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Differential equation problem

  1. Jan 15, 2014 #1
    I'm solving a differential equation to do with quadratic resistance and it seems to be acting very strangely - I get the opposite sign of answer than I should. If anybody could have a quick look through that would be much appreciated.

    For a particle moving downward and taking positive upwards, I have
    mdv/dt=-mg+bv2

    The terminal velocity comes out at vl=-√(mg/b) (negative as it obviously has its terminal velocity downwards).

    Substituting this in gives
    dv/dt=-g(1-(v/vl)2)

    Now make the substitution z=v/vl so dv/dt=vldz/dt. Then we obtain

    vldz/dt=-g(1-z2)

    Noting that 1/1-z2=0.5[(1/1+z)+(1/1-z)] and separating variables we get

    ∫[(1/1+z)+(1/1-z)]dz=-2g/vl∫dt

    Integrating each side then gives (the initial conditions are v=0 at t=0 so the constant of integration is zero)

    ln(1+z/1-z)=-2gt/vl.

    Next let k=vl/2g so that

    ln(1+z/1-z)=-t/k

    e-t/k=1+z/1-z

    e-t/k-ze-t/k=1+z

    z(1+e-t/k)=e-t/k-1

    z=(e-t/k-1)/(e-t/k+1)

    Therefore

    v=vl[(e-t/k-1)/(e-t/k+1)]

    Now, the correct answer should be

    v=vl[(1-e-t/k)/(1+e-t/k)]

    i.e

    v=-vl[(e-t/k-1)/(e-t/k+1)]

    which must be right because as t→∞, v→vl which by the definition of vl is expected.

    I've spent a lot of time trying to work out where I'm going wrong and it's driving me crazy. Thanks in advance :)
     
  2. jcsd
  3. Jan 15, 2014 #2

    pasmith

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    Homework Helper

    Probably better to define [itex]v_0 = (mg/b)^{1/2} > 0[/itex] so that terminal velocity is [itex]-v_0[/itex] and [itex]v[/itex] and [itex]z[/itex] increase in the same direction.

    I believe this is correct.

    I think you have confused yourself. You've defined [itex]k = v_l/2g[/itex] and you initially defined [itex]v_l[/itex] to be negative, so [itex]k[/itex] is negative. It follows that [itex]-1/k[/itex] is positive, so that [itex]e^{-t/k} \to \infty[/itex] as [itex]t \to \infty[/itex]. Thus your answer
    [tex]
    v=v_l \frac{e^{-t/k}-1}{e^{-t/k}+1} \to v_l < 0
    [/tex]
    as required.

    On the other hand, if you define [itex]v_l[/itex] to be positive then [itex]-1/k[/itex] is negative, so that your answer tends to [itex]-v_l < 0[/itex] as required.
     
    Last edited: Jan 15, 2014
  4. Jan 15, 2014 #3
    But I defined it as vl=-(mg/b)1/2...
     
  5. Jan 15, 2014 #4

    pasmith

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    Homework Helper

    I only noticed that after replying, since it's extremely unnatural to do that. More normal would be to define [itex]v_l = (mg/b)^{1/2}[/itex].

    I have now edited my post accordingly.
     
  6. Jan 15, 2014 #5
    Ah right, thankyou very much.
     
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