Newton's second law and Springs

1. Jun 24, 2011

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?

2. Jun 24, 2011

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.

3. Jun 24, 2011

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).

4. Jun 25, 2011

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:

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

5. Jun 25, 2011

gkangelexa

makes sense thanks!