(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 damping constant \(\gamma\) should be \(0\) for the system to exhibit a simple undamped harmonic motion.
(b) Find the solution to the homogeneous equation, and comment on the possible
cases. How does the amplitude depend on the frequency?
Homogeneous equation: \[m\ddot{x}+\gamma \dot{x}+kx = 0\]
Try \(x = e^{rt}\).
\[mr^2 e^{rt}+\gamma r e^{rt} + k e^{rt} = 0\]
\[mr^2 + \gamma r + k = 0\]
Root of r is \[-\frac{\gamma}{2m}\pm \sqrt{(\frac{\gamma}{2m})^2-\frac{k}{m}}\],
which makes the general solution
\[x_g = C_1 \cdot exp\{(-\frac{\gamma}{2m}+ \sqrt{(\frac{\gamma}{2m})^2-\frac{k}{m}})\}\]
\[+ C_2 \cdot exp\{(-\frac{\gamma}{2m} - \sqrt{(\frac{\gamma}{2m})^2-\frac{k}{m}})\}\]
Let \(\beta = \frac{\gamma}{2m}\), \(\alpha = \frac{k}{m}\). Then the general solution
\[x_g = C_1 \cdot exp\{(-\beta + \sqrt{\beta^2 - \alpha})t\} + C_2 \cdot exp\{(-\beta - \sqrt{\beta^2 - \alpha})t\}\]
When \(\beta^2 < \alpha\), the solution becomes
\[C_1 \cdot exp\{(-\beta + i\sqrt{\alpha - \beta^2})t\} + C_2 \cdot exp\{(-\beta - i\sqrt{\alpha - \beta^2})t\}\]
\[= e^{-\beta} (C_1+C_2)(cos{\sqrt{\alpha - \beta^2}t}+i sin{\sqrt{\alpha - \beta^2}t})\]
(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, γ = 0.1.
Inhomogeneous equation: \[m\ddot{x}+\gamma \dot{x}+kx = e^{i\omega t}\]
Try particular solution \(x = A e^{i\omega t}\):
\[-mA \omega^2 e^{i\omega t} + \gamma A i\omega e^{i\omega t} + k A e^{i\omega t} = e^{i\omega t}\]
\[A = (-m\omega^2 + i \gamma \omega + k)^{-1}\]
If we substitute \(m = k = 1, \gamma = 0.1 \),
\[A = (-\omega^2 + 1 + i \cdot 0.1\omega)^{-1}\]
Let's plot it!
(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 2π 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?
First definition seems too obscure for me..
Let's look at the second definition first.
(Useful link 1
2
)
Total energy of oscillator is:
\[E = E_{kinetic} + E_{potential} = \frac{1}{2}m {\frac{dx}{dt}}^2 + \frac{1}{2}k x^2\]
\[\begin{align*}\frac{dE}{dt} & = m \frac{d^2x}{dt^2} \frac{dx}{dt} + kx \frac{dx}{dt}
\\& = \frac{dx}{dt} (m\frac{d^2x}{dt^2} + kx)
\\& = -\gamma(\frac{dx}{dt})^2
\end{align*}\]
(e) Now find the solution to the equation by using Laplace transforms.
\[\mathcal{L}\{x(t)\} = X(s) = \int_{0}^{\infty} e^{st}x(t)dt\] \[\mathcal{L}(m\ddot{x}+\gamma \dot{x}+kx) = m \{s^2 X(s) - s x(0) - \dot{x}(0)\} + \gamma \{s X(s) - x(0)\} + k X(s)\] Since the initial condition is \(x(0) = \dot{x}(0) = 0\), \[mX(s) \cdot s^2 + \gamma X(s) \cdot s + k X(s) = \frac{1}{s-i \omega}\] \[\begin{align*} X(s) & = \frac{1}{(ms^2+\gamma s + k)(s-i\omega)} \\& = \frac{1}{m(s+s_1)(s+s_2)(s-i\omega)} \end{align*}\] \[(s_{1,2} = \frac{\gamma}{2m} \pm \sqrt{(\frac{\gamma}{2m})^2 - \frac{k}{m}})\] \[x(t) = \frac{e^{-s_1t}}{(s_1-s_2)(s_1+i\omega)}-\frac{e^{-s_2t}}{(s_1-s_2)(s_2+i\omega)} + \frac{e^{i\omega t}}{(s1+i\omega)(i\omega + s_2)}\]
(f) For an arbitrary potential minimum, work out the form of the lowest-order correction to simple undamped unforced harmonic motion.
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Explicitly solve (and try to simplify) the system of differential equations for two coupled harmonic oscillators, and then find the normal modes by matrix diagonalization.
Using Newton's law and Hooke's law, we get differential equations \[m \ddot{x_1} = -kx_1 - k(x_1-x_2) = -k(2x_1 - x_2)\] \[m \ddot{x_2} = -kx_2 - k(x_2-x_1) = -k(x_1 - 2x_2)\] The corresponding vector eqaution is \[m\frac{d^2\vec{x}}{dt^2} + k \mathbf{A} \cdot \vec{x}\] where \[\vec{x} = \begin{bmatrix}x_1\\x_2\end{bmatrix}, \mathbf{A} = \begin{bmatrix}2 & -1\\-1 & 2\end{bmatrix}.\] Try \(\vec{x} = \vec{v} e^{rt}\): \[mr^2 \vec{v} + k \mathbf{A} \vec{v} = 0\] \[(mr^2 \mathbf{I} + k \mathbf{A}) \cdot \vec{v}= 0\] \[det(mr^2 \mathbf{I} + k \mathbf{A}) = 0\] \[mr^2 + 2k = \pm k\] \[r = i \sqrt{3k/m} \text{ or } i\sqrt{k/m}\] Now we have our general solution \[\vec{x} = C_1 \vec{v} e^{i\sqrt{3k/m}}+ C_2 \vec{v} e^{i\sqrt{k/m}}\]
A common simple digital filter used for smoothing a signal is
y(k) = αy(k − 1) + (1 − α)x(k) ,
where α 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?
\[\begin{align*}\mathcal{Z}\{\alpha y(k-1)\} & = \sum_{k=0}^{\infty}\alpha y(k-1)z^{-k} \\& = \sum_{k'=-1}^{\infty}\alpha y(k')z^{-k'}z^{-1} \\& = \frac{\alpha}{z} \sum_{k'=0}^{\infty} y(k')z^{-k'} + \frac{\alpha}{z} y(-1)z \\& = \frac{\alpha}{z} Y(z) \end{align*}\] \[Y(z) = \frac{\alpha}{z}Y(z) + (1-\alpha)X(z)\] \[(1- \frac{\alpha}{z})Y(z) = (1-\alpha)X(z)\] \[Y(z) = \frac{z(1-\alpha)}{z-\alpha} X(z)\] How do I get the amplitude out of this?