# 2.3: Equations of Motion

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Now that we have set our axioms - Newton’s laws of motion and the various force laws - we are ready to start combining them to get useful results, things that we did not put into the axioms in the first place but follow from them. The first thing we can do is write down equations of motion: an equation that describes the motion of a particle due to the action of a certain type of force. For example, suppose you take a rock of a certain mass m and let go of it at some height h above the ground, then what will happen? Once you’ve let go of the rock, there is only one force acting on the rock, namely Earth’s gravity, and we are well within the regime where Equation 2.2.2 applies, so we know the force. We also know that this net force will result in a change of momentum (Equation 2.1.4), which, because the rock won’t lose any mass in the process of falling, can be rewritten as Equation 2.1.5. By equating the forces we arrive at an equation of motion for the rock, which in this case is very simple:

\[m \boldsymbol{g}=m \ddot{\boldsymbol{x}} \label{rock}\]

We immediately see that the mass of the rock does not matter (Galilei was right! - though of course he was in our set of axioms, because we arrived at them by assuming he was right...). Less trivially, Equation (\ref{rock}) is a second-order differential equation for the motion of the rock, which means that in order to find the actual motion, we need two initial conditions - which in our present example are that the rock starts at height h and zero velocity.

Equation (\ref{rock}) is essentially one-dimensional - all motion occurs along the vertical line. Solving it is therefore straightforward - you simply integrate over time twice. The general solution is:

\[\boldsymbol{x}(t)=\boldsymbol{x}(0)+\boldsymbol{v}(0) t+\frac{1}{2} \boldsymbol{g} t^{2}\]

which with our boundary conditions becomes

\[\boldsymbol{x}(t)=\left(h-\frac{1}{2} g t^{2}\right) \hat{\boldsymbol{z}} \label{soln}\]

where \(g\) is the magnitude of \(g\) (which points down, hence the minus sign). Of course Equation \ref{soln} breaks down when the rock hits the ground at \(t=\sqrt{2h \over g}\), which is easily understood because at that point gravity is no longer the only force acting on it.

We can also immediately write down the equation of motion for a mass on a spring (no gravity at present), in which the net force is given by Hooke’s law. Equating that force to the net force in Newton’s second law of motion gives:

\[-k \boldsymbol{x}(t)=m \ddot{\boldsymbol{x}}(t) \label{spring}\]

Of course, we find another second-order differential equation, so we again need the initial position and velocity to specify a solution. The general solution of Equation \ref{spring} is a combination of sines and cosines, with a frequency \(\omega=\sqrt{k \over m}\) (as we already know from the dimensional analysis in Section 1.2):

\[\boldsymbol{x}(t)=\boldsymbol{x}(0) \cos (\omega t)+\frac{\boldsymbol{v}(0)}{\omega} \sin (\omega t)\]

We’ll study this case in more detail in Section 8.1. In general, the force in Newton’s second law may depend on time and position, as well as on the first derivative of the position, i.e., the velocity. For the special case that it depends on only one of the three variables, we can write down the solution formally, in terms of an integral over the force. These formal solutions are given in Section 2.6. To see how they work in practice, let’s consider a slightly more involved problem, that of a stone falling with drag.