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Spring help-please

  1. Dec 8, 2003 #1
    At t=0, a 650-g(0.65kg) mass at rest on the end of a horizontal spring(k=184n/m) is struck by a hammer which gives it an initial speed of 2.26 m/s. Determine (a) the period and frequency of the motion (b)the amplitude (c)the maximum acceleration (d) the position as a function of time (e) the total energy, and (f)the kinetic energy when x=0.4A where A is the amplitude.

    a) period=2pi*sqrt0.65/184=.37
    Frequency=2.07
    b)v=sqrtk/m(A^2-X^2)=0.13

    C) A-max=184*0.13/0.65=36.8 m/s^2

    d)x=0.13cos(phi)

    e)no idea

    f)no idea
     
  2. jcsd
  3. Dec 8, 2003 #2
    Revisit the formula for Total energy and K.E.

    Total Energy remains Constant
     
  4. Dec 8, 2003 #3
    so the total energy would be

    1/2 KA^2
     
  5. Dec 9, 2003 #4
    yes and what about the K.E.
     
  6. Dec 9, 2003 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Two other points:

    I get that the frequency is 2.70 not 2.07. A typo?

    The position, measured from the rest point, is .13 sin(16.8t).

    There was no "phi" in the problem. What was that?
     
  7. Dec 9, 2003 #6
    So the kinetic energy would be

    1/2*184(0.052)^2=.25J
     
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