Problem 12.6

12.6. a)

We know that the solution must take the form of
$x(t) = A \cos \omega t + B \sin \omega t$
where And B are constants that we have to determine.
If we substitute in the value t=0, we find
x(0) = x₀ = A
so, immediately we know that constant. Now, let's take a derivative of our equation.
$v(t) = -A\omega \sin \omega t + B\omega \cos \omega t$
Again, if we substitute the time t=0, we find
$v(0) = v_{0} = B\omega$
or that
$B = \frac{v_{0}}{\omega}$
substituting again, we get that our equations for x(t) and v(t) reduce to what we were looking for
$x(t) = x_{0}\cos \omega t + (\frac{v_{0}}{\omega})\sin \omega t$
$v(t) = -x_{0}\omega \sin \omega t + v_{0} \cos \omega t$

b)
Now, we are trying to show the identity
v²-ax = 0
We have x(t) and v(t) from part a), now, to find a(t), we differentiate.
$a(t) = -x_{0}\omega^{2} \cos \omega t - v_{0}\omega \sin \omega t$
substituting into the equation, and using the identity that
$sin^{2} x + \cos^{2} x = 1$
we find that
$v^{2} - ax = x_{0}^{2} \omega^{2} + v_{0}^{2}$
Well, since this has no time dependence, we can evaluate this at any time and the answer is independent of that time. Therefore,
v² - ax = v₀² - a₀² x₀²
Now, let's evaluate this at a maximum in position. In other words, at a turning point of the osciallator, then we find
v = 0,   x = A
Then the equation reduces to
$v^{2} - ax = x_{0}^{2} \omega^{2} + v_{0}^{2} ~=~ A^{2}\omega^{2} + 0^{2}$
or $v^{2} - ax = A^{2}\omega^{2}$