# How do you find the position function x(t) if you suppose that the mass in a mass-spring-dashpot system with m=25, c=10 and k=226 is set in motion with x(0)=20 and x'(0)=41?

Aug 9, 2016

$x = \left(21.1062 \cos \left(3 t\right) + 15.0737 \sin \left(3 t\right)\right) {e}^{- \frac{t}{5}} - 1.10619$

#### Explanation:

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

$- f = k x + c \dot{x}$

Joining both equations we have

$m \ddot{x} + c \dot{x} + k x = - m g$

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

$m {\ddot{x}}_{p} + c {\dot{x}}_{p} + k {x}_{p} = - m g$

In this case ${x}_{p} = - \frac{m g}{k}$

The homogeneous equation has the general solution

${x}_{h} = {e}^{\lambda t}$

Substituting this generic solution we get at

$\left(m {\lambda}^{2} + c \lambda + k\right) {e}^{\lambda t} = 0$

The feasible $\lambda$ are given by

$\lambda = \frac{- c \pm \sqrt{{c}^{2} - 4 m k}}{2 m}$. For our case we have

$\lambda = \frac{- 10 \pm \sqrt{100 - 4 \times 25 \times 226}}{50} = - \frac{1}{5} \pm 3 i$

so

${x}_{h} = {e}^{- \frac{t}{5}} {e}^{\pm i 3 t} = {e}^{- \frac{t}{5}} \left(\cos \left(3 t\right) \pm i \sin \left(3 t\right)\right)$

when $\lambda$ is a complex conjugate pair, for each pair the solution proposal is as follows.

${x}_{h} = \left({C}_{1} \cos \left(3 t\right) + {C}_{2} \sin \left(3 t\right)\right) {e}^{- \frac{t}{5}}$

(This can be demostrated but requires a lot of algebra.)

The general solution is given by

$x = \left({C}_{1} \cos \left(3 t\right) + {C}_{2} \sin \left(3 t\right)\right) {e}^{- \frac{t}{5}} - \frac{m g}{k}$ 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

${C}_{1} = \frac{m g}{k} + 20 , {C}_{2} = \frac{m g}{15 k} + 15$ assuming $g = 10$ we have

${C}_{1} = 21.1062 , {C}_{2} = 15.0737$

So finally

$x = \left(21.1062 \cos \left(3 t\right) + 15.0737 \sin \left(3 t\right)\right) {e}^{- \frac{t}{5}} - 1.10619$