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SHM involving a spring

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1.6-kg block on a horizontal frictionless surface is attached to a spring whose force constant is 190 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.40 s is closest to:



    2. Relevant equations



    3. The attempt at a solution

    ω=√k/m=190/1.6=118.75 rad/s

    v=-Aωcos(ωt)

    =-0.080*(10.90)cos(10.90*.40)=.30m/s

    However the answer is .8m/s

    Am I missing a step?

    Thank you
     
  2. jcsd
  3. Apr 21, 2012 #2

    Doc Al

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    Staff: Mentor

    You are using x = Asin(ωt). But that assumes that at t = 0 the mass is at x = 0.

    (Pick a form that starts at maximum displacement.)
     
  4. Apr 21, 2012 #3
    I dont quite understand what you are saying,

    isnt the maximum displacement .080 which is the amplitude, im not sure how to incorporate it.
     
  5. Apr 21, 2012 #4

    Doc Al

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    Staff: Mentor

    Where did this come from?

    When they say "at time t = 0.40 s" they are measuring the time from the moment it was released, which is the point of maximum displacement.
     
  6. Apr 21, 2012 #5
    Which means I use -waSin(wt)

    since the graph of sin starts at zero?
     
  7. Apr 21, 2012 #6

    Doc Al

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    Staff: Mentor

    Yes.
     
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