### suspension dynamics (1/3)

This is the first of a 3 part series of posts on the mathematics of car suspension systems.

The purpose of this is to learn a bit more about coilover suspensions. I will assume only a basic understanding of Newtonion physics and differential equations. We will start with simple harmonic oscillation and work our way up.?

Write down the general equation of motion, Newton’s second law.

$$F = ma$$

Assuming an idealized spring that takes advantage of hooks law, we declare the spring’s force, and then plug the force into (1). k is the spring constant. This assumption holds true in practice in the case of linear springs not undergoing extreme deflection.

$$-kx = F$$

$$m\frac{d^2x}{dt^2} = -kx$$

We know that sin and cos are their own second derivatives, with an additional constant term. With this information we guess that the solution to the differential equation is as follows

$$x(t) = a cos(\omega t) + b sin(\omega t)$$

Taking the solution at time t=0, we find

$$a = x_0$$

Taking the first derivative of (4) yields the following

$$v(t) = \frac{dx}{dt} = -\omega a sin(\omega t) + \omega b cos(\omega t)$$

Setting t = 0, we get

$$b = \frac{v_0}{\omega}$$

Combining the new substitions in (4), our new equation becomes

x(t) = x0 cos(\omega t) + \frac{v0}{\omega} sin(\omega t)

Derive this twice and factor out the constant.

v(t) = -\omega x0 sin(\omega t) + \omega \frac{v0}{\omega} cos(\omega t)

a(t) = -\omega^2 x0 cos(\omega t) - \omega^2 \frac{v0}{\omega} sin(\omega t)

a(t) = -\omega^2 ( x0 cos(\omega t) + \frac{v0}{\omega} sin(\omega t) )

Notice that the term inside of the parenthesis is equal to equation (8). Substituting x(t), we get:

$$a(t) = -\omega^2 x(t)$$

plugging in equation (3)

$$a(t) = -\frac{kx(t)}{m} = -\omega^2 x(t)$$

$$\omega = \sqrt{\frac{k}{m}}$$

Now that we have all of our unknowns covered, we can write our final equation of motion

x(t) = x0 cos( \sqrt{\frac{k}{m}} t ) + \frac{v0}{ \sqrt{\frac{k}{m}} } sin( \sqrt{\frac{k}{m}} t )

We verify our solution by taking the second derivative of (15)

a(t) = - x0 \frac{k}{m} cos( \sqrt{\frac{k}{m}} t ) - \frac{v0}{ \sqrt{\frac{k}{m}} } \frac{k}{m} sin( \sqrt{\frac{k}{m}} t )

Setting that equal to (3)

a(t) = -\frac{kx(t)}{m} = - x0 \frac{k}{m} cos( \sqrt{\frac{k}{m}} t ) - \frac{v0}{ \sqrt{\frac{k}{m}} } \frac{k}{m} sin( \sqrt{\frac{k}{m}} t )

a(t) = -x(t) = - x0 cos( \sqrt{\frac{k}{m}} t ) - \frac{v0}{ \sqrt{\frac{k}{m}} } sin( \sqrt{\frac{k}{m}} t )

We see equation (15) poke its head out. Making the substitution leads to

$$-x(t) = -x(t)$$

Win.