How can I solve this differential equation for quadratic resistance?

[fayled]

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 v_l=-√(mg/b) (negative as it obviously has its terminal velocity downwards).

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

Now make the substitution z=v/v_l so dv/dt=v_ldz/dt. Then we obtain

v_ldz/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/v_l∫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/v_l.

Next let k=v_l/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=v_l[(e^-t/k-1)/(e^-t/k+1)]

Now, the correct answer should be

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

i.e

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

which must be right because as t→∞, v→v_l which by the definition of v_l 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 :)

 

[pasmith]

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 v_l=-√(mg/b) (negative as it obviously has its terminal velocity downwards).

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

Now make the substitution z=v/v_l so dv/dt=v_ldz/dt.

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

Then we obtain

v_ldz/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/v_l∫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/v_l.

Next let k=v_l/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=v_l[(e^-t/k-1)/(e^-t/k+1)]

I believe this is correct.

Now, the correct answer should be

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

ie.

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

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

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

On the other hand, if you define $v_l$ to be positive then $-1/k$ is negative, so that your answer tends to $-v_l < 0$ as required.

 

[fayled]

Not quite. If you're defining $v_l = (mg/b)^{1/2} > 0$ then you should find that $v \to -v_l$. You're measuring displacement upwards but the object is moving downwards so $v < 0$. 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 v_l=-(mg/b)^1/2...

 

[pasmith]

But I defined it as v_l=-(mg/b)^1/2...

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

I have now edited my post accordingly.

 

[fayled]

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

I have now edited my post accordingly.

Ah right, thankyou very much.