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#?

Question

Christopher Agresta · Accepted Answer

#x = (21.1062 cos(3t)+ 15.0737sin(3t))e^{-t/5}-1.10619#

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#

Christopher Agresta · Answer

undefined To find the position function $ x(t) $ for the mass-spring-dashpot system, we first need to solve the second-order linear ordinary differential equation (ODE) governing the system's motion. The equation is given by:

$$ m\frac{{d^2x}}{{dt^2}} + c\frac{{dx}}{{dt}} + kx = 0 $$

Given the values $ m = 25 $, $ c = 10 $, and $ k = 226 $, and the initial conditions $ x(0) = 20 $ and $ x'(0) = 41 $, we substitute these values into the differential equation.

The solution to this second-order ODE is a linear combination of exponential functions. The general form of the solution is:

$$ x(t) = e^{rt} $$

where $ r $ is a constant.

To find $ r $, we substitute $ x(t) = e^{rt} $ into the differential equation:

$$ m\frac{{d^2}}{{dt^2}}e^{rt} + c\frac{{d}}{{dt}}e^{rt} + ke^{rt} = 0 $$

After differentiation, we get:

$$ m(r^2e^{rt}) + c(re^{rt}) + ke^{rt} = 0 $$

This simplifies to:

$$ mr^2 + cr + k = 0 $$

Now, substitute the given values of $ m $, $ c $, and $ k $ into the equation:

$$ 25r^2 + 10r + 226 = 0 $$

Solve this quadratic equation for $ r $. Once you have the values of $ r $, the solution for $ x(t) $ will be:

$$ x(t) = Ae^{r_1t} + Be^{r_2t} $$

where $ A $ and $ B $ are constants determined by the initial conditions, and $ r_1 $ and $ r_2 $ are the roots of the quadratic equation obtained earlier.

Finally, use the initial conditions $ x(0) = 20 $ and $ x'(0) = 41 $ to solve for $ A $ and $ B $, and plug them back into the solution $ x(t) $.

Charles Anctil · Answer

undefined To find the position function $ x(t) $ for the mass-spring-dashpot system, you need to solve the second-order linear differential equation that governs the system. The equation is:

$$ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 $$

Given $ m = 25 $, $ c = 10 $, and $ k = 226 $, and the initial conditions $ x(0) = 20 $ and $ x'(0) = 41 $, you can use these values to solve for $ x(t) $.