An object performs a harmonic motion. If the values of the deflections are x1=2 [cm], x2=3 [cm] and the velocities are v1=5 [m/s], v2=4 [m/s], calculate the amplitude and the angular frequency of the harmonic oscillator.

Here is what I solved so far:
x_1=A_1 cos \omega t_1
x_2=A_2 cos \omega t_2
v_1=\omega A_1 cos \omega t_1 = \omega x_1
v_2=\omega A_2 cos \omega t_2 = \omega x_2

What should I do next to reach:

A = sqrt(\frac{(x_2)^2 (v_1)^2 - (v_2)^2 (x_1)^2}{(v_1)^2 - (v_2)^2})

\omega = sqrt(\frac{(v_1)^2 - (v2)^2}{(x_2)^2 -(x1)^2}) ?

Fist off your equations aren't quite right, for several reasons:

1. Your first equation makes an assumption that x_1 = 0 at t= 0, but unless you've been told this you don't know it. The correct equation is:


x(t)= A \cos(\omega t + \phi)


The \phi constant need to be included in all four equations.

2. The constant for the amplitude should be same for all equations. Call it A,; there is no A_1 or A_2.

3. The velocity is the derivative of the position function, so:

v(t) = \frac {dx} {dt} = A \omega (sin \omega t + \phi)


Once you make these changes you can make use of the trigonometry identity sin^2 (\omega t + \phi) + \cos ^2 (\omega t + \phi) = 1.

Hence:


v_1^2 + (x_1 \omega)^2 = 1 = v_2 ^2 + (x_2 \omega)^2


Can you take it from here?

Yes, thank you.