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Homework Help: Dartmouth Extended Laplace Tables -- Not general enough? item26.a

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data

    http://www.dartmouth.edu/~sullivan/22files/New Laplace Transform Table.pdf
    (see item 26a)

    homogenous solution to underdamped in amplitude phase form: (see attached image)

    2. Relevant info

    - non zero initial conditions: x(t=0) = xo AND dx/dt(t=0) = vo
    - unforced motion: F(t) = 0
    - second order dynamic system (1DOF & in one dimension x)
    - ODE and final solution is written in terms of damping ratios (z), and natural frequencies (wn)
    - damping frequency: wd = wn*sqrt(1-z^2)
    -Underdamped system: |z| < 1

    3. The attempt at a solution

    d^2x/dt^2 + 2*z*wn*dx/dt + wn^2*x = 0

    -Now I take laplace

    [X(s)*s^2 - s*xo - vo] + 2*z*wn*[X(s)*s - xo] + wn^2*X(s) = 0


    X(s) (s^2+2*z*wn*s+wn^2) = s*xo + vo + 2*z*wn*xo

    -Solving for X(s)

    X(s) = [s*xo + vo + 2*z*wn*xo] / [s^2 + 2*z*wn*s + wn^2]

    More algebra so that I may use item 26a in the extended Laplace tables

    X(s) = xo*[ s + (vo+2*z*wn*xo)/xo ] / [s^2+2*z*wn*s+wn^2]

    I will define alpha (as listed in the Extended Laplace Table item 26a):

    alpha = (vo+2*z*wn*xo)/xo

    When I directly apply this formula from the Extended Laplace Tables I get an inconsistent answer (see attached)

    Just looking at the Magnitude (magnitude is off)

    sqrt{ [( alpha/wn - z*wn )^2] / (1-z^2) + 1 }

    Replacing alpha with (vo+2*z*wn*xo)/xo

    = sqrt{ [ [(vo+2*z*wn*xo)/xo)/wn - z*wn]^2/(1-z^2) + 1 }

    Making common denominator, and making 1-z^2 = (wd/wn)^2

    = sqrt{ [vo+2*z*wn*xo]/(xo*wn) - z*wn^2*xo/(xo*wn)]^2/ (wd/wn)^2 + 1}

    Bringing up wd/wn into the square & notice wn and 1/wn cancel

    Factor out 1/xo^2

    = sqrt{ [vo+2*z*wn*xo] - z*wn^2*xo]^2/ (wd*xo)^2 + }

    Making 1 have common denominator with other stuff

    = sqrt{ [vo+2*z*wn*xo] - z*wn^2*xo]^2/ (xo*wd)^2 + 1*(xo*wd)^2/(xo*wd)^2 }

    Factor out 1/(xo*wd)

    = 1/(xo*wd)* sqrt{ [vo+2*z*wn*xo] - z*wn^2*xo]^2 + (xo*wd)^2}

    This magnitude (just above) does not agree with the correct magnitude (next couple of lines) (can also see attached source):

    = 1/(xo*wd)* sqrt{ [vo+2*z*wn*xo - z*wn*xo]^2 + (xo*wd)^2}

    = 1/(xo*wd)* sqrt{ [vo+z*wn*xo ]^2 + (xo*wd)^2}

    I believe that item 26a may not be general enough. As the z*wn term within the square thats within the square root should just be z


    What's going on on here? Thank you.

    Attached Files:

  2. jcsd
  3. Mar 12, 2016 #2


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    Homework Helper
    Gold Member

    I'm actually a little suspicious that the entry 26a in the table might not even be correct at all.

    Unless I'm making a large mistake myself, here is my logic:

    I suspect that the constant [itex] \alpha [/itex] should have dimensionality of [itex] \frac{1}{[\mathrm{time}]} [/itex]. (Again, that is if I'm not mistaken). If so, then one of the terms in 26a doesn't make sense to me: in particular, the part of term that is [itex] \frac{\alpha}{\omega_n} - \zeta \omega_n [/itex].

    [itex] \omega_n [/itex] also has units of [itex] \frac{1}{[\mathrm{time}]} [/itex]. That makes [itex] \frac{\alpha}{\omega_n} [/itex] dimensionless (if I'm correct about [itex] \alpha [/itex]), and [itex] \zeta [/itex] is dimensionless. But that gives [itex] \zeta \omega_n [/itex] dimensionality of [itex] \frac{1}{[\mathrm{time}]} [/itex]. So that operation is subtracting a [itex] \frac{1}{[\mathrm{time}]} [/itex] value from a dimensionless number.

    With that I conclude (or at least suspect) that something doesn't look right with 26a to me.
  4. Mar 12, 2016 #3
    Yes I agree the dimensions do look off
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