Newton's second law and Springs

[gkangelexa]

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?

 

[WannabeNewton]

There is no restriction that Newton's second law must have constant acceleration. You can have a varying acceleration and if you solve F = m\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}} you can get the resulting varying force. For a spring, the potential is v(x) = \frac{1}{2}kx^{2} and F = -\frac{\partial V}{\partial x} = m\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}} so m\frac{\mathrm{d} ^{2}x}{\mathrm{d} t^{2}} = -kx. 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.

 

[Matterwave]

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:

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

Which is a second order differential equation that we have to solve (we need 2 initial conditions to completely solve this).

 

[Superstring]

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

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:

x(t)=x_0~cos(\omega t+\phi )

Where:

\omega^2=\frac{k}{m}

 

[gkangelexa]

makes sense thanks!