Damped Oscillation 12-4-2020

Substituting the given parameters in the given differential equation, we obtain,

$\ddot{y} + 0.5 \dot{y} + 5 y = 10$

Taking the Laplace transform of the both sides, and keeping in mind that $y(0) = \dot{y}(0) = 0$ ,

$s^2 Y(s) + 0.5 s Y(s) + 5 Y(s) = \dfrac{10}{s}$

So that,

$Y(s) = \dfrac{ 10 } { s (s^2 + 0.5 s + 5 ) } = \dfrac{ 10 }{ s ( (s + a)^2 + b^2 )}$

where $a = 0.25$ , $b^2 = 5 - a^2$

Using partial fractions, $Y(s)$ can be expressed as follows:

$Y(s) = \dfrac{10}{ s( (s + a)^2 + b^2 ) } = \dfrac{A}{s} + \dfrac{ B (s + a) }{ (s + a)^2 + b^2 } + \dfrac{ C b } { (s + a)^2 + b^2 }$

And this means that,

$10 = A ( (s + a)^2 + b^2 ) + B s (s + a) + C b s$

Equating equal powers of $s$ on both sides of the equation, we deduce that,

$0 = A + B$

$0 = 2 A a + B a + C b$

$10 = A (a^2 + b^2)$

Hence,

$A = \dfrac{10}{a^2 + b^2}$

$B = - \dfrac{10}{a^2 + b^2}$

$C = - \dfrac{a}{b} (2 A + B ) = - \dfrac{10 a}{b(a^2 + b^2)}$

Finally, we Laplace invert the partial fraction expansion of $Y(s)$ , to get,

$y(t) = A + B e^{-a t} \cos( b t ) + C e^{-at} \sin(b t )$

Using the values of $a , b$ , we find that $A = 2 , B = -2 , C = -0.22501758$ , hence, substituting $t = 7$ , we get,

$y(7) \approx \boxed{ 2.337 }$

It's been a long, long time. It's good to be back.

Here's the differential equation:

$\ddot{y} = 10-5y-0.5 \dot{y}$

Solve using RK4 till $t = 7$ :

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#4th order Runge-Kutta numerical solution of second order ODE 

import math 

#initialisation 

def sin(x): 
    return math.sin((math.pi/180)*x) 

def cos(x): 
    return math.cos((math.pi/180)*x) 

time = 0 
deltaT = 10**-4


u_1 = 0
u_2 = 0

def differentialEquation(t,x,xdot):  #applying RK4 by splitting 2nd order ODE into 2 first order ODEs 
    return (10-5*x-0.5*xdot)


while time <= 7: 

    #evaluating slope at various points 


    k0 = deltaT*u_2 
    L0 = deltaT*differentialEquation(time,u_1,u_2) 

    k1 = deltaT*(u_2 + 0.5*L0) 
    L1 = deltaT*differentialEquation(time + 0.5*deltaT, u_1 + 0.5*k0, u_2 + 0.5*L0) 

    k2 = deltaT*(u_2 + 0.5*L1) 
    L2 = deltaT*differentialEquation(time + 0.5*deltaT, u_1 + 0.5*k1, u_2 + 0.5*L1) 

    k3 = deltaT*(u_2 + L2) 
    L3 = deltaT*differentialEquation(time + deltaT, u_1 + k2, u_2 + L2) 


    #RK4 approximation of values 

    u_1 = u_1 + (1/6)*(k0 + 2*k1 + 2*k2 + k3) 
    u_2 = u_2 + (1/6)*(L0 + 2*L1 + 2*L2 + L3) 

    time += deltaT 

print(u_1)