Harmonic Oscillator with Friction

  • #1
Fascheue

Homework Statement



I don’t have a specific problem to solve, and I’m not sure I would be able to correctly find one, but I need to know how to solve a harmonic Oscilator problem with Friction. I believe I should be starting with F = -kx -Ff, and that I will be given some information about the frictional force. I’m guessing the frictional force will just be constant. If that’s the case I believe the force could just be written as F = -kx - b, where b is some constant? Can anybody confirm that this is correct and help me solve the differential equation

Homework Equations



F = -kx - b

The Attempt at a Solution


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mx’’= -kx - b

This is where I have to solve the differential equation that I’m unsure how to solve. I’m not quite sure where to start, but I believe it might have something to do with guessing a solution.
 
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  • #2
Why do you think the force of friction is a constant? When it's sitting at the equilibrium position, what should the net force be? What does your equation say it is?
 
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  • #3
The friction force can't be a constant, it's some function of velocity ##x'##, and in the simplest case it's a linear function of velocity as in

##mx''(t) = -kx(t) - bx'(t)##
 
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  • #4
RedDelicious said:
Why do you think the force of friction is a constant? When it's sitting at the equilibrium position, what should the net force be? What does your equation say it is?
I’m imagining a mass attached to a spring on a surface with friction. I thought friction would be constant because Ff = uFn where u is constant and Fn wouldn’t change. I guess that only gives the value for maximum possible friction though, I see that it’s wrong.

If friction is -bx’(t) (in a simple case) as hilbert2 said, then I’m a bit confused. In the case of the mass on a spring, I don’t believe friction can exceed uFn. If that is the case, shouldn’t the force of friction no longer be dependent on x’(t) at some point when the mass is moving quickly enough?
 
  • #5
Fascheue said:
If friction is -bx’(t) (in a simple case) as hilbert2 said, then I’m a bit confused. In the case of the mass on a spring, I don’t believe friction can exceed uFn. If that is the case, shouldn’t the force of friction no longer be dependent on x’(t) at some point when the mass is moving quickly enough?

What does uFn mean? Linear models usually work only under some limiting conditions, the air resistance on a cannonball or a jet plane is likely to be a more complicated function of velocity than a linear one.
 
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  • #6
hilbert2 said:
What does uFn mean? Linear models usually work only under some limiting conditions, the air resistance on a cannonball or a jet plane is likely to be a more complicated function of velocity than a linear one.
The coefficient of friction multiplied by the normal force. Wouldn’t that equation describe the maximum frictional force possible on a mass attached to a spring on a frictional surface?
 
  • #7
Fascheue said:
The coefficient of friction multiplied by the normal force. Wouldn’t that equation describe the maximum frictional force possible on a mass attached to a spring on a frictional surface?

Yes, that's usually true. I was interpreting friction as air resistance here, hence the confusion.

If you assume that the spring is exerting enough force on the mass to make it move in the first place, and that the motion is not too fast, you can probably find a constant ##b## that is relatively well consistent with the friction force being of the form ##bx'(t)##.
 
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  • #8
hilbert2 said:
Yes, that's usually true. I was interpreting friction as air resistance here, hence the confusion.

If you assume that the spring is exerting enough force on the mass to make it move in the first place, and that the motion is not too fast, you can probably find a constant ##b## that is relatively well consistent with the friction force being of the form ##bx'(t)##.
Alright that makes sense. If F = -kx -bx’(t) is the equation for the simplest harmonic oscillator with friction, that’s probably what I’ll need to be able to work with. What exactly is the process for find x(t) from that equation?

I think I should probably use Newton’s second law.

x’’(t) = (-kx -bx’(t))/m

My knowledge of differential equations is very limited, but it looks like a second order homogenous differential equation, and I think that means that I’m supposed to guess the answer somehow.
 
  • #9
You can try to read this:

https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/modes-and-the-characteristic-equation/MIT18_03SCF11_s12_2btext.pdf

Basically, the idea is to substitute functions of the type ##x(t)=Ae^{Bt}## or ##x(t)=Ate^{Bt}##, where ##B## can be a complex number, in the differential equation, and find the general form of the solution (in the end, complex-valued solutions are ignored by taking the real part of ##x(t)##).
 
  • #10
As long as the mass is moving in one direction, a Coulomb friction force is constant. It is when it reverse direction that the difficulties begin, but this particular problem was solved and published many years ago by J.P. den Hartog.
 
  • #11
hilbert2 said:
You can try to read this:

https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/modes-and-the-characteristic-equation/MIT18_03SCF11_s12_2btext.pdf

Basically, the idea is to substitute functions of the type ##x(t)=Ae^{Bt}## or ##x(t)=Ate^{Bt}##, where ##B## can be a complex number, in the differential equation, and find the general form of the solution (in the end, complex-valued solutions are ignored by taking the real part of ##x(t)##).

F = -kx - bx’mx’’ = -kx - bx’x’’ +kx/m + bx’/m = 0r^2 + b/mr + k/m = 0r = (-b/m +/- sqrt(b^2/m^2 -4k/m)/2x(t) = Ae^((-b/m + sqrt(b^2/m^2 -4k/m)t/2) + Be^((-b/m - sqrt(b^2/m^2 -4k/m)t/2)Does it seem like I did this correctly? Also, I just found some information about oscillators with friction in my lecture notes. I have the same equation that I used above except instead of b I have mΓ. Would mΓ just be a specific case of b?
 
  • #12
Who knows what m*capGamma means? This is not a common notation, so it is hard to interpret.
 
  • #13
Dr.D said:
Who knows what m*capGamma means? This is not a common notation, so it is hard to interpret.
I believe it represents the mass multiplied by some coefficient of friction. I guess I’m just a bit curious why mass is involved in that equation but not the one mentioned in this thread.
 

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