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Harmonic oscillator with friction

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Spring with spring constant k=2000N/m has an object with mass 10kg attached to it. When it is pulled 0.1m away from the equilibrium state it starts oscillating and came to a stop. The coefficient of kinetic friction is 0.2 and the coefficient of static friction is 0.5. Find the time the object took to come to a stop and where it came to a stop.

    2. Relevant equations
    ##F=-kx##
    ##F=μmg##



    3. The attempt at a solution
    ##\frac{1}{2}kx_n^2=\frac{1}{2}kx_{n+1}^2+f_kx_n+f_kx_{n+1}##
    This is the formula that I took by denoting the nth amplitude (because the amplitude keeps on decreasing) and ##mw^2x=kx-20## solving this gets me the period T but my friend says period T is ##2π\sqrt{m/k}## regardless as to whether the friction exists or not. Which is correct?
     
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  3. Mar 27, 2016 #2

    Simon Bridge

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    It does not have a proper period since it does not return to it's initial position (see definition of "period").
    You could say the time between the peaks is the period ... if you check the math for damped HO, is the frequency of the damped oscillations the same as the natural frequency? What about the case for critically damped?
     
  4. Mar 27, 2016 #3

    Nathanael

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    I like your recursive formula for the semi cycle amplitudes, but Im not sure where the next equation came from; you mean ω to be frequency? Explain that part please.

    Remember the goal: under what condition will the object stop?
     
  5. Mar 27, 2016 #4
    Damped and critically damped situations have additional accelerations that are proportional to ##v##. So they do not have the same frequency as the natural frequency. But here friction is constant. How about the time it takes to move between the two outermost points?

    The second formula is where I put the centripetal force and the other two forces(spring and friction) as equals since Harmonic oscillator can be thought of as a shadow of circular motion
     
  6. Mar 27, 2016 #5

    ehild

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    The frequency changes with the damping when it depends on the speed. Here the force of friction has constant magnitude, only the direction changes. The position of the body is described between two extremes as x=Acos(wt+a)+B, where w is the same as that of the un-damped oscillator. The amplitude and phase change at each turn and also the direction of the force of friction.
     
  7. Mar 27, 2016 #6

    Nathanael

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    Harmonic oscillatoin can be thought of as the shadow of circular motion, ok, but that does not mean you can use the centripetal acceleration... If you wanted to extend the analogy you would also need to take the shadow of the centripetal acceleration vector.
    We don't even have simple harmonic motion anyway, so there would be radial motion in such an analogy. Such a perspective becomes more complicated than the differential equations which describe the motion.

    Speaking of which, what are those equations? You haven't written them yet, or even said when the block will stop. If you show your mathematics, it's easier to help.
     
  8. Mar 27, 2016 #7
    When the block is at the maximum displacement it needs acceleration higher than the maximum of static friction force. The objects stops when the force from Hooke's law is smaller thatn this maximum of static friction force. I found out that after 5 movements (here a movement is when the object moves from one maximum displacement to the equilibrium state) the force is smaller than MSFF.
     
  9. Mar 27, 2016 #8

    ehild

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    How did you get it?
    The object stops at an extreme position, when its velocity is zero. Those 5 "movements" mean that the object stops at the equilibrium position which can not be true, as it has maximum speed there.
     
  10. Mar 28, 2016 #9
    No the five movements, since the objects starts from an extreme position, gets the object to stop at an extreme position.
     
  11. Mar 28, 2016 #10

    ehild

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    Do you mean the five movements as shown in the figure?
    The object starts from position x1, and reaches x2. It turns there, and reaches x3, and so on. |xi+1|<|xi| , by how much? At what position does the object stop? How long does it move?

    upload_2016-3-28_15-46-12.png
     
  12. Mar 28, 2016 #11
    The object stops at point ##x_5## and moves 2T seconds. (##T=2π\frac{\sqrt{m}}{\sqrt{k}}##)
     
  13. Mar 29, 2016 #12

    ehild

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    That is correct.
     
  14. Mar 29, 2016 #13
    Why doesn't the period change from the formula ##T=2π\frac{\sqrt{m}}{\sqrt{k}}##
    shouldn't it change since friction is now on the line of forces acting on the object?
     
  15. Mar 29, 2016 #14

    ehild

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    A constant force does not change the angular frequency. Think of a suspended spring with a weight attached to it. The force of friction is the same during the motion, μmg, only its direction changes, always opposite to the velocity. The object moves from x1 to x2 , x2 to x3, .... according to the formula x=Acos(wt) +B, with w= √(k/m). But the motion is not periodic, as A and B changes from one stage to the other. At each extreme, wt is integer number of pi, therefore the time between two consecutive extremes is always the same.
     
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