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Magnitude of the maximum acceleration spring problem

  1. Nov 18, 2004 #1

    ---A 29.0kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350m, then lets go. The mass undergoes simple harmonic motion with a period of 5.30s. What is the position of the mass 4.293s after the mass is released?

    Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion? --------------

    Um so....what I tried to do for the first part was find the x postion using the equation x=Acos(wt)....but then I realized I don't have enough information. I know for the second part that Max acceleration is a= Aw^2 ...soo anyway, in the first question I think A = 0.35m...and w= k/m --> ?/29kg .... and t= 4.293s....so basically I'm lost and I'd be very thankful for help. Hope some of that made sense! Thanks :)
  2. jcsd
  3. Nov 18, 2004 #2
    [tex]\omega = \frac{2*\pi}{T}[/tex]

    T is the time period.
    So, you can directly calculate [tex]\omega[/tex]

  4. Nov 18, 2004 #3

    Andrew Mason

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    Use the equation of shm to determine the k of the spring (ie. [itex]\omega = \sqrt{k/m}[/itex]). Of course, maximum acceleration is the point of maximum force. Since F = -kx, this is at the point or maximum displacement, which is the amplitude.

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