Chapter 4.8, Problem 50E

To determine

To write: an equation that describes the motion of the buoy where the high point corresponds to the time $t=0$ .

Answer to Problem 50E

The equation that describes the motion of the buoy is

  $d=1.75\cos \left(0.2\pi \omega t\right)$

Explanation of Solution

Given information:

The buoy moves a total of 3.5 feet from its low point to its high point and it returns to its high point every 10 seconds.

Given picture,

  

Calculation:

Consider the following figure

  

A point that moves on a coordintate is said to be in simple Harmonic Motion if its distance $d$ from the origin at time $t$ is given by the equations

  $d=a\sin \omega tord=a\cos \omega t$

Where $a$ and $\omega$ are real numbers such that $\omega \ne 0$

The motion has amplitude $\left|a\right|$ , period $\frac{2\pi }{\omega }$ and frequency $\frac{\omega }{2\pi }$ .

Since the buoy at its high point, when $t=0$ , use the equation

  $d=a\cos \omega t$

Since the distance between highest and lowest point is 3.5 feet

Therefore

  $\begin{array}{c}a=\frac{3.5}{2}\\ =1.75\end{array}$

Since the buoy returns to the highest point every 10 seconds, period is $T=10$

That is

  $\begin{array}{c}\frac{2\pi }{\omega }=10\\ \omega =\frac{2\pi }{10}\\ =0.2\pi \end{array}$

Thus the equation that describes the motion of the buoy is

  $d=1.75\cos \left(0.2\pi \omega t\right)$