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Homework Help: Why a spring-mass assembly is a SHM?

  1. Sep 4, 2011 #1
    I was trying to understand why a spring -mass assembly and a simple pendulum is a simple harmonic movement.

    I've done like this:

    In a SHM that starts in -A, we have x = -A.cos(wt)

    In a spring-mass we have that the maximum velocity is [itex] \frac{mv²} {2} = \frac{kA²}{2} -> v = A\sqrt{\frac{k}{ m}} [/itex]

    [itex] w = v/A = \sqrt{\frac{k}{m}} [/itex]

    [itex] v = \int a.dt = \int \frac{-k.x}{m} dt [/itex]

    But in a SHM, [itex] v = -w .x.tan(wt) -> -w .x.tan(wt) = \int \frac{-k.x}{m} dt [/itex]

    If we substitute x = -A.cos(wt), we see that the expression is true, so spring-maass is a SHM.

    But I don't know how to demonstrate why pendulum is a SHM?
  2. jcsd
  3. Sep 4, 2011 #2


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    Well at an angle θ, what the torque about the center of rotation equal to? (use m,g, and L for parameters)

    This torque is also equal to I(d2θ/dt2). 'I' in this case just the inertia of the mass.
  4. Sep 4, 2011 #3


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    The condition for SHM is that the restoring force is proportional to displacement.
    That is pretty easy to do with a spring/mass set-up.

    For a pendulum, when the bob is off to one side, the component of the mass, in the direction of travel, is the restoring force. This involves a trig function, but we can show that for small deflections, the restoring force is pretty much proportional to displacement.
  5. Sep 4, 2011 #4

    If we calculte we get F = mg sin(y)cos(y)

    where sin(y) = x/L if cos(y) ~ 1, we get a SHM

    So in big angles, pendulum is not a SHM?
  6. Sep 4, 2011 #5


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    Not sure where that cos function came from?? but yes for small angles

    Tan(x) = x = sin(x) [x in radians of course]

    So this is SHM for "small" angles. The conjecture then becomes "what is small?" And that can make the subject of a great little practical exercise.
  7. Sep 4, 2011 #6
    mg sin(y) would be the force tangent to the movement, so the force in the horizontal (x) axes is mgsin(y)cos( y)
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