Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spring equation.

  1. May 15, 2010 #1

    MathematicalPhysicist

    User Avatar
    Gold Member

    There is a chain of springs, such that qk is the displacement of the kth mass in this chain.

    We have the next equation of Newton second law, where T is the tension force:
    [tex]\frac{d^2q_k}{dt^2}=T(q_{k+1}-q_k)-T(q_k-q_{k-1})[/tex]
    Let Qk=qk+1-qk

    I showed that if Q(t)k=Q(k-ct)=Q(s)

    then we get the next ODE:
    (1)[tex] c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)[/tex]
    I need to show that the solution is:
    [tex]Q(s)=\int_{s-1}^{s+1} (1-|s-z|)T(z)dz[/tex]
    by direct integration i.e integrating (1) or by fourier transform.

    I tried both but I get stuck, by direct integration I get:

    [tex]c^2Q(s)= \int^{s} \int^{x} [T(z+1)-2T(z)+T(z-1)] dz dx[/tex]
    don't know how to proceed from here, can anyone give a helping hand? (-:
     
  2. jcsd
  3. May 22, 2010 #2

    MathematicalPhysicist

    User Avatar
    Gold Member

    I also tried the fourier transform I got to:
    [tex]Q(s)=FT(FT^-1(T(s))*\frac{sin^2(w/2)}{(w/2)^2})[/tex]

    I tried mathematica to solve this, but I am not sure how to write it with an implicit function as FT^-1(T(s)).
     
  4. Jun 15, 2010 #3
    I wonder about equation (1). Since:

    [tex]\frac{d^2Q_k}{\text{dt}^2}=T\left(Q_{k+1}\right)-2T\left(Q_k\right)+T\left(Q_{k-1}\right)[/tex]

    So, we have:

    [tex]c^2 Q''(s)=T(Q(s+1))-2T(Q(s))+T(Q(s-1))[/tex]

    instead of

    [tex]c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)[/tex]

    Please clarify.
     
  5. Jun 16, 2010 #4

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What are the equations if you have k = 2 springs? Using your equations you would have displacements q0, q1, q2, and q3 involving 4 displacements and only two equations.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook