Finding velocity from equilibrium

Question

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 ?

Answer 1

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.