Chapter 13.5, Problem 76SR

a.

To determine

To write: an equation for the motion of the buoy. Assume that it is at equilibrium at t = 0 and that it is on the way up from the normal water level.

Answer to Problem 76SR

The equation for the motion of the buoy: $y=2\sin \frac{\pi }{5}t$ .

Explanation of Solution

Given information: A harbor buoy bobs up and down with the waves. The distance

between the highest and lowest points is 4 feet. The buoy moves from its highest point to its lowest point and back to its highest point every 10 seconds.

Calculation:

The motion of the buoy can be modeled by the equation:

  $\begin{array}{l}y=A\sin \omega t\\ A=\frac{4}{2}=2.\\ T=10.\\ T=\frac{2\pi }{\omega }\\ \omega =\frac{2\pi }{T}=\frac{2\pi }{10}=\frac{\pi }{5}\\ y=2\sin \frac{\pi }{5}t.\end{array}$

Therefore, the equation for the motion of the buoy: $y=2\sin \frac{\pi }{5}t$ .

b.

To determine

To draw: a graph showing the height of the buoy as a function of time.

Answer to Problem 76SR

Explanation of Solution

Given information: A harbor buoy bobs up and down with the waves. The distance

between the highest and lowest points is 4 feet. The buoy moves from its highest point to its lowest point and back to its highest point every 10 seconds.

Calculation:

The graph of the equation for the motion of the buoy: $y=2\sin \frac{\pi }{5}t$ is shown below.

  

c.

To determine

To find: the height of the buoy after 12 seconds.

Answer to Problem 76SR

The height of the buoy after 12 seconds is $y\approx 1.9\text{feet}.$

Explanation of Solution

Given information: A harbor buoy bobs up and down with the waves. The distance

between the highest and lowest points is 4 feet. The buoy moves from its highest point to its lowest point and back to its highest point every 10 seconds.

Calculation:

The equation for the motion of the buoy: $y=2\sin \frac{\pi }{5}t$

  $\begin{array}{c}y=2\sin \frac{\pi }{5}t.\\ t=12\\ y=2\sin \frac{\pi }{5}\times 12=2.\sin 7.5360=2\times 0.94986839\approx 1.9\\ y\approx 1.9\text{feet}.\end{array}$

Therefore, the height of the buoy after 12 seconds is $y\approx 1.9\text{feet}.$