Solving ode of forced oscillator with dumping

[Cypeq]

Hi i have to solve this ODE which descirbes motion of forced oscillator with dumping and constant friction :p

I'm already solving it numerically with Runge-Kutta 4 yet I'm totaly puzzeled how to do it analytically.

equation:

mx'' + kx' + w^2_0x + F_f = A cos(\delta t) Ff delta k and w are constant

moving acceleration x'' to one side we get

x'' = \frac 1 m (- kx' - w^2_0x - F_f + A cos(\delta t))

i need to solve this equation twice to get velocity x' than position x. Yet i have no clue i know only how to solve x' = f(x) first order ODE :/

 

[elibj123]

first of all rearrange the equation

mx''+kx'+w^{2}_{0}x=Acos(\delta t)-F_{f}

Divide the solution into two steps.

1) Homogenous solution
This is the behaviour of the system without the external force acting on it

x''+\frac{k}{m}x'+w^{2}_{0}x=0

This is a constant coefficient equation, and solved with the characteristic polynomial:

r^{2}+\frac{k}{m}r+\frac{w^{2}_{0}}{m}=0

The roots are given by:

r_{1,2}=\frac{-k\pm \sqrt{k^{2}-4mw^{2}_{0}}}{2m}

Now I'm going to assume 4mw^{2}_{0}>k^{2} and define:

\alpha = \frac{k}{2m}; \omega=\sqrt{-\frac{k^{2}}{4m}+w^{2}_{0}}

so
r_{1,2}=\alpha\pm i\omega

Therefore the set of homogenous solutions is
x_{1}(t)=e^{-\alpha t}cos(\omega t); x_{2}(t)=e^{-\alpha t}sin(\omega t)

The general homogenous solution is given by superposition.

2) Particular Solution
The response of the system to the external force

I'm going to use "guessing", but first we will divide the response into two:

2.1) Response to the constant force F_{f}
The easiest guess is that x itself will be constant. So if we choose x=c, we'll have

x_{p,1}=c=-\frac{F_{f}}{w^{2}_{0}}

2.2) Response to the harmonic force
Now we going to choose some x=Ccos(\delta t)+Dsin(\delta t)

Substituting this into the equation gives:

-mC\delta^{2} cos(\delta t) -mD\delta^{2}sin(\delta t)-kC\delta sin(\deltat) + kD\delta cos(\delta t)+w^{2}_{0}Ccos(\delta t)+w^{2}_{0}Dsin(\delta t)=Acos(\delta t)

Comparing coefficient of cosines and sines gives:

-mC\delta^{2}+kD\delta+w^{2}_{0}C=A
-mD\delta^{2}-kC\delta+w^{2}_{0}D=0

which gives

C=\frac{A(w^{2}_{0}-m\delta^{2})}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}; D=\frac{Ak\delta}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}

So x_{p,2}=\frac{A(w^{2}_{0}-m\delta^{2})}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}cos(\delta t)+\frac{Ak\delta}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}sin(\delta t)Now finally we compose all of the results together to get the general solution of the ODE:

x(t)=c_{1}e^{-\alpha t}cos(\omega t)+c_{2}e^{-\alpha t}sin(\omega t)-\frac{F_{f}}{w^{2}_{0}}+\frac{A(w^{2}_{0}-m\delta^{2})}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}cos(\delta t)+\frac{Ak\delta}{(w^{2}_{0}-m\delta^{2})^{2}+k^2\delta^{2}}sin(\delta t)

The first & second term are superposition of the homogoneous solutions.
The third & fourth term are sum of the responses to the external force.

The coefficients c1 & c2 are arbitrary but given initial conditions you can find them.

 

[HallsofIvy]

Nicely done.

 

[Cypeq]

big thanks it took me some time how to get this results but u helped me a lot thanks mate :)