: Problem of the week 4a
« Reply #1 on: January 31, 2013, 06:19:20 PM »
The characteristic equation for (2) is as follows
\begin{equation}
r^2 + \alpha r + \omega^2 = 0
\end{equation}
which has the solutions
\begin{equation}
r_{1,2} = \frac{-\alpha \pm \sqrt{\alpha^2 - 4\omega^2}}{2}
\end{equation}
We consider three cases:
• $\alpha^2 > 4\omega^2$, i.e., $\alpha > 2\omega$ (assuming $\omega$ is always positive). In this case $r_1, r_2 \in \mathbb{R}$ and the general solution to (2) is given by $y = A e^{r_1 t} + B e^{r_2 t}$. Indeed, both roots are negative so all solutions decay monotonically to zero as $t \to \infty$ (unless $\omega = 0$ in which case we have constant solutions).
• $\alpha^2 = 4\omega^2$, i.e., $\alpha = 2\omega$. Here $r_1 = r_2 = -\alpha/2$ so the general solution is given by $y = (A + Bt)e^{-\alpha t/2} = (A+Bt)e^{-\omega t}$. Again, all solutions decay to zero as $t \to \infty$.
• $\alpha^2 < 4\omega^2$, i.e., $\alpha < 2\omega$. In this case we have two complex roots, $r_1 = -\alpha/2 + i \sqrt{4\omega^2 - \alpha^2}/2$ and $r_2 = -\alpha/2 - i \sqrt{4 \omega^2 - \alpha^2}/2$. The solutions therefore have both a time-decaying exponential factor $e^{-\alpha t/2}$ and a sinusoidal factor with angular frequency $\sqrt{4 \omega^2 - \alpha^2}/2$. The general solution is $y = e^{-\alpha t/2} (A \cos (\sqrt{4 \omega^2 - \alpha^2}t/2) + B \sin (\sqrt{4\omega^2 - \alpha^2}t/2))$.

(continued)

#### Brian Bi ##### Re: Problem of the week 4a
« Reply #2 on: January 31, 2013, 07:25:00 PM »
Note that the greater of the two roots in case 1 is always greater than the root $-\alpha/2 = -\omega$ in case 2. We call case 1 overdamped because $\alpha$ is large and the greater root (i.e., less negative) leads to a slower exponential decay. In contrast, case 2 is called critically damped because solutions decay with time constant $\alpha/2 = \omega$, which is maximal. Case 3 is called underdamped because $\alpha$ is small and the damping force is insufficient to prevent oscillation. In this case the decay will also be slow since the time constant is $\alpha/2 < \omega$.

The solution to equation (3) is given by the sum of the general solution to the homogeneous equation (1) and a particular solution to (3) which we will solve by the method of undetermined coefficients. As hinted, let $y_p = A \cos (\beta t) + B \sin(\beta t)$. Then

\begin{align}
y'' + \alpha y' + \omega^2 y
&= (A \cos(\beta t) + B \sin(\beta t))'' + \alpha (A \cos(\beta t) + B \sin(\beta t))' + \omega^2 (A \cos (\beta t) + B \sin(\beta t)) \\
&= -A\beta^2 \cos(\beta t) - B\beta^2 \sin(\beta t) + -A\alpha\beta\sin(\beta t) + B\alpha\beta\cos(\beta t) + A\omega^2 \cos(\beta t) + B\omega^2 \sin(\beta t) \\
&= (-A\beta^2 + B\alpha\beta + A\omega^2)\cos(\beta t) + (-B\beta^2 -A\alpha\beta + B\omega^2)\sin(\beta t)
\end{align}

By matching coefficients we find
\begin{align}
(-\beta^2 + \omega^2) A + \alpha\beta B &= 1 \\
(-\alpha\beta) A + (-\beta^2 + \omega^2) B &= 0
\end{align}
This is a linear system in two equations and two unknowns, which is therefore easy to solve. The solution is
\begin{align}
A &= \frac{\omega^2 - \beta^2}{(\omega^2-\beta^2)^2 + (\alpha \beta)^2} \\
B &= \frac{\alpha\beta}{(\omega^2 -\beta^2)^2 + (\alpha \beta)^2}
\end{align}
so the particular solution to (3) is $y_p = A \cos(\beta t) + B\sin(\beta t)$ with $A, B$ given above and the general solution to (3) is given by adding $y_p$ and the general solution to (1) as found above.

(continued)

#### Brian Bi ##### Re: Problem of the week 4a
« Reply #3 on: January 31, 2013, 07:40:19 PM »
The amplitude $\sqrt{A^2 + B^2}$ is then:
\begin{align}
\sqrt{A^2+ B^2} &= \left[\left(\frac{\omega^2 - \beta^2}{(\omega^2-\beta^2)^2 + (\alpha \beta)^2}\right)^2 + \left(\frac{\alpha\beta}{(\omega^2 -\beta^2)^2 + (\alpha \beta)^2}\right)^2\right]^{1/2} \\
&= \left[\frac{(\omega^2 - \beta^2)^2 + (\alpha\beta)^2}{((\omega^2-\beta^2)^2+(\alpha\beta)^2)^2}\right]^{1/2} \\
&= ((\omega^2 - \beta^2)^2 + (\alpha \beta)^2)^{-1/2}
\end{align}
This will be maximized if we minimize the expression $(\omega^2 - \beta^2)^2 + (\alpha \beta)^2$, which we expand to give
$\omega^4 + (\alpha^2 - 2\omega^2) \beta^2 + \beta^4$. This is a quadratic in $\beta^2$ with positive leading coefficient, so it has
a global minimum at $\beta^2 = -\frac{1}{2}(\alpha^2 - 2\omega^2)$, or $\beta = \sqrt{\omega^2 - \alpha^2/2}$. This value of $\beta$, the resonance frequency, is real and nonzero only if $\alpha < \sqrt{2} \omega$, so resonance occurs only in underdamped systems (and only those that are sufficiently underdamped).

#### Victor Ivrii ##### Re: Problem of the week 4a
« Reply #4 on: February 01, 2013, 04:47:13 AM »
Very nice. The only remark: in computer calculations or sketching overdumpness starts earlier

http://weyl.math.toronto.edu/MAT244-2011S-forum/index.php?topic=48.msg159#msg159