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Damped Oscillators

  1. Jun 14, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-6-14_22-0-8.png

    2. Relevant equations


    3. The attempt at a solution

    After the release the block will move towards right and friction will be towards the left.

    ##M\ddot x = f - kx##

    Solving for ##x##,

    ##x = A\cos (\omega t) + B\sin(\omega t) + f/k##

    Initial conditions are ##x(0) = x_0, \dot x(0) = 0##

    ##\therefore x(t) = \left(x_0 - \dfrac fk\right)\cos (\omega t) + f/k##

    But now how shall I show that amplitude decreases with each oscillation at a constant rate ?
     
    Last edited: Jun 14, 2017
  2. jcsd
  3. Jun 14, 2017 #2

    scottdave

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    One error that I see is you have Sum of forces = f - kx, which means that you are assuming f is always pointing in the same direction. This is not the case. The friction force (f) will be in the opposite direction of motion.
     
  4. Jun 14, 2017 #3

    scottdave

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    Here is a way to approach it. You can derive (or get formulas from the textbook) that energy of a spring is (1/2)*k*x2, where x is the displacement from equilibrium. Could you make an equation that takes into account the work done by friction (f*d)? Sorry I forgot the squared after x, which has been corrected now.
     
    Last edited: Jun 14, 2017
  5. Jun 14, 2017 #4
    I took the left side of the equilibrium point as +ve axis and other side as negative. Since the spring force is pulling towards right, it is -ve and since friction is towards left it is +ve, but I guess the signs will change with each oscillation.

    Ok I will try.
     
  6. Jun 14, 2017 #5
    Here is what I did,

    ##\dfrac 12 kx_0^2 = -\int_0^{x_0 + A} f\cdot dr + \dfrac 12k A^2 \implies kx_0^2 = -2 f(x_0 + A) + k A^2##

    Solving for ##A## I got, ##A = - x_0## or ##A = \dfrac {2f}k + x_0##,

    Second solution say amplitude increased. Where I am wrong ?
     
  7. Jun 14, 2017 #6

    haruspex

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    The work done against friction is ##+
    \int_0^{x_0 + A} f\cdot dr##
     
  8. Jun 14, 2017 #7
    I felt that should be the case but think of it like this, if friction was not there then the enegry would be ##1/2kA^2##, friction is taking some energy out of it and converting it into heat. So I thought it should be -ve.
     
  9. Jun 14, 2017 #8

    haruspex

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    As far as I can see, you have not defined A. x0 is the initial amplitude. From your integral, I deduced A represents the amplitude after one half cycle, i.e. from max displacement one way to max the other way.
    The initial energy is ##\frac 12 kx_0^2##. After one half cycle, ##\int _{-x_0}^Af.dr## has gone to friction and the remainder is ##\frac 12kA^2##.
     
  10. Jun 15, 2017 #9
    Ok so after one full oscillation the amplitude will be ##A = x_0 - \dfrac{4f}k##.
     
  11. Jun 15, 2017 #10

    haruspex

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    Yes.
     
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