Differential equation problem

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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 :)
 

Answers and Replies

  • #2
pasmith
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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.

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.

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)]

I believe this is correct.

Now, the correct answer should be

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

ie.

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 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:
  • #3
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Not quite. If you're defining [itex]v_l = (mg/b)^{1/2} > 0[/itex] then you should find that [itex]v \to -v_l[/itex]. You're measuring displacement upwards but the object is moving downwards so [itex]v < 0[/itex]. That's consistent with the answer you got yourself, and there is as far as I can see no error in your calculation.

But I defined it as vl=-(mg/b)1/2...
 
  • #4
pasmith
Homework Helper
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But I defined it as vl=-(mg/b)1/2...

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.
 
  • #5
177
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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.

Ah right, thankyou very much.
 

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