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Homework Help: Write different equation from physical system

  1. Dec 9, 2009 #1
    Hi to all,

    I've to write differential equation for desciribe system in attached file.

    The system is intended: from force f as input to position x2 as output.

    Initial condition are the following x1=0 and x2=0.

    In the attached file I've included all formula for descrive relationship between force and position for components: Spring, suspension.

    I'd like some suggestion about ho to resolve this kind of problem.

    I've found result for this exercise, I've try to computate the valur, but I made samo mistake, cold you suggest if my starting equation are correct, if they're not could you explain my where I made error?

    Starting Equation:
    f + k(x1-x2) + BD(x1-x2)=D^2mx1
    -k(x1-x2)-BD(x1-x2)=D^2mx2

    Expand First and Second Equation
    f +kx1 -kx2 +BDx1 -BDx2 = D^2mx1
    -kx1 +kx2 -BDx1 +BDx2 = D^2mx2

    Aggregate for common factor x1
    ( k + BD - D^2m )*x1 = -f + kx2 +BDx2
    ( -k -BD )*x1 = -kx2 -BDx2 +D^2mx2

    Computate x1
    x1 = ( -kx2 -BDx2 +D^2mx2 ) / ( -k -BD )

    Replace x1 in first equation:
    ( k + BD - D^2m )*( -kx2 -BDx2 +D^2mx2 ) / ( -k -BD ) = -f + kx2 +BDx2

    ( k + BD - D^2m )*( -kx2 -BDx2 +D^2mx2 ) = (-f + kx2 +BDx2) * ( -k -BD )

    -k^2x2 -kBDx2 + kD^2mx2 -BDkx2 -B^2D^2x2 +BD^3mx2 +kD^2mx2 + BD^3x2m D^4m^2x2
    = +kf -fBD -k^2x2 -kBDx2 -kBDx2 -B^2D^2x2

    +kD^2mx2 +BD^3mx2 +kD^2mx2 + BD^3x2m -D^4m^2x2
    +kf -fBD

    My result:
    -D^4m^2x2 + BD^3x2m + BD^3mx2 +kD^2mx2 +kD^2mx2 = +kf -fBD

    Correct Result:
    +D^4m^2x2 + BD^3x2m + BD^3mx2 +kD^2mx2 +kD^2mx2 = +kf +fBD

    Thank you very much!

    Bye
    Maurizio
     

    Attached Files:

    Last edited: Dec 9, 2009
  2. jcsd
  3. Jan 3, 2010 #2
    Your mistake is in the multiplication (-f + kx2 +BDx2) * ( -k -BD ). On the right hand side rather than -fBD you should get +fBD
     
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