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Simple Harmonic Motion of a table

  1. May 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A table moves up and down in SHM of period 1 second. Find the amplitude it can have at which objects on the table just comence to lose contact with the table.

    3. The attempt at a solution

    How do I need to solve this problem? I'm only given the period from which I can work out only the frequency [tex]f[/tex] and angular frequency [tex]\omega[/tex].

    I know that the position of a particle that exhibits simple harmonic motion is given by

    [tex]x(t)=A cos (\omega t + \phi)[/tex]

    And I know the maximum value cosine can have is 1. But this is all I know, I don't see how I can approach this problem. Any help here is very much appreciated.
  2. jcsd
  3. May 14, 2010 #2


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    Homework Helper
    Gold Member

    Here's a hint.

    [tex] v(t) = \frac{\partial}{\partial t}x(t) [/tex]

    [tex] a(t) = \frac{\partial}{\partial t} v(t) = \frac{\partial^2}{\partial t^2} x(t) [/tex]

    Special bonus hint: Hmmm. How does g fit into all of this. Hmmm. :wink:
  4. May 15, 2010 #3
    What is this? I'm not looking for the velocity or the acceleration. The question asks for the maximum amplitude!
  5. May 15, 2010 #4

    Doc Al

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    Staff: Mentor

    Collinsmark has given you a big hint. The problem states "objects on the table just comence to lose contact with the table". Under what conditions will that happen?
  6. May 15, 2010 #5
    Think about it like this: what's the maximum acceleration the table can have for particles to remain on it? :wink:
  7. May 15, 2010 #6
    This is exactly what I don't understand. You can explain it to me, this is not a homework problem. I already know that the answer must be 0.25 m.
  8. May 16, 2010 #7
    Try experimenting.
    Hold a large book or something with something smaller on it. Raise it up and down, getting increasingly faster. If you do it fast enough, the smaller object will break away. Why? :wink:
  9. May 21, 2010 #8
    I know that this is empirically true. When we increase the frequency the objects start to lose contact with the surface. But I don't know the reasoning behind it, and I'm not sure how to explain it.

    Also I don't have the formula above (posted by collinsmark) in my textbook, could you please show me how to use it? What do I need to do with all the partial derivative signs?
  10. May 21, 2010 #9

    Doc Al

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    Staff: Mentor

    See if you can answer this question. Imagine you are standing on a scale in an elevator. Under what conditions will the scale read zero? (Note that when the scale reads zero is the point at which you begin to lose contact with the floor of the elevator.)

    Does your book describe the position, velocity, and acceleration of a body in SHM in terms of trig functions? If so, that's all you need.
  11. May 21, 2010 #10
    You don't need any partial derivatives. You can solve the problem using the general definition of SHM, i.e [tex]a = -w^2[/tex]X
  12. May 22, 2010 #11
    [tex]\sum F_y = N -mg = ma_y[/tex]

    [tex]N = ma_y +mg[/tex]

    When the elevator cable breaks and the elevators is in freefall, then the acceleration is [tex]a_y=-g[/tex]. Therefore the person will be weightless and the scale will read zero. Right?

    Okay, the velocity & acceleration of an object in simple harmonic motion are respectively

    [tex]v=\frac{dx}{dt} = - \omega A sin (\omega t + \phi)[/tex]

    [tex]v=\frac{d^2x}{dt^2} = - \omega^2 A sin (\omega t + \phi)[/tex]

    But I think since we are looking for the "maximum values" here, the trig functions will have to be 1:

    [tex]v_{max}= \omega A[/tex] .....(1)

    [tex]a_{max}= \omega^2 A[/tex] .....(2)

    And I can find the angular frequency

    [tex]T=1=\frac{2 \pi}{\omega}[/tex]

    [tex]\omega = 2 \pi[/tex]

    I can't find the velocity, so I can't use equations (1) and (2). Any help?
  13. May 22, 2010 #12
    He's not weightless, he appears weightless. But you have the principals correct. Both the elevator and the person are falling with the same acceleration, so relative to the elevator he has no acceleration, so relative to the elevator there are no forces acting on him.

    So, what is the maximum acceleration the person can have to just remain on the surface of the elevator?
    You've just answered it.

    Okay, [tex]\omega = 2 \pi[/tex], and [tex]a_{max}= \omega^2 A[/tex].
    Try and answer the question above, and apply this.
  14. May 22, 2010 #13
    Thank you very much! :biggrin:
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