Resolving Forces Homework: 4\frac{d^{2}x}{dt^2} + 12\frac{dx}{dt} + 9x = 0

[FeDeX_LaTeX]

Homework Statement

A particle P of mass 4 kg moves along a horizontal straight line under the action of a force directed towards a fixed point O on the line. At time t seconds, P is x metres from O and the force towards O has magnitude 9x Newtons. The particle P is also subject to air resistance, which has magnitude 12v Newtons when P is moving with speed v m/s.

Show that the equation of motion of P is:

4\frac{d^{2}x}{dt^2} + 12\frac{dx}{dt} + 9x = 0

Homework Equations

F = ma

The Attempt at a Solution

The only way you get this is if you take 12v and 9x to be acting in the same direction. But how can they be? Surely they must be opposite? If I just resolve it assuming they act in opposite directions I get a sign error. Can anyone help me?

 

[DryRun]

Your equation is a homogeneous 2nd order constant coefficient linear ODE. Just solve the auxiliary equation. Since there is only one distinct root: t = -3/2
I think the general solution would be: x=Ae^{-3t/2}

 

[FeDeX_LaTeX]

Thanks for the reply, but I know how to solve the ODE... the issue is setting it up in the form they want.

 

[tiny-tim]

Hi FeDeX_LaTeX!

The only way you get this is if you take 12v and 9x to be acting in the same direction. But how can they be? Surely they must be opposite?

If the motion is towards O, yes.

If the motion is away from O, no.

(and sharks, a 2nd order equation has two independent solutions … in this case, (A+Bt)e^-3t/2)

 

[FeDeX_LaTeX]

Hi, thanks for the reply. So, the motion is away from O then? How come?

 

[tiny-tim]

hmm … why not? …

A particle P of mass 4 kg moves along a horizontal straight line under the action of a force directed towards a fixed point O on the line. At time t seconds, P is x metres from O and the force towards O has magnitude 9x Newtons. The particle P is also subject to air resistance, which has magnitude 12v Newtons when P is moving with speed v m/s.

 

[FeDeX_LaTeX]

So we're just supposed to guess?

 

[tiny-tim]

well, mister glass-half-empty, they did give you the answer!

you might even pick up an extra mark … or at least a smiley-face … for sarcastically pointing out the missing condition!

 

[FeDeX_LaTeX]

I'm confused... so if they didn't give you the equation of motion, then it would be impossible to tell what the directions of the forces were?

 

[tiny-tim]

you'd have to put |dx/dt| into the equation

 

[FeDeX_LaTeX]

Ah okay thanks, that makes sense. This also made me think about F = ma... for a scenario where a single force causes an acceleration in the same direction, F and a are both positive, so F + ma = 0 leading to F = -ma? Or is that wrong?

 

[tiny-tim]

i can't make any sense of that

a single force always causes acceleration in the same direction!

(it's only the velocity that can be in any direction )

get some sleep! :zzz:​