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SHM with initial velocity

  1. Aug 18, 2014 #1
    Does anyone know how the mathematics/formulae of SHM (say a particle in a force field, with the standard a=-d) changes when the particle is given an initial velocity independent of the force causing SHM?

    For example in 1D say we have a graph of acceleration-displacement in the standard form y=-x for x from -inf to +inf, but this force is turned off. A particle is then accelerated, and arrives at point (10,-10) with an initial velocity of -200. Then the SHM causing force is turned on, and the other off.

    How big would the osciallations be, and which point would be the equilibrium position?

    Many Thanks,
     
  2. jcsd
  3. Aug 18, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The equilibrium position is where the restoring force is zero; that is independent of initial velocity.

    You can figure out the amplitude using conservation of energy. Find the point where all the energy is potential and that's the amplitude of the oscillation.
     
  4. Aug 18, 2014 #3

    Dale

    Staff: Mentor

    Regardless of the initial conditions, you will get SHM about the same equilibrium as Doc Al mentioned. If you are at rest at position x0 at t0 then the solution is ##x_0 \; \cos(\omega(t-t_0))##. If you are at the equilibrium position with some velocity v0 at t0 then the solution is ##v_0/\omega \; \sin(\omega(t-t_0))##. If you have a position x0 and a velocity v0 at t0 then you just add up those two solutions.
     
  5. Aug 19, 2014 #4
    Kl, thanks

    So for initial speed of √200 and mass=1, k=1,

    KE=1/2*1*200 = 100J.

    From the position (I assume we need to shift to -10 and not use 0)

    PE=1/2*k*(-10)^2 = 50J

    So total is always 150J and PE on its own is 150J when x=+/- 17.3

    However, the specific question this is from is attached, where infact the linear force always remains. So does the particle venture back into the linear force area, and undergo a slightly distorted SHM?
     

    Attached Files:

  6. Aug 19, 2014 #5

    Dale

    Staff: Mentor

    This would not be simple harmonic motion in general. The equilibrium position will be -10, and it will undergo SHM only if the amplitude is small (10 or less). If the amplitude is large then large displacements will go from SHM to parabolic motion. So it would be something like a cosine with the tops chopped off and replaced with parabolas.
     
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