Chapter 4, Problem 96RE

To determine

The equation which describe the motion of the bobber where the high point corresponds to the time t =0.

Answer to Problem 96RE

The equation of the motion is given as:

  $d=0.75\cos \left(\frac{2\pi }{3}t\right)$

Explanation of Solution

Given:

The fishing bobber oscillate in the simple harmonic motion because of the wave in the lake. The bobber moves total of 1.5 inches from its low point to the high point and return to its high point after every 3 seconds.

When the point moving on the coordinate is known as the simple harmonic motion, it the distance d from the origin at the time t is give as:

  $\begin{array}{l}d=a\sin \omega t\text{or }d=a\cos \omega t\\ \text{where,}\\ a\text{and }\omega \text{ are the real numbers}\end{array}$

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

Since, the bobber is at high point, t =0, using the equation:

  $d=a\cos \omega t$

The distance between the highest point and the lowest point is 1.5 in.

  $\begin{array}{c}a=\frac{1.5}{2}\\ =0.75\text{in}\end{array}$

When the bobber return to the highest point every 3 second, the time period will be 3s.

Hence,

  $\begin{array}{c}T=\frac{2\pi }{\omega }\\ =3\\ \omega =\frac{2\pi }{3}\text{rad/s}\end{array}$

Therefore, the equation of the motion is given as:

  $d=0.75\cos \left(\frac{2\pi }{3}t\right)$