In those considerations that follow, the positive axis is oriented from down-up for forces, velocities and displacements.
Supposing the mass m vertically placed over the spring-dashpot, the mass is actuated by
f-m g = m ddot x
where f is the force exerted by the spring-dashpot.
The spring-dashpot reacts according to
-f = kx + c dot x
Joining both equations we have
m ddot x + c dot x + k x = -mg
This is a linear non-homogeneous differential equation whose solution is given by
x = x_h + x_p
where x_h is the homogeneous solution
m ddot x_h + c dot x_h + k x_h = 0
and
x_p is a particular solution for
m ddot x_p + c dot x_p + k x_p = -mg
In this case x_p = -(mg)/k
The homogeneous equation has the general solution
x_h = e^{lambda t}
Substituting this generic solution we get at
(m lambda^2+c lambda + k)e^{lambda t} = 0
The feasible lambda are given by
lambda=(-c pm sqrt(c^2-4mk))/(2m). For our case we have
lambda = (-10 pm sqrt(100-4 xx 25 xx 226))/(50) = -1/5 pm 3i
so
x_h = e^{-t/5} e^{pm i 3 t} = e^{-t/5}(cos(3t)pm i sin(3t))
when lambda is a complex conjugate pair, for each pair the solution proposal is as follows.
x_h = (C_1 cos(3t)+ C_2sin(3t))e^{-t/5}
(This can be demostrated but requires a lot of algebra.)
The general solution is given by
x = (C_1 cos(3t)+ C_2sin(3t))e^{-t/5}-(m g)/k The constants C_1,C_2 are determined according to the movement initial conditions. So
{(x(0) = C_1-(m g)/k = 20),
(dot x(0) = -C_1/5+3C_2=41)
:}
From this system, we obtain the constants
C_1 = (m g)/k + 20, C_2 = (m g)/(15k) + 15 assuming g = 10 we have
C_1 = 21.1062, C_2 = 15.0737
So finally
x = (21.1062 cos(3t)+ 15.0737sin(3t))e^{-t/5}-1.10619
