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Homework Help: Please help with simple harmonic motion

  1. Nov 9, 2004 #1
    1a)A simple pendulum is 4.35 m long. What is the period of simple harmonic motion for this pendulum if it is hanging in an elevator that is accelerating upward at 5.90 m/s2?

    b)What is the period of simple harmonic motion for this pendulum if it is placed in a truck that is accelerating horizontally at 5.90 m/s2?

    2) A large block P executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency of f = 1.52 Hz. Block B rests on it, as shown in the figure below, and the coefficient of static friction between the two is μs = 0.630.

    What maximum amplitude of oscillation can the system have if block B is not to slip?

    1)l=4.35 m
    a= 5.90 m/s/s

    w^2= 2.252873563

    then i have to find T.. but i don't know what formula to use...

    2) w=2(pie)f
    w-2(pie)(1.52 Hz)
    w= 9.5504
    and i don't know what else to do
  2. jcsd
  3. Nov 9, 2004 #2

    T= 2pi * sqrt(L/g)
    T= 2pi* sqrt(4.35 m/ [9.8m/s^2-5.90m/s^2])


    "as shown in the figure below" <- picture?
  4. Nov 9, 2004 #3
    it was a picture with a big block on the bottom.. a small block on the top.. with a spring attached to it..

    i got the first part of the question.. it was T=2pi sqrt(l/g+a)

    but i cannot get the second part of it... with it acclerating horizontally..
  5. Nov 9, 2004 #4
    springs? I thought we were talking about pendulums?! I can't visualize #2
  6. Nov 9, 2004 #5
    #2 is different from #1... it's.. a small block on a big block.. and the big block has a spring attached to it.. the coefficient of static friction was..0.630.. they want to know what's the maximum amplitude of the oscillation so that the little block doesn't slip..

    f= 1.52 Hz

    AND for #1.. if they change the a from upward.. to horizontal.. how does that change the answer...
  7. Nov 9, 2004 #6
    spring attached to it where? On the bottom? sides? Need to be specific

    I dont think the answer for #1 changes
  8. Nov 9, 2004 #7
    i've got it.. thanks.. :)
  9. Nov 20, 2004 #8
    Are you sure it's 2 pi ( sqrt ( L / (g - a ) ) ? Or is it g + a?
  10. Nov 21, 2004 #9

    Doc Al

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    Staff: Mentor

    In a non-accelerating frame, the period of a simple pendulum is:
    [tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
    In an accelerating frame, the effective "g" is different.

    For case (a), the apparent acceleration is [itex]-g\hat{y} -a\hat{y} = -(g+a)\hat{y}[/itex], so [itex]g_{eff} = g + a[/itex].

    For case (b), the apparent acceleration is [itex]-g\hat{y} -a\hat{x}[/itex], so [itex]g_{eff} = \sqrt{g^2 + a^2}[/itex].
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