Finding velocity when mass is $8.0$ cm from equilibrium I have the question "A mass is tethered between two springs. It is displaced by $10$ cm then released. The mass oscillates with $SHM$ with a frequency of $0.55$ Hz. Calculate the velocity when the mass is $8.0$ cm from the equilibrium position."
I used the formula 
Velocity = total displacement / total time.
$$V = 0.08 / 1.82$$
And the answer I get is $V = 0.04$ m/s.
Is this correct ? 
 A: This is simple harmonic motion.
Hence, the displacement is given by the equation:
$$s=A \cos{(\omega t)} \tag{1}$$
Where $A$ is the amplitude, and $\omega$ is the angular frequency.
If we differentiate both sides with respect to $t$, we obtain:
$$v=-A \omega \sin{(\omega t)} \tag{2}$$
Where $v$ is the velocity.
Reminder:
$$\omega = 2\pi f=1.1\pi$$
Where $f$ is the frequency.
We evaluate a time $t$ where the displacement $s=8 \text{ cm}=0.08 \text{ m}$ using equation $(1)$.
$$0.08=0.1\cos{(\omega t)}$$
$$\frac{4}{5}=\cos{(1.1 \pi \cdot t)}$$
$$t=\frac{\arccos\left(\frac{4}{5}\right)}{1.1\pi}$$
One solution is:
$$t \approx 0.186212 \text{ s}$$
We do not need to know all solutions, since for each of these solutions, the velocity is the same (Simple Harmonic Motion).
We now can substitute this value of $t$ into our expression for the velocity $v$ on equation $(2)$.
$$v=-0.1\cdot 1.1\pi \cdot \sin(1.1\pi \cdot t) \approx -0.207345 \text{ ms}^{-1}$$
Hence, the velocity when displaced by $8.0 \text{ cm}$ from the equilibrium position is:
$$|v| \approx 0.21 \text{ ms}^{-1}$$
