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}$