FEB 14, 2017
$ m\frac{d^2x}{dt^2} + \gamma\frac{dx}{dt} + 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?
When $ \gamma = 0 $
(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\frac{d^2x}{dt^2} + \gamma\frac{dx}{dt} + kx = 0 $
characteristic equation:
$ m\lambda ^2+\gamma\lambda + k = 0 $
solve for $ \lambda $:
$ \lambda_1 = \frac{-\gamma+\sqrt{\gamma^2-4mk}}{2m} $ and $ \lambda_2 = \frac{-\gamma-\sqrt{\gamma^2-4mk}}{2m} $, $\Delta=\gamma ^2-4mk$
(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.
$ x = Ae^{i\omega t}$
plugging in the homogeneous equation:
$ A(i\omega)^2+\frac{Ai\omega}{10}+A=1 $
$ A=\frac{1}{-\omega^2+\frac{i\omega}{10}+1} $
a particular solution:
$ x=\frac{1}{-\omega^2+\frac{i\omega}{10}+1}e^{i\omega t}=\frac{1}{-\omega^2+\frac{i\omega}{10}+1}(cos\omega t+isin\omega t) $
amplitude = $ abs(\frac{1}{-\omega^2+\frac{i\omega}{10}+1}) = (\sqrt{(1-w^2)^2+(\frac{\omega}{10})^2})^{-1} $
import numpy as np
import scipy as sp
from matplotlib import pyplot as plt
omega = np.arange(0, 3, 0.05)
amplitude = 1 / np.sqrt((1-omega**2)**2 + (omega/10)**2)
phase = np.angle(1/(-omega**2+1+(0.1j*omega)))
plt.plot(omega, amplitude)
plt.xlabel('Frequency')
plt.ylabel('Amplitude')
plt.show()
plt.plot(omega, phase)
plt.xlabel('Frequency')
plt.ylabel('Phase')
plt.show()
I found a very clear animation about the amplitude and the phaseof a driven harmonic oscillator.
Click "An animation" in the contents.
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