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Newton's second law and Springs

  1. Jun 24, 2011 #1
    Spring Force is F = -kx
    So, because the force is not constant, the acceleration is also not constant..

    How then can we apply Newton's second law: F = mass x acceleration to springs?
     
  2. jcsd
  3. Jun 24, 2011 #2

    WannabeNewton

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    There is no restriction that newton's second law must have constant acceleration. You can have a varying acceleration and if you solve [itex]F = m\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}}[/itex] you can get the resulting varying force. For a spring, the potential is [itex]v(x) = \frac{1}{2}kx^{2}[/itex] and [itex]F = -\frac{\partial V}{\partial x} = m\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}}[/itex] so [itex]m\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}} = -kx[/itex]. Acceleration doesn't have to be a constant; it could be given as some varying function of time and the varying spring force would follow suit.
     
  4. Jun 24, 2011 #3

    Matterwave

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    We note that acceleration is the second derivative of the position function. Our position function is x(t), so that v(t)=dx(t)/dt and a(t)=dv(t)/dt. Now Newton's law tells us:

    [tex]F=-kx(t)=m\frac{d^2 x(t)}{dt^2}[/tex]

    Which is a second order differential equation that we have to solve (we need 2 initial conditions to completely solve this).
     
  5. Jun 25, 2011 #4
    [tex]F=-kx(t)=m\frac{d^2x(t)}{dt^2}[/tex]

    If you were to bring the mass on the spring to an initial displacement and let it go, you could figure out the position function x(t) by solving that differential equation:

    [tex]x(t)=x_0~cos(\omega t+\phi )[/tex]

    Where:

    [tex]\omega^2=\frac{k}{m}[/tex]
     
  6. Jun 25, 2011 #5
    makes sense thanks!
     
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