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endlesslove230 · Oct 17, 2011, 07:15 AM

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:

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

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

ebaines · Oct 17, 2011, 10:39 AM

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?

endlesslove230 · Oct 17, 2011, 11:08 AM

Yes, thank you.