How Do Two Masses and Three Springs Interact with a Dashpot?

[a.man]

Homework Statement

Two particles, each of mass M, are hung between three identical springs. Each spring is massless and has spring constant k. Neglect gravity. The masses are connected as shown to a dashpot of negligible mass.

The dashpot exerts a force of bv, where v is the relative velocity of its two ends. The force opposes the motion. Let x₁ and x₂ be the displacement of the two masses from equilibrium.


a. Find the equation of motion for each mass.

b. Show that the equation of motion can be solved in terms of the new dependent variables y₁ = x₁ + x₂ and y₁ = x₁ - x₂.

c. Show that if the masses are initially at rest and mass 1 is given initial velocity v₀, the motion of the masses after a sufficiently long time is
x₁=x₂
= (v₀/2ω)* sin(ωt)

Evaluate ω.

Homework Equations

x^.. + γx^.+ ω²x = 0

x = Ae^-γt/2cos(ωt + ∅)

The Attempt at a Solution

I think I managed to get the first two by just using different F = -kx formulae and adding -bv to it.

The second part I just added and subtracted the equations to get:
y^..₁ + ω²y₁ + $\frac{2bv}{m}$ = 0

and

y^..₂ + 3ω²y₂ = 0

I think these are right, as far as I know. The thing is, I'm not sure how to proceed from here. I tried using:

x = Ae^-γt/2cos(ωt + ∅)

I can get it very close to the answer, but I don't think that's how you're supposed to do this... using what I got in part 2 would be more relevant, I think.

 

[voko]

Seeing the diagram or at least its description would be helpful.

 

[a.man]

Sorry, this is the image:

 

[voko]

Could you show your equations for a) and the derivation for b)?

 

[a.man]

ma₁=-kx₁-k(x₁-x₂) - bv

ma₂=-kx₂-k(x₂-x₁) - bv

These were my first two equations of motion.

I just added and subtracted them, using k = ω²m to get the equations for 2.

Quick question: can I take relative velocity v as equal to the derivative of y₂ with respect to time?

 

[voko]

But what is v in your equations? And how come it is the same in both equations?

 

[a.man]

"... where v is the relative velocity of its two ends. The force opposes the motion."

So yeah, it should be the same for both cases – both magnitude and direction.

 

[voko]

"Relative velocity" is relative to the velocity of the opposite end. It is the same (= zero) only of the ends have the same velocity. Express it in terms of x1 and x2 for both ends.

 

[a.man]

"Relative velocity" is relative to the velocity of the opposite end. It is the same (= zero) only of the ends have the same velocity. Express it in terms of x1 and x2 for both ends.

Yeah, it's the relative velocity of one end of the dashpot to the other (thus the relative velocity of one mass to another).

So for the first mass it'll be derivate of x₂ minus that of x₁ with respect to time... and for the second mass it'll be negative of that? But the force opposes the motion. Therefore, for the first mass that is displaced by x₁, the force will be negative (to the left). Similarly, if the second mass is displaced by x₂, the force will be negative as well. Are my signs wrong?

If relative velocity is in terms of x₂ - x₁, then I could express it in terms of y₂, right?

On another note, in the third part, if x₁ = x₂ then relative velocity should be 0.

 

[voko]

Yeah, it's the relative velocity of one end of the dashpot to the other (thus the relative velocity of one mass to another).

So for the first mass it'll be derivate of x₂ minus that of x₁ with respect to time... and for the second mass it'll be negative of that? But the force opposes the motion.

I am with you till this point.

Therefore, for the first mass that is displaced by x₁, the force will be negative (to the left). Similarly, if the second mass is displaced by x₂, the force will be negative as well. Are my signs wrong?

This is less clear. The net force acting on either mass is the sum of three forces, two due to springs and one due to damping. I think should write down the equations in terms of x₁ and x₂ just for clarity.

If relative velocity is in terms of x₂ - x₁, then I could express it in terms of y₂, right?

Yes, but, again, you should write down the equations in terms of x₁ and x₂ first.