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Force and the sin wave

  1. Apr 11, 2006 #1

    daniel_i_l

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    This is just something I was thinking about the other day:
    Lets say that we have an object with the mass of 1kg and at t=0: x=0m and v=0m/s. Now we apply a force over time t with the force: F=sin(t)N.
    that means that a=sin(t). To get v we integrate and get: v = 1-cos(t).
    Then we integrate for x and get x = t - sin(t).
    That means that after a few seconds the mass will be way on the x+ side. But why should this be if on average the amount of force in that direction is 0?!
     
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  3. Apr 11, 2006 #2

    Kurdt

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    You're forgetting constants of integration when you integrate each time. Think about the answer you should get and what the constants could be.
     
  4. Apr 11, 2006 #3

    daniel_i_l

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    I got the answers I got when I included the constant:

    v(t) = -cos(t) + C (t=0, v=0)
    0 = -1 + C
    C = 1 => v(t) = 1 - cos(t)

    Then for x:
    x(t) = t - sin(t) + C (t=0, x=0)
    0 = 0 - 0 +C => C=0

    And I get:
    x(t) = t - sin(t) ?!
     
  5. Apr 11, 2006 #4

    Kurdt

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    ahh so you have. Sorry I'm just used to that as being such a common mistake that i didn't think to check (much embarrassment).

    The error occurs in assuming an acceleration and initial conditions and working back over. If you assume a=sin(t) and then derive v=1-cos(t) you'll see that at t =0 acceleration is 0 while the velocit is max which cannot be a true statemnet since you've said the velocity is 0 to start off with. Thus a=cos(t) to give max acceleration and zero initial velocity.
     
  6. Apr 11, 2006 #5

    daniel_i_l

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    Why is the velocity max?
    v(t) = 1 - cos(t) => v(0) = 1 - cos(0) = 1-1 = 0?
     
  7. Apr 11, 2006 #6

    Hootenanny

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    I'm sorry, I may be missing something but when we integrate sin(t) do we not simply get -cos(t), not 1-cos(t)?
     
  8. Apr 11, 2006 #7

    daniel_i_l

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    The 1 comes from the constant C.
     
  9. Apr 11, 2006 #8

    Hootenanny

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    Ahh yes, didnt read the initial conditions.
     
  10. Apr 11, 2006 #9

    Kurdt

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    Argh I'm having a nightmare today I apologise to everyone. If you start from rest and begin to accelerat that acceleration will go through half a cycle, at which time you will be at your maximum speed then the rest of the acceleration is used to slow whatever it is to zero again all the while it has been travelling a certain distance x. When the acceleration repeats the cycle it travels further in the positive x direction.

    I think you were thingin the velocity would oscillate between positive and negative values like the acceleration and thus the velocity and displacement should average zero.

    Third time lucky i hope. Apologies again it has been a long day!
     
  11. Apr 11, 2006 #10

    daniel_i_l

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    Thanks, that sounds right but why isn't the work 0J? I mean, if we look at the sum of the force over t after n cycles the it is 0, and if the work is 0 then how can it effect the kinetic energy and cause the mass to speed up?
     
  12. Apr 11, 2006 #11

    Kurdt

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    For a general force work is the integral of the force in the direction it is applied between two points. As the force is a sine wave the integral will be a cosine and over one cycle the work done is zero because you've acceleratedthe block to a max speed then done the same amount of work slowing it down only in the opposite direction.

    I guess its best to look at work as being done over half cycles or if you're interested in how much energy has been expended at the modulus of a cycle multiplied by n cycles.
     
  13. Apr 11, 2006 #12

    Integral

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    Your velocity, v = 1 - cos(t) is always postive, therefore you displacement is always increasing.
    Here is a plot. Yellow is the force, blue is velocity, pink is displacement.
     
    Last edited: Apr 11, 2006
  14. Apr 12, 2006 #13

    daniel_i_l

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    Now I understand, from the graph (thanks Integral) it's clear that whenever the work is 0, so is the speed.
     
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