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How to find amplitude in SHM problem

  1. Jul 1, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass of 120g rolls down a frictionless hill reaching a speed of 4.2 m/s and collides with another mass 3.00g attached to a spring of constant 30 N/m. The two masses stick together and enter into periodic motion. What is the equation for the motion?

    2. Relevant equations
    m1v1+m2v2=mtV
    x(t)=Acos(wt+phi)
    w=k/m^1/2
    Vmax=wA

    3. The attempt at a solution
    I found Vfinal=1.2 m/s
    Also found w=8.45 rad by w=k/m^1/2
    I thought A would be 0.142 but in the answer A=0.266
    the answer is: x(t)=0.266sin(8.44t)
     
  2. jcsd
  3. Jul 1, 2016 #2

    TSny

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    Gold Member

    I guess you meant to type 300 g rather than 3.00 g for the second mass.

    Your work looks good to me. I agree that A = 0.142 m.

    [EDIT: I'm assuming that the first mass reaches the bottom of the hill and then moves horizontally before hitting the spring as shown below.]

    (If there is no friction on the hill, the 120 g mass would slide rather than roll down the hill.)
     

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    Last edited: Jul 1, 2016
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