Hey,
I am trying to model the second order differential equations for the shock absorber. I would like to get the funtion of the movement of the mass. I am getting errors from the Solve Block in MathCAD, and I cannot understand where is it coming from. I suppose the reason is the sin function x(t), but I have no idea how can I write it differently. Without units the solver was working fine, but with units I get always an error. I am uploading the MathCAD file.
I would be glad for some hints and tipps! Any help is welcome! 
Best regards,
Jolanta
ValeryOchkov, thank you for your answer 
It is veeery helpful, I was stuck with it for many hours...
But can you elaborate it a little bit, for example why should I write it like this:
I do not get this t over s 🙂
Thanks!
You can't do this exactly in Prime. But with a little handwork, and using the laplace transform, you should be able to work this out...
(Note that the expression for ys() stretches much further to the right...)
Note that we get the exact solution here (not a numerically approximated one).
I wonder if, when you take smaller time steps in the odesolve, the bump at about 2 seconds shows there as well.
And with those values, for a, b...v0 (all SI units) the function ys() is:
Success!
Luc
Correction.
I messed up a few variables and got an incorrect result (that didn't match the Odesolve graph.).
Here's a corrected symbolic solution:
(now the bump at about 2 seconds is gone)
The full solution (for the given values of constants) is:
Luc
I was wondering why you get a sinh and cosh ( as multiplicands of e^...), where I get sin and cos (non-hyperbolic).
The reason is that you define
is you calculate this you get:
The sinh of a pure imaginary argument is a sin (multiplied with i) and the cosh of a pure imaginary argument is a cos.
That explains the difference between the two symbolic solutions.
Luc
They are equivalent representations of the same function. How to say: instead of writing x I can also write exp (ln (x)) does not change anything, the result is always the same. Further, but not the last, simplification:
" instead of writing x I can also write exp (ln (x)) does not change anything, the result is always the same"...
Careful: that may be true for all x except a few...or at least one.
Only in the limit does x=exp(ln(x) for x->0.
Luc
