Finding velocity from equilibrium I have the question "A mass is oscillating vertically on a spring, with a maximum amplitude of 10.0 cm. If it has a frequency of oscillation of 5.0 Hz. 
Find its velocity at 5.0 cm from equilibrium."
Here is my attempt. Is this correct ? 


For a harmonic oscillator, the total energy $E$ is the following:
$$E={1\over2}mv^2+{1\over2}m\omega^2x^2$$
so $$v=\pm\sqrt{{2E\over m}-\omega^2x^2}$$
Now we know that $v=0$ when $x=A$, i.e. at the points of max amplitude, hence ${2E\over m}=\omega^2A^2$ and we get:
$$v=\pm\sqrt{\omega^2A^2-\omega^2x^2}=\pm\omega\sqrt{A^2-x^2}$$
that is your formula.
Replacing the data of the problem we get $v=\pm2,720699046$ m/s $\approx2,72$ m/s.
