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Basic harmonic oscilator problem (but I'm having troble solving it)

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data
    I have small block of mass m=1kg on top of a bigger block mass M=10kg
    The friction coefficient between the blocks μ=0.40
    No fricton between the big block and the ground.
    There is a spring with k=200N/m attached to the bigger block.

    The problem asks what is the maximum amplitude A the system can have in an harmonic oscilation without relative motion between the blocks.


    2. Relevant equations

    [itex]x=A\cos(ωt+\phi)[/itex]
    [itex]v=-ωA\sin(ωt+\phi)[/itex]
    [itex]a=-ω^2A\cos(ωt+\phi)[/itex]



    3. The attempt at a solution
    well, first I noticed that the maximum static force possible between the two blocks will be:
    mgμ which gives: [itex](1kg)(9.8m/s^2)(0.40) = 3.92N[/itex]
    so far no problems I guess.

    using F=ma, I calculate the acceleration of the system in the case of maximum friction:
    [itex]a= F/m = (3.92N)/(11kg)[/itex] which gives [itex]a = 0.356m/s^2[/itex]

    using [itex]ω^2=k/m=(200N/m)/(11kg)[/itex] which gives [itex]ω^2= 18.18rad/s^2[/itex]


    Now here is where I got stuck
    I plug this to [itex]a=-ω^2Acos(ωt+\phi)[/itex] and try solving for A

    the thing is:
    I don't know the time
    I don't know if there is a phase angle involved or not. or how to find out.
     
    Last edited: Oct 31, 2011
  2. jcsd
  3. Oct 31, 2011 #2

    Simon Bridge

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    When is the acceleration a maximum?
     
  4. Oct 31, 2011 #3
    Period of a spring is given by [itex] T = 2\pi \sqrt{\frac{k}{m}} [/itex]
    Since the problem wants you to solve for the two blocks moving synchronously the phase angle is zero, otherwise they would reach the same points in oscillation at different times (ie one reaches its maximum amplitude as the other is already moving back).
     
  5. Oct 31, 2011 #4
    well, according to the textbooks the maximum positive acceleration occurs when the body is in its maximum negative position...

    then it explains that "combining" the equations for position and acceleration will give [itex]a(t)=-ω^2x(t)[/itex]
    which I honestly did not understand why. (I seem to have a problem understanding "combinations" of equations. I'm never sure if they mean adding, dividing or whatever)


    Anyway, using this info and the fact explained above by JHamm I find

    [itex] x= a/-ω^2 = (0.356 m/s^2)/(18.18 rad/s^2) = (0.0196 m)[/itex]

    which is 1.96cm and the answer in the book is 21.6cm ....

    so I'm still missing something....
     
  6. Oct 31, 2011 #5

    HallsofIvy

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    You were told that [itex]v=-A sin(\omega t+\phi)[/itex]. All you need to know to find the maximum value for that is the maximum and minimum possible values for sine. What are they?
     
  7. Oct 31, 2011 #6
    The maximum acceleration of the system is [itex] a = \frac{\kappa x}{M+m} [/itex] and the maximum acceleration of the small block is [itex] a = \frac{mg\mu}{m} = g\mu [/itex]. You want these two to equal each other.
     
  8. Nov 1, 2011 #7
    That's right!

    equaling the acceleration of the system with the acceleration of the small block gives
    x = 0.2156 m which rounding up gives 21.6 cm


    I guess I was fixed in the harmonic oscillator part and didn't quite notice I could do something like this.

    Thanks a lot.
     
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