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Simple harmonic motion block and spring

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A block, attached to a spring, is set into vertical oscillation with an amplitude of 25.0 cm and a frequency of 5.00 Hz. At t = 0.00 s, the block is 14.0 cm below the equilibrium position and is moving upwards. Assume that the displacement from equilibrium is given by
    y = ym cos(ω t + φ ) , with y positive upwards.
    Calculate the phase angle, φ.
    (a) 5.31 rad
    (b) 0.976 rad
    (c) 4.12 rad
    (d) 2.17 rad
    (e) 1.87 rad

    the answer is c....
    2. Relevant equations

    ive been using the equation that was given to me in the question, using 0.14m as my y, 0.25m as my ym, 0 as my t therefore w isnt needed? and i have a frequency??? im lost....please help lol

    i am arriving at (b).....if this helps at all



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 9, 2009 #2

    turin

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    Homework Helper

    Did you use the fact that the block is moving upwards? Here's a hint: what is the numerical difference between b and c? Does it look familiar?
     
  4. Feb 9, 2009 #3
    pi lol.....do i have to add pi? confusion
     
  5. Feb 10, 2009 #4

    turin

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    Homework Helper

    That was only a hint. Forget about [itex]\pi[/itex].

    EDIT: Your y is wrong. Can you tell me why? (haha)

    EDIT: EDIT: Wait a minute! THEIR y is wrong! Are you sure you copied every word of the problem correctly?
     
    Last edited: Feb 10, 2009
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