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Homework Help: Stiffness in mass spring system

  1. Nov 16, 2012 #1
    1. The problem statement, all variables and given/known data
    What happens to the frequency of oscillation if stiffness increases and why?

    2. Relevant equations


    3. The attempt at a solution
    Frequency increases but trying to figure out why.
  2. jcsd
  3. Nov 16, 2012 #2
    By stiffness I'm assuming you mean the spring constant, often known as k. It's found in Hooke's law F = -kx where F is the force the spring is exerting and x is the distance from the equilibrium point.

    Don't you have some equations relating the frequency of the oscillations of the spring-mass system and the spring constant of the spring?
  4. Nov 16, 2012 #3
    a=[2pie f]^2x. So if k increases f increases which increases a(acceleration) and according to the formula given at the start frequency increases. If that make sense:blushing:
  5. Nov 16, 2012 #4
    [itex]a = \left( 2 \pi f\right)^{2x}[/itex] ?

    I don't recognize that equation >.> I'm very sorry

    do you recognize

    [itex]x \left( t \right) = Acos \left( \omega t - \phi \right)[/itex]


    [itex]\omega = \sqrt{ \frac{k}{m}}[/itex]

    is the angular frequency of the oscillation, A is the amplitude and [itex]\phi[/itex] is the phase shift?
  6. Nov 16, 2012 #5
    Its a = \left( 2 \pi f\right)^{2} x.
    I dint knew the 2nd equation but i get it now. stiffness is proportional to frequency according to that.but does the natural frequency of spring increases as well?
  7. Nov 16, 2012 #6
    [itex]a = \left( 2 \pi f\right)^{2}x[/itex] makes much more sense :)

    angular frequency is just the natural frequency times 2π


    [itex] \omega = 2 \pi f[/itex]
  8. Nov 16, 2012 #7
    Thanks alot:biggrin:
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