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Differentiating a function of x wrt time

  1. Dec 11, 2013 #1
    Hi,

    I figured out the only redundancy to my problem is this:

    I'll start off with a simple case, where w1,w2 are the displacements at intervals of one third along a beam.

    w = 3w1/L.x (Note, x is in the numerator for all cases)

    To differentiate this with respect to time, I use the chain rule as follows:

    ∂w/∂t = ∂w/∂x.∂x/∂t = 3w1/L.w1'

    Now for the second function, which depends on two displacement terms w1,w2

    w = (2 - 3/L.x)w1 + (-1 + 3/L.x)w2

    ∂w/∂t = ∂w/∂x.∂x/∂t = -3/L.w1.w1' + 3/L.w2.w2'

    I'd imagine this wrong, any hints?
     
  2. jcsd
  3. Dec 11, 2013 #2

    Mark44

    Staff: Mentor

    One third of what? If it's 1/3 of a unit, then there's no need to give them names.
    This is extremely difficult to read, with '.' apparently being used to indicate multiplication. The asterisk character, *, is typically used for multiplication.

    What does x represent? What does w represent?
     
  4. Dec 12, 2013 #3
    Thanks for your reply.

    w = (3w1/L)*x for 0<x<L/3
    w = [ 2 - (3/L)*x ]*w1 + [ -1 + (3/L)*x ]*w2 for L/3<x<2L/3
    w = 3*w2*[ 1 - x/L ] for 2L/3<x<L

    w is a displacement function for a simply supported beam satisfying w=0 at x=0 and x=L, w=w1 at x=L/3 and w=w2 at x=2L/3

    w is the displacement function of the beam (breaking up a beam of length L into L/3 sized intervals, such that it can displace vertically). x is the horizontal axis along the beam

    Now i'm trying to find ∂^2w/∂t^2, hope this is clear, to solve the integral ∫m*(∂^2w/∂t^2)*wdx, to get the acceleration term for the total potential energy functional (not important at this stage)
     
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