Why Does My Integration Approach Fail in the Voigt Model Dashpot Equation?

  • Thread starter Thread starter yosimba2000
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
yosimba2000
Messages
206
Reaction score
9
Problem:
So I already know how to correctly find the formula for a Voigt model, but why is my proposed way incorrect?

Equations and Assumptions:
for dashpot: F = n(dx/dt), where n is the dashpot constant and dxdt is change in distance/change in time
for spring: F = kx

assuming: arm connecting dashpot and spring doesn't bend or rotate, then F applied is equal to Force on Spring + Force on Dashpot. Also, because the arm doesn't bend, displacement of Dashpot and Spring are equal.

the correct equation: X = F(1-exp(-kt/n))

My Proposed Method:
Force on spring + Force on dashpot = total Force (A)
F + F = F
kx + n(dx/dt) = F

n(dx/dt) is Force on dashpot.
why can't I separate to get: (ndx) = (Force on dashpot)dt
then integrate to get, both from 0 to respective final values, to get (nx) = (Force on dashpot)(t)
Then solve for Force on dashpot to get nx/t
Then plug this in for Force on dashpot in equation (A)
kx +nx/t = total Force
Solve for X
X = (F)/[(k+n)/t]

Essentially, what I am asking is: why do I HAVE to leave the dx/dt in the dashpot equation as is? Why can I not just say F = n(dx/dt), then separate and integrate to get F = nx/t, and then just substitue this in equation (A)?
 
Physics news on Phys.org
yosimba2000 said:
Force on spring + Force on dashpot = total Force (A)
F + F = F
kx + n(dx/dt) = F

n(dx/dt) is Force on dashpot.
why can't I separate to get: (ndx) = (Force on dashpot)dt
Because it isn't. How do you justify ignoring kx there?

I'm no maths expert, but it does seem like it must be:

kx. dt + n. dx = F. dt

I'm assuming that the dashpot and spring are arranged so that their forces are summed, but not necessarily equal?

Anyway, I've given you something to mull over while we wait for an expert to stroll by. :)