Forced Damped Harmonic Oscillator with
Arbitrary Initial Conditions
Introduction
The motion
of the subject harmonic oscillator is a result that reflects very many
real-life mechanisms. Typically we
expect a drag force that is proportional to the speed of the oscillator and a
restoring force that is proportional to its displacement. The drag coefficient will be labeled b here
and the restoring force coefficient will
be labeled k. Of course, the coefficient
of the acceleration will be the mass, m, of the oscillating object as expected
from Newton's law of motion. The frequency of oscillation of the free
oscillator will be complex since there is a drag force and this frequency will
be labeled w. Since ω is complex, it will have a real part which I
will name ω0 and an imaginary part
labeled α. For initial conditions, we may have
arbitrary displacement, y(0), as well
as arbitrary initial speed where 0 here indicates time=0, and time is
labeled t here. These are the parameters of the free
oscillator.
The parameters of the sinusoidal force are the peak force
value, F, and the frequency of the force, ωF.
Calculating the Motion of the Oscillator.
The values of ω0 and α come from the solution to a quadratic equation
of the damped and un-forced (free)
harmonic oscillator:
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(1.1)
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Assuming that 4mk>b2
and separating ω into real and imaginary parts we have:
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(1.2)
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For the sinusoidally forced harmonic oscillator we
have the equation:
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(1.3)
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(1.4)
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(1.5)
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Combining these the general solution for both the
transient and steady state is:
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(1.6)
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The time derivative of the general solution is:
(1.7)
To match initial conditions, we use the values of both y and
at t=0.
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(1.8)
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where both cc and cs have to be solved
for.
For generality, I will assume that both y(0) and are non-zero and will now solve for the values
of constants cc and cs.
Setting up a matrix multiplier for these
constants and using equation (1.8) we obtain:
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(1.9)
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The inverse of a 2x2 matrix is computed by swapping
the diagonal elements and negating the off-diagonal elements and then dividing
all elements by the determinant of the original matrix.
Multiplying equation (1.9)
by the inverse of the matrix on its left we get:
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(1.10)
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(1.11)
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We can also express the general solution as cosines only by
applying the phase of the ω0 and the ωF terms separately.
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(1.12)
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where:
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(1.13)
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(1.14)
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and
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(1.15)
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(1.16)
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