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Homework Help: Simple harmonic motion and friction

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data

    A large block P executes horizontal simple harmonic motion by sliding across a frictionless surface with a frequency f = 1.5 Hz. Block B rests on it, and the coefficient of static friction between the two is [tex]\mu_s=0.60[/tex]. What maximum amplitude of oscillation can the system have if the block is not to slip?

    2. Relevant equations

    [tex]F=-kx=ma[/tex]

    [tex]a=-\frac{k}{m}x[/tex]


    3. The attempt at a solution

    I'm not positive that i need the equations above, but I do know that the amplitude will be given by [tex]\frac{\mu_sg}{4\pi^2f^2}[/tex] according to the textbook answer. The numerical value comes out to be 0.662 m. But I'm not sure how to associate the forces with the harmonic motion exactly. I was thinking that somehow the force must not be greater than than [tex]\mu_smg[/tex] since that is what the maximum friction force allowed would be. But I'm not sure if I need to work with the motion equations for harmonic motion or if I need to use some method involving the period, forces, etc. Any pointer in the right direction would be much appreciated.
     
    Last edited: Apr 2, 2008
  2. jcsd
  3. Apr 2, 2008 #2
    I suppose that my two biggest things that I can't get straight in my mind are where does the pi come from, and why is there no k needed to find the answer. Also, there should be an exponent of 2 on the pi in the denominator, sorry about that.
     
  4. Apr 2, 2008 #3

    Dick

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    Given the amplitude and frequency of the motion you can figure out the maximum acceleration of the block, a. Then set F=ma=mu*m*g, right? Since the frictional force must produce the needed acceleration or the block will slide.
     
  5. Apr 2, 2008 #4

    Dick

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    The displacement d(t)=A*sin(2*pi*f*t) is an equation for simple harmonic motion. To get the acceleration you need to differentiate it twice. Can you see where the 4*pi^2 might come from?
     
  6. Apr 2, 2008 #5
    I see...but I don't have an amplitude, I need to find the maximum amplitude. I do see how the friction force is equal to ma though.
     
  7. Apr 2, 2008 #6

    Dick

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    Just call the amplitude 'A'. What is the maximum acceleration of d(t)=A*sin(2*pi*f*t)?
     
  8. Apr 2, 2008 #7
    In my book, it says that [tex]a=-\omega^2Acos(\omegat+\phi)[/tex]

    So if I put in 2*pi*f for w, I get

    [tex]a=-4\pi^2f^2Acos(2ft\pi+\phi)[/tex]

    OK, thanks now I can see where those quantities come from. OK, so the frictional force must not be greater than an acceleration given by this, right? I think I'm getting it now.

    [tex]ma=\mu_smg[/tex] reduces to

    [tex]\mu_smg=a[/tex]

    I think I need to subsitute something in for a, would I put in the whole quantity above that I wrote?
     
    Last edited: Apr 2, 2008
  9. Apr 2, 2008 #8

    Dick

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    I think you are getting it too. But I can't quite read the conclusion. Just solve for A.
     
  10. Apr 2, 2008 #9
    Alright...thank you for the help!
     
  11. Jun 7, 2010 #10
    Hello, I am stuck on this same problem and I've tried following the steps provided here, but it was not explained all the way to the end.

    Here is what I've done;
    a= -(omega^2)Acos(omega(t) + phi)
    substituting omega = (2)(pi)(f)
    a= (-4)(pi^2)(f^2)Acos[(2)(pi)(f)(t) + phi)

    also, a= (umg / m) ---> a=ug

    I equated these two expressions in an attempt to solve for A, but there are still several unknowns: the time t, and phase constant phi.
    Any help is greatly appreciated.
     
  12. Jun 8, 2010 #11
    So you have
    [tex]a=-4\pi^2f^2Acos(2\pi ft+\phi)[/tex]

    What is the maximum value that this whole thing can equal to? Think about how cosine behaves.
     
  13. Jun 8, 2010 #12
    That value would be 1, right? Because that is the maximum y value of cos?

    Should I just equate cos[(2)(pi)(f)(t) + phi) to 1? aka, ignore it? (Sorry about my notation.. I'm new to this board so I'm not used to how to get the symbols)
     
  14. Jun 9, 2010 #13
    Yes, that's right. Cosine would oscillate between -1 and 1. So what would be the maximum value of acceleration?

    Also, you don't really ignore it. It's always there; however, at some point in time, it is equal to that maximum value (or minimum).
     
    Last edited: Jun 9, 2010
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