Chapter 6.4, Problem 56E

(a)

To determine

To Find: The equation of the motion.

Answer to Problem 56E

  $y=-1.5\sin \frac{\pi }{4}t$

Explanation of Solution

Given:Equilibrium point is at $t=0$ and buoy on its way down at a time.

Given: $t=0$

  $\begin{array}{l}\left|A\right|=\left|\frac{A-(-A)}{2}\right|\frac{2\pi }{k}=8,k=\frac{\pi }{4}\\ \left|A\right|=\left|\frac{3}{2}\right|\\ \left|A\right|=\pm 1.5\\ downfirst\\ thus\\ A=-1.5\end{array}$

Thus

The equation of the motion is $y=-1.5\sin \frac{\pi }{4}t$ .

(b)

To determine

To Find: The height of the buoy at $3\sec$ .

Answer to Problem 56E

  $-1.1ft$

Explanation of Solution

Given:Equilibrium point is at $t=0$ and buoy on its way down at a time.

Given:For $3\sec$

  $t=3$

The equation of the motion is $y=-1.5\sin \frac{\pi }{4}t$ .

  $\begin{array}{l}y=-1.5\sin \frac{\pi }{4}t\\ y=-1.5\sin \frac{\pi }{4}(3)\\ y=-1.1ft\end{array}$

Hence, the height is nearly equal to $-1.1ft$ .

(c)

To determine

To Find: The height of the buoy at $12\sec$ .

Answer to Problem 56E

  $0ft$

Explanation of Solution

Given:Equilibrium point is at $t=0$ and buoy on its way down at a time.

Given:For $12\sec$

  $t=12$

The equation of the motion is $y=-1.5\sin \frac{\pi }{4}t$ .

  $\begin{array}{l}y=-1.5\sin \frac{\pi }{4}t\\ y=-1.5\sin \frac{\pi }{4}(12)\\ y=0ft\end{array}$

Hence, the height is equal to $0ft$ .