Newton's second law and Springs

  • Thread starter gkangelexa
  • Start date
  • #1
81
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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?
 

Answers and Replies

  • #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.
 
  • #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).
 
  • #4
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[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]
 
  • #5
81
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makes sense thanks!
 

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