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Nonlinear Retarding force

  • Thread starter Mindscrape
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  • #1
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Homework Statement


A particle moves in a medium under the influence of a retarding force equal to [tex]mk(v^3 + a^2 v)[/tex], where k and a are constants. Show that for any value of the initial speed the particle will never move a distance greater than pi/2ka and that the particle comes to rest only for [tex]t \rightarrow \infty[/tex]


Homework Equations


Legrangian seems overkill, so I used Newton's.


The Attempt at a Solution


[tex]\frac{dv}{dt} = k (v^3 + a^2 v)[/tex]

then seperate the ODE and integrate

[tex] \frac{lnv}{a^2} - \frac{ln(a^2 + v^2)}{2a^2} = kt + C[/tex]

multiply by 2a^2

[tex] 2lnv - ln(a^2 + v^2) = 2kta^2 + C[/tex]

use log properties and combine the natural logs

[tex]ln ( \frac{v^2}{a^2 + v^2} ) = 2kta^2 + C[/tex]

exponentiate and carry constant down

[tex]\frac{v^2}{a^2 + v^2} = Ce^{2kta^2}[/tex]

add and subtract a^2 in numerator to simplify, and then subtract the one

[tex] \frac{-a^2}{v^2+a^2} = Ce^{2kta^2}[/tex]

use algebra to isolate v

[tex] v^2 = a^2 (1 - \frac{1}{1-Ce^{2kta^2}})[/tex]

Now I am at the point where I can solve for the integration constant, but I don't see it giving me anything close to revealing a max distance of pi/2ka. Also, in my solution as t approaches infinity v = a, and not zero. Maybe I made an algebraic mistake?
 
Last edited:

Answers and Replies

  • #2
dextercioby
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The error is clearly in the integration of the ODE.
 
  • #3
HallsofIvy
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Presumably, though you didn't show it, you used partial fractions to write
[tex]\frac{dx}{k(v^3+ a^2v)}= \frac{A}{v}+ \frac{Bv+C}{v^2+ a^2}[/tex]

How did you integrate [tex]\frac{Bv+C}{v^2+ a^2}[/tex]?
 
  • #4
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I used mathematica, though partial fractions is the obviously the way to do it by hand.
 
  • #5
AlephZero
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The error is clearly in the integration of the ODE.
The first error is integrating the wrong ODE.

[tex]\frac{dv}{dt} = - k (v^3 + a^2 v)[/tex] would be better, since the form of the answer implies k is positive.
 
  • #6
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The first error is integrating the wrong ODE.

[tex]\frac{dv}{dt} = - k (v^3 + a^2 v)[/tex] would be better, since the form of the answer implies k is positive.
Yep, that was actually the problem. I went through it again and figured out what I did wrong. You end up with v = +/- a/sqrt(-1-Ce^(2kta^2).
 

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