How do you prove that a function shows that something is in SHM if the equation isn't in the form:
x = Acos (kt + b)
Is a graph would be acceptable or is there a mathematical way to do this?
(Edited version)
The equation you have written is one way to define harmonic motion. So if your trial function cannot be put into this form, then it does not represent a SHO.
On the other hand you can define simple harmonic motion as $\displaystyle x"(t) = -Ax(t)$. By taking small displacements from equilibrium for many physical systems you will find it is a SHO. (Usually you need to do a Taylor expansion to prove this. The linear coefficients in the series needs to be 0 and the quadratic coefficient needs to be negative. The condition on the linear term is automatic around an equilibrium point, so many systems have this kind of behavior.)
For your function
$\displaystyle sin(t) + A~cos(t)$
you can easily see that it is the (linear) sum of two SHO functions, so it is also a harmonic oscillator.
-Dan
$\displaystyle x = sin(t) + A~cos(t)$
$\displaystyle x' = cos(t) - A~sin(t)$
$\displaystyle x" = -sin(t) - A~cos(t) = -x$
So this is a harmonic oscillator.
Or
$\displaystyle sin(t) + A~cos(t) = \sqrt{1 + A^2}~sin(t + \phi)$
(where $\displaystyle \phi$ is a constant) which is a sine function and hence a harmonic oscillator.
(This is from the more general:
$\displaystyle a~sin(t) + b~cos(t) = \sqrt{a^2 + b^2}~sin(x + \phi)$
where
$\displaystyle \phi = tan^{-1} \left ( \frac{b}{a} \right )$
for $\displaystyle a \geq 0$
and
$\displaystyle \phi = tan^{-1} \left ( \frac{b}{a} \right ) + \pi$
for $\displaystyle a < 0$)
-Dan