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Simple harmonic motion period with velocity defined

  1. Jan 14, 2015 #1
    1. The problem statement, all variables and given/known data
    A mass m is sliding back and forth in a simple harmonic motion (SHM) with an amplitude A on a horizontal frictionless surface. At a point a distance L away from equilibrium, the speed of the plate is vL (vL is larger than zero).

    2. Relevant equations
    What is the period of the SHM?

    3. The attempt at a solution
    a_x=-kx/m -> vX= (-kx^2)/(2m)
    k = (-vX*2m)/(x^2)
    T=2π*√(m/k)=2π*√(m/((-vX*2m)/(x^2))
    T=2π√((-x^2)/vX))
    Filling in point at distance L from equilibrium, I get:

    T=2π√((-L^2)/vL))

    The correct answer is T=2π√((A^2-L^2)/vL)), but I cannot imagine where the A^2 comes from.

    Any help is appreciated!
     
  2. jcsd
  3. Jan 14, 2015 #2
    For simple harmonic motion,
    x=Asin(wt+Φ)
    and v=dx/dt = Awcos(wt+Φ)

    Can you try relating x and v somehow?
     
  4. Jan 14, 2015 #3
    Yes, x=∫vdt, or is that not what you meant with relating x and v?
     
  5. Jan 14, 2015 #4
    Nope, I meant try substituting the value of (wt+Φ) from the first equation into the second one.
     
  6. Jan 14, 2015 #5
    I am terribly sorry, but my native language is not English and I do not know what you mean with the value of (wt+Φ). I do not have the frequency in the data?
     
  7. Jan 14, 2015 #6
    (wt+Φ) = arcsin(x/A)
    What is cos(wt+Φ)?
    After figuring out cos(wt+Φ), substitute it into v=Awcos(wt+Φ). You'll get an expression that relates x and v.
     
  8. Jan 15, 2015 #7
    Ik think you have to apply conservation of energy, it will result in:
    1/2*k*L^2 + 1/2*m*v^2=1/2*k*A^2
    k =(m*v^2)/(A^2 -L^2)
     
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