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Timing is Everything!

If the equation of motion is the ``question,'' what constitutes an ``answer''? Surely the most convenient thing to know about any given problem is the explicit time dependence of the position,  x(t),  because if we want the velocity   $v(t) \equiv \dot{x}$,  all we have to do is take the first time derivative - which may not be entirely trivial but is usually much easier than integrating! And if we want the acceleration   $a(t) \equiv \dot{v} \equiv \ddot{x}$,  all we have to do is take the time derivative again. Once you have found the acceleration, of course, you also know the net force on the object, by NEWTON'S SECOND LAW. A problem of this sort is therefore considered ``solved'' when we have discovered the explicit function  x(t)  that ``satisfies'' the equation of motion.

For example, suppose we know that

$$x(t) = x_0 \cos(\omega t) ,$$ (12.5)

where $\omega$ is some constant with units of radians/unit time, so that $\omega t$ is an angle. The time derivative of this is the velocity

$$\dot{x} \equiv v(t) = -\omega \, x_0 \sin(\omega t)$$

[look it up if needed] and the time derivative of that is the acceleration

$$\ddot{x} \equiv \dot{v} \equiv a(t) = -\omega^2 \, x_0 \cos(\omega t) .$$

Note that the right-hand side of the last equation is just $-\omega^2$ times our original formula for x(t), so we can also write

$$\ddot{x} = -\omega^2 \; x .$$ (12.6)

Multiplying through both sides by the mass m of the object in motion gives

$$m a = F = -m \omega^2 \; x ,$$

which ought to look familiar to you: it is just HOOKE'S LAW with a force constant   $k = m \omega^2$. Rearranging this a little gives

$$\omega = \sqrt{k/m} ,$$

which may also look familiar . . . . More on this later. Note, however, that we can very easily deduce what is going on in this situation, including the type of force being applied, just from knowing  x(t). That's why we think of it as ``the solution.''