Chapter 4, Problem 189RE

To determine

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

Answer to Problem 189RE

The equation of motion of the buoy is $d=6\cos \frac{2\pi }{15}t$ .

Explanation of Solution

Given information:

A buoy oscillates in simple harmonic motion as waves go past. The buoy moves a total of $6$ feet from its high point to its low point, returning to its high point every $15$ seconds.

Calculation:

Simple harmonic motion of a body is given by

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

Total distance that buoy moves from its high point to its low point : $\left|a\right|=6$

Total time that buoy takes to returns to its high point is $15$ seconds.

  $\begin{array}{l}T=\frac{2\pi }{\omega }\\ 15=\frac{2\pi }{\omega }\\ \omega =\frac{2\pi }{15}\end{array}$

The equation contain cosine term since the high point of the buoy is at $t=0$

Therefore, equation of motion of the buoy is given by

  $d=6\cos \frac{2\pi }{15}t$