# A few examples from the kinematics

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It is quite common that we know the rate of change of something, but not the ting itself.  Say for example we know that the acceleration is a constant, a. We know that the acceleration is the rate of change of the rate of change of position, i.e that $s'' = a$

where s is the distance (position) and the second derivative is the derivative with respect to time twice.

The derivative of the position (the rate of change of the distance), so the antiderivative of the above will give us the velocity. We get $s' = v = at + C$

But the constant must be the initial velocity (put t=0) so we have $s' = v = at + {v_0}$

A second antiderivative gives us $s = \frac{{a{t^2}}}{2} + {v_0}t + {s_0}$

Here we could reason that the constant must be the initial position.

Simple harmonic motion

Say we have a mass hanging in a spring that obeys Hooke´s law, $F = - k\Delta y$

The total force will thus be $\begin{gathered} F = - mg - k\Delta y = - mg - k(y - {y_0}) \hfill \\ \quad = - mg - ky + k{y_0} = - ky \hfill \\ \end{gathered}$

Here up is in the positive direction and y0 is the height at which the spring force and gravity balance each other (the equilibrium point).

We have that $F = ma = my'' = - ky$

or $F = ma = y'' = - \frac{k}{m}y$

We thus want to find a function such that its second derivative is (except for a constant factor) equal to the negative of itself. We have that $\frac{{{d^2}}}{{d{t^2}}}\sin (t) = (\sin (t))'' = - \sin (t)$

To make it work we may multiply t with a constant, ω, called the angular velocity. This gives us $(\sin (\omega t))'' = - {\omega ^2}\sin (\omega t)$

So with $y = \sin (\omega t)$

substituted into our original equation we get $y'' = - \frac{k}{m}y = - {\omega ^2}\sin (\omega t) = - \frac{k}{m}\sin (\omega t)$

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

And thus that $y = \sin \left( {\sqrt {\frac{k}{m}} t} \right)$

This is a particular solution, and it tells us that the motion of the mass can be like a sine wave. This we have derived from basic physics. The general solution is $= A\sin \left( {\sqrt {\frac{k}{m}} t - \theta } \right)$

where A is the amplitude of the motion, and θ is the so called phase angle.  You can verify that this is a solution yourself by finding the second derivative of the above and verify that it satisfies the original differential equation. Up a level : Differential Equations
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