How to find equation of simple harmonic motion from positional information at 3 different times? Given a particle at three distinct position $x_1, x_2 \ and \ x_3 $from equilibrium position at different times $ t_1, t_2 \ and \ t_3 $ how can we find the amplitude, frequency and initial phase?
It seems to me that I lack the mathematical knowledge necessary to solve the equations involved. What are such equations called and where can we learn about solving them?
$
x_1 = A \sin(\omega \times t_1) + B \cos(\omega \times t_1)$
$x_2 = A \sin(\omega \times t_2) + B \cos(\omega \times t_2)$
$x_3 = A \sin(\omega \times t_3) + B \cos(\omega \times t_3)$
where $ \omega \times t_1 , \ \omega \times t_2, \omega \times t_3 $ is less than $ 2\pi$ but greater than 0.
 A: Apologies I am ignorant of the nomenclature and sources. If you were interested in a quick homing in and won't falter at issues of uniqueness and optimization of solutions, here is a direct way to get you where you might like to go.
First rearrange your real coefficients to polar form,
$$
A\equiv r \cos \phi , \qquad B\equiv r \sin\phi ,\\
r=\sqrt{A^2+B^2} , \qquad \phi=\arctan (B/A).
$$
It follows that your three equations are transcribable as 
$$
x_i/r=\sin (\omega t_i+\phi),
$$
whence 
$$
y_i\equiv \arcsin (x_i/r)= \omega t_i+\phi ~.
$$
One may eliminate the unknowns $\phi,\omega$ among these three equations to consider the (transcendental ?) equation 
$$
\frac{t_2-t_1}{t_3-t_1}= \frac{y_2-y_1}{y_3-y_1},
$$
where the only unknown is r on the r.h.s. Graphing the implied rhs-lhs equation versus r should yield allowable rs from its zeroes. (I gather there are more efficient ways to do this...)
Given r, you may proceed to determine
$$
\omega=\frac{y_2-y_1}{t_2-t_1}, \qquad \phi=y_1-\omega t_1,
$$
where the actual values of the 6 inputs might suggest better choices of i or pairs thereof for these last two equations. You then plug in your variables to determine A,B.
It is possible you might experiment with toy values such as 
$$
t_i=(\pi/8,3\pi/8,11\pi/8) ~sec, \qquad   x_i=(0.71,1,-1)~cm
$$
to obtain
$$
\phi=\pi/8, \qquad r=1 ~cm,  \qquad \omega=1~sec^{-1}.
$$
