,differential equation for the 2nd order system

Question

I would like to know which differential equation I have to solve in order to get the answer to a unitary step like the one report in figure:

tell me if I'm wrong: starting from :

$m\frac{d^2y_1}{dt}+b\frac{dy_1}{dt}+k y_1= 0$

I can compute the substitution $\frac{dy}{dt} = y_2 $

so the system of first order equation became : \begin{cases} y'_1 = y_2 \\ y'_2 = -\frac{b}{m}y_2 -\frac{k}{m} \\ \end{cases} right ? how can I set the constant ? which initial value should I use ?

EDIT Maybe I'm not well explain! I don't wanna know the output but which value and boundary condition I have to use in the system that I wrote in order to get the set of response in function of the parameter $\zeta$ (and also know what is $\zeta$) I know the soultion but I wanna solve the differential equations and get this set of curve !! Consider my question like: which is the function to integrate .... with wich value or like :"I have to give an exercise: using a euler method find the solution of a unit step response for different value of $\zeta$ and $\omega_o$ "

Thanks for your precious support ... what is $x(t)$ ?

Answer 1

From the image you can read of a period of about $6$ which confirms the label of a frequency $ω_0=1$. Thus the basis is a harmonic oscillator $y''+(y-1)=0$. In addition there is a friction term that leads to the represented exponential decay, where one can guess that the friction coefficient is $2\zeta$ (as per the normalized form in the answer of Mostafa Ayaz). In total this makes an equation $$y''+2\zeta y'+y=x.$$ As per usual in such situations of exploring system responses it is assumed that all functions are zero for $t<0$, so that the initial conditions are $y(0)=0$ and $y'(0)=0$. The unit step function is $x(t)=1$ for $t\ge 0$, and of course $x(t)=0$ for $t<0$. Then solve $$y''+2\zeta y'+y=1$$ for $t\ge 0$.

t = np.linspace(0,30,1501)
for zeta in [0.1,0.2,0.5,1,2,4]:
    sol = odeint(lambda y,t:[y[1], 1-y[0]-2*zeta*y[1]],[0,0], t)
    plt.plot(t,sol[:,0], label="$\zeta=$%.2f"%zeta);
plt.grid(); plt.legend(); plt.show()

to get

Answer 2

The output has such a Laplace transform$$Y(s)=\dfrac{1}{s^2+2\zeta\omega_0s+\omega_0^2}$$which is the output of the following ODE$$y''(t)+2\zeta\omega_0y'(t)+\omega_0^2y(t)=x(t)$$with initial rest

Answer 3

The differential equation you propose, when disturbed, will oscillate around y=0. The differential equation you should consider is $$m\frac{d^2y}{dt^2}+b\frac{dy}{dt}+ky=F(t)$$ where F(t) is a simple step function, namely, F=0 for t<0 and F=1/k for t>0. This will make the displacement oscillate around 1 as above. The solution to this equation can be achieved by several standard methods such as the Laplace transform listed @Mostafa Ayaz but in this case, the Laplace transform of the forcing function, F(t), must also be included.