Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 35

Suppose that a mass on a spring vibrates horizontally on a smoooth level surface. Its equation of motion is $x(t) = 8 \sin t$. where $t$ is in seconds and $x$ in centimeters.
a.) Determine the velocity and acceleration at time $t$
b.) Find the position, velocity and acceleration of the mass at time $\displaystyle t= \frac{2\pi}{3}$. In what direction is it moving at that time?


a.) $x(t) = 8 \sin t$

$
\begin{equation}
\begin{aligned}
\text{velocity } = x'(t) &= 8 \frac{d}{dt} \sin t\\
&= 8 \cos t (1)\\
&= 8 \cos t
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\text{acceleration} = x''(t) &= 8 \frac{d}{dt} \cos t\\
&= 8 (-\sin t) (1)\\
&= -8 \sin t
\end{aligned}
\end{equation}
$


b.) The position at $\displaystyle t = \frac{2\pi}{3} \text{ is } x \left( \frac{2\pi}{3}\right) = 8 \sin \left( \frac{2\pi}{3}\right) = 4 \sqrt{3} cm$

The velocity at $\displaystyle t = \frac{2 \pi}{3} \text{ is } x' \left( \frac{2\pi}{3}\right) = 8 \cos \left( \frac{2\pi}{3}\right) = -4 \frac{cm}{s}$
A negative velocity indicates that the mass on the spring is vibrating to the left assuming that the direction to right has a positive velocity.
The acceleration of $\displaystyle t = \frac{2 \pi}{3} \text{ is } x''\left( \frac{2 \pi}{3} \right) = - 8 \sin \left( \frac{2\pi}{3} \right) = -4\sqrt{3} \frac{cm}{s}$
A negative acceleration indicates that the spring has decreasing acceleration at $\displaystyle t = \frac{2\pi}{3}$