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Spring Stiffness and Amplitude, not understanding

  1. Aug 22, 2012 #1
    Hi,

    For an undamped mass, spring system subject to a harmonic force, the equation is: ma+kx=Fsin(ωt)
    where a=d^2/dx^2

    After solving the diff eqns, the steady state amplitude is:
    X=(F/m)/(ω^2-ω0^2)
    where ω is the frequency and ω0 is the natural frequency =sqrt(k/m)

    according to the amplitude equation, the amplitude will increase if you increase the stiffness.

    I am struggling to understand why this is true, as i understand it, the spring should be resisting the motion of the mass and hence the stiffer the spring the less it moves. Think about if you increase the stiffness to infinity, equivelant to placing the mass on a rigid floor, the displacement of the mass should surely reduce to zero. Am I missing something?

    Anyone able to explain?
     
  2. jcsd
  3. Aug 22, 2012 #2

    Simon Bridge

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    You mean this equation:
    ... lets test this out ... a stiff spring would mean a higher k, so the relationship between ampltude and stiffness would be:

    [tex]X=\frac{F}{m}\left ( \omega^2 - \frac{k}{m} \right )[/tex]

    Initially it looks like the bigger k, the smaller the term in brackets due to the subtraction. However, it kinda looks like if k is very big, then [itex]\omega_0 > \omega[/itex] ... making the amplitude negative: what could this mean?

    [later] Taking a further look:
    http://en.wikipedia.org/wiki/Harmonic_oscillator#Sinusoidal_driving_force
    ... your system has a damping ratio of zero, which simplifies the solution.

    Using this, their amplitude comes out different from yours.
    I'd check your working.
     
    Last edited: Aug 22, 2012
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