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Simple Harmonic Motion oscillation amplitude

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass on a spring has an angular oscillation frequency of 2.56 rad/s. The spring constant is 27.2 N/m, and the system's kinetic energy is 4.16 J when t = 1.56 s. What is the oscillation amplitude? Assume that the mass is at its equilibrium position when t = 0.


    a. 63.1 cm
    b. 47.7 cm
    c. 73.4 cm
    d. 55.3 cm
    e. 84.0 cm


    2. Relevant equations


    3. The attempt at a solution
    w^2=k/m => m= 27.2/6.5536 = 4.15kg mg= - kx => x=1.497

    K=(1/2)mv^2 => v=1.416 m/s
    x= Acos(2.56t + φ)
    v(1.56) = 1.416 = -2.56Asin(3.994+φ)
     
  2. jcsd
  3. Feb 18, 2015 #2

    Nathanael

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    I would use x=Asin( ... ) that way φ = 0 because the function sin(ct) starts out at zero (equilibrium) at t=0

    Anyway that's not important.

    At 1.56 seconds, a certain fraction of the total energy will be kinetic. This fraction is said to be equal to 4.16 joules, so you can then find the total energy. Then you can use that total energy to find the displacement amplitude, because they told you the spring constant.
     
  4. Feb 18, 2015 #3
    But how would you propose I find the total energy because at that time, I don't know what position it is in.
     
  5. Feb 18, 2015 #4

    Nathanael

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    You don't know the position, but can you solve for the speed at that time in terms of A? Then try to find the ratio of kinetic energy at that time to the kinetic energy when it is maximum. The mass and the amplitude should cancel.
     
  6. Feb 18, 2015 #5
    Now I get it, thank you so much for answering all of my questions! I really appreciate it!
     
  7. Feb 19, 2015 #6
    good question, nice explaination, it was e right?

    get v at time t in terms of A from the first derrivative, then solve for A with the Kinetic energy at time t given

    v=Awcos(wt) or Awsin(wt+pi/2) if you prefer

    K=.5mv^2
     
    Last edited: Feb 19, 2015
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