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Derivation: Simple Harmonic Motion

  1. Jan 17, 2012 #1
    I know how to arrive at

    a = -(k/m)x

    for a SHM system.

    However, most derivations i found did an unexplained step from here onward.

    They simply let k/m = ω2.

    And later on then mention ω as the angular frequency and link to the period of oscillation.

    My question is,

    Why did that substitution take place? Why must it be ω2?
    Why can't k/m = v3 or anything else?

    Please help as I couldn't find any explanation for this step!
  2. jcsd
  3. Jan 17, 2012 #2
    a = d2y/dt2 = v*(dv/dx)
    Also you have kx + ma = 0, then substitute a with v*(dv/dx), then solve the first ODE.
  4. Jan 17, 2012 #3
    hi, thanks for the respond.

    could u please explain how you arrive at

    a = d2y/dt2 = v*(dv/dx)?

    why is there 'y' and 'x' and why dv/dx is multiplied by v?
  5. Jan 17, 2012 #4
    A physics definition of SHM is the motion that results when an object experiences a restoring force when it is displaced.
    SHM is when the restoring force is proportional to the displacement.
    i.e F proportional to x, F= kx (k is a constant)
    This means that k = F/x which has units N/m and is known as the STIFFNESS (same term is used for springs)
    If the object has mass, m then acceleration = F/m = (kx)/m
    By making comparisons with circular motion where a = ω^2 A (A is radius which becomes amplitude in oscillations) it is possible to show that ω^2 = k/m
    The period of the circular motion is the same as the period of the SHM
    This means that ω = √k/m or 2πf = √k/m
    so you see that straight away you can calculate the frequency of the SHM from the stiffness and mass of the system.
  6. Jan 17, 2012 #5
    sorry but I believe your explanation is flawed.

    the ω in my equation refers to angular frequency, but in circular motion, the ω is angular velocity. They are two different things and cannot be compared as such.
  7. Jan 17, 2012 #6
    OK, have you looked in any text books to see the similarities between SHM and circular motion?
    Good luck in your search for an explanation
    Last edited: Jan 17, 2012
  8. Jan 17, 2012 #7
    similarity is not a concrete explanation.
    it may be coincidence.
  9. Jan 17, 2012 #8
    a = d2y/dt2 = v*(dv/dx) should be 'x' instead of 'y'
    So, a = d2x/dt2 = v*(dv/dx).
    d(dx/dt)/dt = dv/dt = dv/dx * dx/dt = dv/dx * v.
  10. Jan 17, 2012 #9
    From Halliday Resnick, Fundamentals of Physics 8E.

    That should explain why they are comparable.
  11. Jan 17, 2012 #10

    Ken G

    User Avatar
    Gold Member

    It's not coincidence, all you have to do is analyze the circular motion one component at a time. In circular motion, you have a force of constant magnitude but changing direction, and when you project such a force onto any one fixed direction, you will immediately get the force law of simple harmonic motion (try the trig, or just look at the Cartesian coordinates of a force of constant magnitude). Indeed, I find it quite useful to go the opposite direction-- take a 1D simple harmonic oscillator and introduce a second dimension, even if there is really only one. Now imagine the oscillator moves in both dimensions, but just give the motion in the second dimension a 90 degree temporal phase shift (it does everything the 1D solution does, just a quarter period later), thinking of it as a kind of "lagged solution" perpendicular to the first. With a little effort, you can see that the combined motion is in a circle at constant velocity, and circular motion is much easier to solve, because the only thing varying in time is the direction of motion, and you know just how that varies (like a clock). When you've solved the steady state in that situation (even if there is driving and damping and the whole 9 yards), it is a simple matter to remove the "lagged" solution that you originally added, and you recover the 1D solution (by the linearity of the equations).

    Note also that springs can really do this kind of motion (you can make a stretched spring go around in a circle), and in fact this is always the easiest device to imagine what springs do. We should imagine that the "natural" thing for a spring to do is go in a circle, and think of 1D oscillation as a special case of that (either project the circular motion, or average it with the circular motion in the opposite direction), rather than the usual way springs are taught as objects that naturally oscillate in 1D. In other words, we understand systems by considering their steady-state response, and the "steady-state" thing that springs do is go in a circle.
    Last edited: Jan 17, 2012
  12. Jan 17, 2012 #11
    Consider a point in uniform circular motion.
    The projection of this point onto a diameter produces a point that is executing SHM
    The link could not be stronger.
  13. Jan 17, 2012 #12
    ok i get where you all are coming from for the case of comparison.

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