Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 76

Suppose that the equation of motion of a particle is given by $s = A \cos (\omega t + \delta)$, the particle is said to undergo simple harmonic motion.

a.) Determine the velocity of the particle at time $t$.

Recall that velocity = $s'(t)$ so...


$
\begin{equation}
\begin{aligned}

\text{velocity } = s'(t) =& A \frac{d}{d(\omega t + \delta)} [\cos (\omega t + \delta)] \cdot \frac{d}{dt} (wt + \delta)
\\
\\
\text{velocity } = s'(t) =& A(- \sin (\omega t + \delta)) (\omega t + \delta)
\\
\\
\text{velocity } = s'(t) =& -A \omega \sin (\omega + \delta)

\end{aligned}
\end{equation}
$



b.) At what time is the velocity 0?

The velocity is zero when...


$
\begin{equation}
\begin{aligned}

0 =& - \cancel{A} \omega \sin (\omega t + \delta)
\\
\\
\omega t + \delta =& \sin^{-1} [0]
\\
\\
\omega t + \delta =& n \pi \qquad ; \text{where $n \pi$ corresponds to succeeding periods and $n$ is an integer}
\\
\\
\omega t =& n \pi - \delta
\\
\\
t =& \frac{n \pi - \delta}{\omega} \qquad ; \text{where $n$ is any integer}

\end{aligned}
\end{equation}
$