Chapter 5.3, Problem 89E

To determine

To find: the times when the weight is at the point of equilibrium $\left(y=0\right)$ for $0\le t\le 1$ .

Answer to Problem 89E

The times when the body will be in equilibrium is given by,

  $t=0.0402+\frac{n\pi }{8}$ .

Explanation of Solution

Given information:

A weight is oscillating on the end of a spring. The displacement from equilibrium of the weight relatives to the point of equilibrium is given by

  $y=\frac{1}{12}\left(\cos 8t-3\sin 8t\right)$

Where $y$ is the displacement and $t$ is the time.

  

Calculation:

Let us consider the following figure

  

Let us consider that,

  $y=\frac{1}{12}\left(\cos 8t-3\sin 8t\right)$

Here $y$ is the distance in meters and $t$ is the time in seconds.

Let us find the equilibrium posting by equating $y=0$ .

  $\begin{array}{c}\frac{1}{12}\left(\cos 8t-3\sin 8t\right)=0\\ \cos 8t-3\sin 8t=0\\ \frac{\sin 8t}{\cos 8t}=\frac{1}{3}\\ \tan 8t=\frac{1}{3}\end{array}$

Hence $t$ for equilibrium is given by,

  $\begin{array}{c}t=\frac{1}{8}\arctan \frac{1}{3}\\ =0.0402\end{array}$

Since, the period of $\tan 8t$ is given by $\frac{\pi }{8}$ ,

Hence, the times when the body will be in equilibrium is given by,

  $t=0.0402+\frac{n\pi }{8}$

Here, $n$ is always a positive integer.