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Acceleration of Mass on Spring

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A small block is oscillating on a spring with a period of 3.05 s. At t = 0 the mass has zero speed and is at x = 4.35 cm. What is the magnitude of the acceleration at t = 4.30 s?


    2. Relevant equations
    a=-Aω2cos(ωt)
    ω=2∏/T

    3. The attempt at a solution
    ω=2∏/3.05=2.06
    x(0)=Acos(0)=4.35=A(1)=A
    so:a=(t)=-4.35(2.06)2cos(2.60(4.30))=-18.3 m/s2

    which I've been told is wrong.

    I have no idea what I'm doing wrong, I'm sure it's something painfully obvious but I just can't figure it out.
     
  2. jcsd
  3. Sep 21, 2012 #2
    F=-k x[t]
    x''[t]=-(k/m)x[t]
    =-(w^2)x[t]
    x[t]=A*cos[w*t]+B*Sin[w*t]
    v[t]=w*(-A*sin[w*t]+B*cos[w*t])
    a[t]=-w^2*(A*cos[w*t]+B*Sin[w*t])

    v[0]=w*B=0, B=0
    x[t]=A*cos[w*t]
    x[0]=A=4.35

    a[t]=-w^2*(4.35*cos[w*t])
    w=2Pi/T

    As long as you did the calculator work correctly you should have the right answer.
     
  4. Sep 21, 2012 #3
    I got 15.58 m/s^2 with my calculator.
     
  5. Sep 21, 2012 #4
    You used 2.60 in your Cosine, rather than 2.06 ;)
     
  6. Sep 21, 2012 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Note that the angle within the cosine is in radians. (That might be the issue if your calculator was set to degree mode.) And it should be 2.06, not 2.6. (Perhaps that was just a typo.)
     
  7. Sep 21, 2012 #6
    Thank you! I knew I must have done something ridiculous like that.
     
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