(3.1)¶
Consider the motion of a damped, driven oscillator (such as a mass on a spring, a ball in a well, or a pendulum making small motions):
$m \ddot{x} + \gamma \dot{x} + kx = e^{i \omega t}$
(a) Under what conditions will the governing equations for small displacements of a particle around an arbitrary 1D potential minimum be simple undamped harmonic motion?
The conditions that allow for simple undamped harmonic motion are when there are no outside forces of friction or gravity being accounted for or affecting the system.
(b) Find the solution to the homogeneous equation, and comment on the possible cases. How does the amplitude depend on the frequency?
$m \ddot{x} + \gamma \dot{x} + kx = 0$
$x = Ae^{i \omega t},\ \dot{x} = Ai \omega e^{i \omega t},\ \ddot{x} = -A \omega^2 e^{i \omega t}$
$-Am\omega^2 e^{i\omega t} + A\gamma i\omega e^{i\omega t} + Ake^{i\omega t} = 0$
$Ae^{i\omega t}(-m\omega^2 + \gamma i\omega + k) = 0$
$-m\omega^2 + \gamma i\omega + k = 0$
$m\omega^2 - \gamma i\omega - k = 0$
$\omega = \frac{\gamma i \pm \sqrt{\gamma^2 + 4mk}}{2m}$
The amplitude does not depend on frequency.
(c) Find a particular solution to the inhomogeneous problem by assuming a response at the driving frequency, and plot its magnitude and phase as a function of the driving frequency for $m = k = 1, \gamma = 0.1$.
$m\ddot{x} + \gamma \dot{x} + kx = e^{i\omega t}$
$x = Ae^{i \omega t},\ \dot{x} = Ai \omega e^{i \omega t},\ \ddot{x} = -A \omega^2 e^{i \omega t}$
$Ae^{i\omega t}(-m\omega^2 + \gamma i\omega + k) = e^{i\omega t}$
$A(-m\omega^2 + \gamma i\omega + k) = 1$
$A = \frac{1}{k + \gamma i\omega -m\omega^2}$
import numpy as np
import matplotlib.pyplot as plt
m = 1
k = 1
gamma = 0.1
omega = np.linspace(0, np.pi, 1000)
A = 1 / (-m*omega**2 + 1j*gamma*omega + k)
A_mag = np.absolute(A)
plt.plot(omega, A_mag)
plt.xlabel("Frequency")
plt.ylabel("Magnitude")
A_phase = np.angle(A)
plt.plot(omega, A_phase)
plt.xlabel("Frequency")
plt.ylabel("Phase")
(d) For a driven oscillator the $Q$ or $Quality factor$ is defined as the ratio of the center frequency to the width of the curve of the average energy (kinetic + potential) in the oscillator versus the driving frequency (the width is defined by the places where the curve falls to half its maximum value). For an undriven oscillator the $Q$ is defined to be the ratio of the energy in the oscillator to the energy lost per radian (one cycle is $2 \pi$ radians). Show that these two definitions are equal, assuming that the damping is small. How long does it take the amplitude of a 100 Hz oscillator with a $Q$ of $10^9$ to decay by $1/e$?
$\langle E \rangle = \langle K \rangle + \langle P \rangle$
$E = \frac{1}{2}m \langle \dot{x}^2 \rangle + \frac{1}{2}k \langle x^2 \rangle$
Because $x=A^{i\omega t}$ from part (c) and $\langle sin^2 \rangle = \langle cos^2 \rangle = \frac{1}{2}$, we get
$\langle E \rangle = \frac{1}{2}m \omega^2 A^2 \langle sin^2 (\omega t) \rangle + \frac{1}{2} k A^2 \langle cos^2 (\omega t) \rangle$
$\langle E \rangle = \frac{1}{4} A^2 (k + m \omega^2)$
$A = \frac{1}{k + \gamma i\omega -m\omega^2}$ (from part (c)), so
$\langle E \rangle = \frac{1}{4} \frac{k + m \omega^2}{(k + \gamma i\omega -m\omega^2)(k - \gamma i\omega -m\omega^2)}$
$\langle E \rangle = \frac{1}{4} \frac{k + m \omega^2}{(k - m \omega^2 )^2 + \gamma^2 \omega^2}$
$\langle E \rangle = \frac{1}{4m} \frac{\omega_0^2 + \omega^2}{(\omega_0^2 - \omega^2 )^2 + \omega^2 \gamma^2 / m^2}$
By substituting $\omega = \omega_0 + \delta$, we get
$\langle E \rangle \approx \frac{2}{4 \delta^2 + \gamma^2 / m^2}$
Which falls to half its peak when $4 \delta^2 = \frac{\gamma^2}{m^2}$, or when $\delta = \pm \frac{1}{2} \frac{\gamma}{m}$
This means that $Q = \frac{\omega}{\Delta \omega} \approx \frac{\omega_0}{\gamma / m}$
We know that the amplitude for an undriven oscillator falls off as $A = e^{frac{-\gamma t}{2m}}$ and for light damping, the frequency is $\omega \approx \omega_0$.
Also, we know that $Q = \frac{E}{\Delta E} = \frac{\omega_0}{\gamma / m}$, so
$e^{\frac{-\gamma t}{2m}} = e^{\frac{-\omega t}{2Q}}$
Therefore, an e-folding time is $t = \frac{2Q}{\omega}$
Plugging in our numbers, we get $t = \frac{2 \times 10^9}{100 Hz} = 2 \times 10^7$ seconds.
(e) Now find the solution to equation (3.58) by using Laplace transforms. Take the initial condition as $x(0) = \dot{x}(0) = 0$.
$m\ddot{x} + \gamma \dot{x} + kx = e^{i\omega t}$
$\ddot{x} \to s^2X(s) - sx(0) - \dot{x}(0) = s^2X(s)$
$\dot{x} \to sX(s) - x(0) = sX(s)$
$x \to X(s)$
$ms^2X(s) + \gamma sX(s) + kX(s) = \frac{1}{s-i\omega}$
$X(s)(ms^2 + \gamma s + k) = \frac{1}{s-i\omega}$
$X(s) = \frac{1}{s-i\omega} \frac{1}{ms^2 + \gamma s + k} = \frac{1}{s-i\omega} \frac{1}{s^2 + \frac{\gamma}{m} s + \frac{k}{m}} \frac{1}{m}$
Using the quadratic formula, we get $s^2+\frac{\gamma}{m}s+\frac{k}{m} = (s+\alpha+\beta)(s+\alpha-\beta)$
Where $\alpha = \frac{\gamma}{2m}$ and $\beta = \frac{1}{2}\sqrt{(\frac{\gamma}{m})^2-\frac{4k}{m}}$
Using Table 3.1, we can now find the inverse Laplace transform to solve for $x(t)$
$X(s) = \frac{1}{s-i\omega} \frac{1}{(s+\alpha+\beta)(s+\alpha-\beta)} \frac{1}{m}$
$x(t) = \frac{e^{i\omega t}}{m(-i\omega-\alpha-\beta)(-i\omega-\alpha+\beta)} - \frac{e^{-(\alpha+\beta)t}}{m(-i\omega-\alpha-\beta)(\alpha+\beta-\alpha+\beta)} - \frac{e^{-(\alpha-\beta)t}}{m(-i\omega-\alpha+\beta)(\alpha-\beta-\alpha-\beta)}$
After quite a bit of expansions/simplifications (but not all of them), we derive the following answer:
$x(t) = \frac{e^{i\omega t}}{-m\omega^2+\gamma i\omega + k} - \frac{e^{-(\alpha+\beta)t}}{(-i\omega-\alpha-\beta)(2\beta m)} - \frac{e^{-(\alpha-\beta)t}}{(-i\omega-\alpha+\beta)(-2\beta m)}$
(f) For an arbitrary potential minimum, work out the form of the lowest-order correction to simple undamped unforced harmonic motion.
Starting with the Taylor series expansion of simple undamped harmonic motion, we have
$V(x) = V_0 + \frac{1}{2} V_2 x^2 + \mathcal{O}(x^3)$
The next term is a cubic, which gives us
$V(x) = V_0 + \frac{1}{2} V_2 x^2 + \frac{1}{6} V_3 x^3 + \mathcal{O}(x^4)$
Substituting $V_2 = k$ and $V_3/2 = \epsilon$, the equation of motion is
$m\ddot{x} + kx + \epsilon x^2 = 0$
We can expand $x = x_0 + \epsilon x_1 + \mathcal{O}(\epsilon^2)$ to get
$m\ddot{x_0} + k x_0 = 0$
The lowest-order correct equation is
$m\ddot{x_1} + k x_1 + x_0^2 = 0$
We can take the initial conditions to give an unperturbed solution of $x_0 = A^{i \omega_0 t}$ where $\omega_0 = \sqrt{k/m}$, which yields a correction equation
$m\ddot{x_1} + k x_1 + A^2 e^{2i\omega_0 t} = 0$
Which is solved by
$x_1 = \frac{A^2}{3m\omega_0^2} e^{2i\omega_0 t}$
(3.2)¶
Explicitly solve (and try to simplify) the system of differential equations for two coupled harmonic oscillators (don’t worry about the initial transient), and then find the normal modes by matrix diagonalization.
We start by adding up the forces on each mass:
$m\ddot{x_1} = -k x_1 -k(x_1 - x_2)$
$m\ddot{x_2} = -k x_2 -k(x_2 - x_1)$
Next, we rearrange terms and substitute $\omega_0^2 = k/m$:
$\ddot{x_1} + 2 \omega_0^2 x_1 - \omega_0^2 x_2 = 0$
$\ddot{x_2} + 2 \omega_0^2 x_2 - \omega_0^2 x_1 = 0$
We can try solutions $x_1 = e^{i \omega t}$ and $x_2 = Ae^{i \omega t}$ to reduce everything to two equations and two unknowns:
$-\omega^2 + 2 \omega_0^2 - \omega_0^2 A = 0$
$-A\omega^2 + 2 \omega_0^2 A - \omega_0^2$
Which simplifies to:
$(\omega^2 - 2 \omega_0^2) + \omega_0^2 A = 0$
$\omega_0^2 + (\omega^2 - 2 \omega_0^2) A = 0$
Solving the first equation for $\omega^2$, we find that $\omega^2 = \omega_0^2 (2 - A)$, which we can plug into the second equation to give us $A^2 = 1$, so $A = \pm 1$.
This gives us two possible values for $\omega$ by plugging in $A$ back into the first equation: $\omega = \omega_0$ and $\omega = \sqrt{3} \omega_0$. Each of these solutions correspond to either both masses moving in the same direction or in opposite directions.
With this knowledge, we will define new variables $x_1 = y_1 + y_2$ and $x_2 = y_1 - y_2$, which changes the governing equations to:
$\ddot{y_1} + \ddot{y_2} + 2 \omega_0^2 (y_1 + y_2) - \omega_0^2 (y_1 - y_2) = 0$
$\ddot{y_1} - \ddot{y_2} + 2 \omega_0^2 (y_1 - y_2) - \omega_0^2 (y_1 + y_2) = 0$
By adding and subtracting the two equations, the resulting equations are:
$\ddot{y_1} + \omega_0^2 y_1 = 0$
$\ddot{y_2} + 3 \omega_0^2 y_2 = 0$
We can find these same equations by diagonalizing the coupling matrix:
$\frac{d^2 \overrightarrow{x}}{dt^2} + \bigl(\begin{array}{cc} 2\omega_0^2 & -\omega_0^2 \\ -\omega_0^2 & 2\omega_0^2 \end{array}\bigr) \cdot \overrightarrow{x} = 0$
Where $\textbf{A}$ is the matrix $\bigl(\begin{array}{cc} 2\omega_0^2 & -\omega_0^2 \\ -\omega_0^2 & 2\omega_0^2 \end{array}\bigr)$
We find the eigenvalues by solving $0 = |\textbf{A} - \lambda \textbf{I}|$
$= \bigl|\begin{array}{cc} 2\omega_0^2 - \lambda & -\omega_0^2 \\ -\omega_0^2 & 2\omega_0^2 - \lambda \end{array}\bigr|$
$= \lambda^2 - 4 \omega_0^2 \lambda + 3 \omega_0^4$
Which gives us:
$\lambda = \omega_0^2, 3 \omega_0^2$
We get the corresponding eigenvectors: (1 1) and (1 -1), respectively, for our values of $\lambda$. Therefore, the coordinate transformation matrix is:
$\textbf{M} = \bigl(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\bigr)$
We can find the inverse of $\textbf{M}$ using basic linear algebra techniques, which give us:
$\textbf{M}^{-1} = \bigl(\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{array}\bigr)$
(3.3)¶
A common simple digital filter used for smoothing a signal is
$y(k)=\alpha y(k - 1) + (1 - \alpha)x(k)$ ,
where $\alpha$ is a parameter that determines the response of the filter. Use $z$-transforms to solve for $y(k)$ as a function of $x(k)$ (assume $y(k < 0) = 0$). What is the amplitude of the frequency response?
We will rearrange the z-transform of the initial equation
$Y(z) = \alpha [z^{-1} Y(z) + y(-1)] + (1 - \alpha) X(z)$
Which gives us the transfer function
$H(z) = \frac{Y(z)}{X(z)} = (1 - \alpha) \frac{z}{z - \alpha}$
We can invert this (using the formulas from Table 3.2) to give us the impulse response
$h(k) = (1 - \alpha) \alpha^k$
The output for an arbitrary input is then found from the convolution with the impulse response
$y(k) = (1 - \alpha) \sum_{n=0)^k \alpha^n x (k - n)$
We find the amplitude of the frequency response by plugging in $z = e^{i\omega \delta_t}$ to the transfer function
$|H(e^{i\omega \delta_t})| = |(1 - \alpha) \frac{e^{i\omega \delta_t}}{e^{i\omega \delta_t} - \alpha}|$
$= (1 - \alpha) [\frac{e^{i\omega \delta_t}}{e^{i\omega \delta_t} - \alpha} \frac{e^{-i\omega \delta_t}}{e^{-i\omega \delta_t} - \alpha}]^{1/2}$
$=(1 - \alpha) [\frac{1}{1 - \alpha e^{-i \omega \delta_t} - \alpha e^{i \omega \delta_t} + \alpha^2}]^{1/2}$
$= \frac{1 - \alpha}{\sqrt{1 + \alpha^2 - 2 \alpha cos \omega \delta_t}}$