Damped and forced oscillations
Department of Physics CTU FEE
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Problem 1: The Laplace transform
Apply Laplace transform
to the following functions and equations without using tables, formulate conclusions.
Parameters A, a, b, ω, μ, Ω do not depend on time, the apostrophe means the time derivative. Function
function at point C:
=
f(t)= 1
f(t)= 1
g(t)=A f(t)
indicates the unit step
g(t)=f' (t) ; f(0)=a
g(t)=f'' (t)=h' (t);f' (t)=h (t) ; f(0)=a; f'(0)=b
; f(0)=a; f'(0)=0
g(t)=
; f(0)=0; f'(0)=b
g(t)=
; f(0)=a; f'(0)=b
g(t)=
; f(0)=a; f'(0)=0
g(t)=
; f(0)=0; f'(0)=b
g(t)=
; f(0)=a; f'(0)=b
Problem 2: The harmonic oscillator
Use the Laplace transform to find the general solution of the differential equation describing the harmonic oscillator
Using the general solution find a specific solution for the following initial conditions:
Formulate conclusions.
Problem 3: Damped linear oscillator
Use Laplace transform to find the general solution of the differential equation describing the damped linear oscillator
Display graphs for different values a, b, μ, ω
Problem 4: Forced Oscillations
Apply Laplace transform to the differential equations describing damped linear oscillator forced by a harmonic signal
Familiarize yourself with the general solution and draw the resonance curve for the amplitude of the forced oscillation.
The harmonic oscillator
We transform the left side of the equation
Substituting into the equation for
variable Y and initial conditions, we solve the resulting equation for Y
The reverse transformation will find the solutions of differential equations in general form:
Note that the numerator of the term reflects the initial conditions and the denominator reflects the differential equation. (The denominator is the characteristic
equation.)
The harmonic oscillator
Correctness of your initial conditions substitution make comparison with the analytical solution
:
Damped linear oscillator
Transforming the left side of the equation
Substituting into the equation for
, we obtain
variable Y and initial conditions, we solve the resulting equation for Y
Note that here the denominator is the characteristic equation and the form of the numerator is adjusted by the initial conditions.
Damped linear oscillator
If μ>ω, denominator can be written as
and reverse transformation will find the solutions of differential equations in general form:
If μ<ω, denominator can be written as
and reverse transformation will find the solutions of differential equations in general form:
In case μ=ω critical damping denominator can be written in the form
general form:
and reverse transformation will find the solutions of differential equations in
Damped linear oscillator
The results compared with the analytical solution
Note that this solution does not expect equality μ = ω.
The equation for the critical damping
In case of subcritical damping, we assume that μ = 2, ω = 3
In case of supercritical damping, we assume that μ = 4, ω = 3
Damped linear oscillator - a graphic representation
To plot the graph we can use the expression in the form of harmonic functions multiplied by the
exponential
Damped linear oscillator - a graphic representation
Representation using a complex exponential helps to better understand why the critical damping system quickly approaches to zero without overshoot.
Forced oscillations
Transforming the left side of the equation
we obtain
Transforming the right side of the equation A sin(Ω t)
Substituting into the equation for
variable Y and initial conditions, we solve the resulting equation for Y
After the reverse transformation of this expression we get a complex expression that, like in the case of damped linear oscillator was difficult to interpret.
Forced oscillations
For simplicity, we choose zero initial conditions.
This result can be written as the sum of two fractions
+
Reverse transformation
is a linear combination of terms
solution damped harmonic oscillator.
Simplify the situation prerequisite μ =
and find a reverse transformation:
, where K, L, M a N are constants depending on Ω, μ a ω.
. Reverse transformation
corresponds to the
Note that the transition process, described by a fraction
generally occurs even at zero initial conditions.
The same two solutions, transient and forced oscillations appear even at zero initial conditions.
Forced oscillations - analytical solutions
The general solution of the equation of forced oscillations:
The last two terms describe transient. Note that the steady oscillations are not affected by the initial conditions i.e. by integration constants
Steady-state oscillations are described in terms
First simplify the denominator of the expression:
a
.
Then we find the amplitude of the resulting oscillations:
Finally, we find maximum amplitude of conditions
Forced oscillations - resonance: amplitude
Created with Wolfram Mathematica 8.0