Simple Damped Harmonic Oscillator with friction

[khfrekek92]

Homework Statement

a block of mass m=.5kg is sliding on a horizontal table with coefficients of static and kinetic friction of .8 and .5 respectively. It is attached to a wall with a spring of unstretched length l=.13m and force constant 200 n/m. The block is released from rest at t=0 when it is x=.23m from the wall. Where will it end up at long times?

Homework Equations

f=ma

The Attempt at a Solution

I've drawn my diagram and set z=x-l, so z'=x' and z''=x''.
next, using f=ma, I got that:
mx''=-k(x-l)-[mu]k*mg
z''=-kz/m-[mu]kg
z''=-(sqrt(k/m))^2*z-[mu]k*g

Then I found the homogeneous and particular solutions to get:
z=Acos(sqrt(k/m)t+[phi])-.5[mu]k*gz^2
x=Acos(sqrt(k/m)t+[phi])-.5[mu]k*g(x-l)^2 +l

Now I am lost.. Where do I go from here? Is this solution even right so far?
For hints our teacher gave us that "at long times" means to just pay attention to the particular solution.. And that using the static friction, we need to check the block at EACH turn-around to see if it has enough energy to turn around using:
k(x-l)>=[mu]s*mg, so
x>=[mu]s*mg/k + l

How do I check this? I don't know what phi or A is so how can I check x at each turn around?

All I know is that at t=0, x=.23 and x'=0

Thanks so much in advance! I really really appreciate it!

 

[AlexChandler]

hmm... at long times it should end up at x=.13 right?

 

[khfrekek92]

Why .13m?

 

[AlexChandler]

Energy is being constantly dissipated through heat... so it should end up at the equilibrium position of the spring right? Which is at x=.13m

 

[khfrekek92]

Ohhh yeah at long times it will end up there, But they want the actual equation of where it will end up at any point in time :/

 

[AlexChandler]

Well the equation will be complicated. You will need a term like sgn(dx/dt) to have the friction always pointed in the right way.

m \ddot x = - k x -\mu_k mg * sgn( \dot x )

This is the only way I can see to set up the equation correctly.

I don't know how you could solve an equation like this... Maybe there is a way that I am unaware of, but maybe there isn't.

 

[AlexChandler]

or you could also write it as

m \ddot x = - k x -\mu_k mg * \frac { \dot x}{\mid \dot x \mid}

but I don't think this will be much better

Or you could even say

K = E - U = \frac{1}{2}kA^2 - \frac{1}{2}kx^2 = \frac{1}{2}mv^2

v = \sqrt{ \frac{k}{m}(A^2 - x^2) } = \mid \dot x \mid

where A is the maximum extension. So

m \ddot x = - kx - \mu_k mg \frac{ \dot x }{ \sqrt{\frac{k}{m}(A^2 - x^2) }}

that is probably no easier to solve. But at least it doesn't have any weird functions.

 

[AlexChandler]

And maybe what you can do is to take an average value of

\sqrt{\frac{k}{m}(A^2 - x^2) }

over half a period, and replace the square root term with just its average value. This would at least be a fair approximation. If you do this, you will just have a normal damped harmonic oscillator equation that is easy to solve.

 

[khfrekek92]

Wow I never even would have thought of any of those! I'm going to go try these now and see if it works :) Thanks so much for your help!

 

[AlexChandler]

Actually I am sorry I've just realized the equation for velocity we got by considering energy conservation is incorrect. Do you see why?

 

[khfrekek92]

No I don't see why :/ I tried it also and got the same thing, do we need to add friction in too? It is also dissipating energy isn't it?

 

[AlexChandler]

No I don't see why :/ I tried it also and got the same thing, do we need to add friction in too? It is also dissipating energy isn't it?

Yes this is exactly why the energy equation was wrong. It should be something more like

E = K.E. + P.E. - \mu_k mg * (\Delta S)

Where Delta S is the total distance traveled so far. However, I think this will just make things very complicated, so the energy consideration will probably not help.

Perhaps one of the earlier equations will work. Or perhaps there is a simpler way of handling this that I am unaware of.