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Spring Harmonic Motion

  1. Jun 23, 2012 #1
    Hi there, SalsaOnMyTaco here again.
    1. The problem statement, all variables and given/known data

    A 44 gram mass is attached to a massless spring and allowed to oscillate around an equilibrium according to:
    y(t) = 1.2*sin( 3.1415*t ) where y is measured in meters and t in seconds

    -What is the spring constant in N/m ?

    HELP: Simple harmonic motion with the amplitude A is equivalent to the motion on a circle with the radius A, and the same angular frequency omega.
    The force acting on an object with the mass m moving on a circle with the radius A with the angular frequency ω is
    F_circ=m*A*ω2.
    The force exerted by a spring with the constant k is equal to
    F(x)=k*x, where x is the displacement.
    Due to the analogy mentioned above, the two forces are equal at the point of maximum displacement (amplitude), that is,
    F_circ=F(A).
    You can solve this equation for omega in terms of k and m.
    2. Relevant equations

    T=2∏√(m/k)

    3. The attempt at a solution
    I have no idea how to approach this problem. Should I start from figuring out the Period to then solve for K on the above equation?
     
  2. jcsd
  3. Jun 23, 2012 #2
    Nvm, problem solved. Since i was given w=2pi/t, i solved for T and used T on the equation from OP to solve for K.
     
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