Newton's second law and Springs

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
Gold Member
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).

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

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:

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

makes sense thanks!