Chapter 4.8, Problem 53E

To determine

To Find:

1. The maximum displacement
2. The frequency
3. Value of $d$ when $t=5$ and
4. The least positive value of $t$ for which $d=0$

Answer to Problem 53E

1. The maximum displacement will be when t=0, therefore, $d=9.\cos \frac{6\pi }{5}.0,d=9$
2. The frequency will be of $d=9.\cos \frac{6\pi }{5}.0,d=9$ is w, therefore

  $d=9.\cos \frac{3}{5}.2\pi t,\omega =\frac{3}{5}$

3. The value of $d$ when $t=5$ is, $d=9.\cos \frac{6\pi }{5}.5=9.\cos 6\pi =9$
4. The least value of $t$ when $d=0$

  $\begin{array}{l}\cos \frac{6\pi }{t}=0\\ \frac{6\pi }{t}={\cos }^{-1}0\\ \frac{6\pi }{t}=\frac{\pi }{2}\\ t=\frac{5}{12}\end{array}$

Explanation of Solution

Given information:

For simple harmonic motion $d=9.\frac{\cos 6\pi }{5}t$

To find under the categorized as follows:

1. The maximum displacement
2. The frequency
3. Value of $d$ when $t=5$
4. The least positive value of $t$ for which $d=0$

Formula used:

  $\begin{array}{l}frequency=\omega \\ d=a.\cos \omega t\end{array}$

Maximum displacement is when t=0

Calculation:

Consider the simple harmonic motion:

  $\begin{array}{l}d=a.\cos \omega t\\ d=9.\cos \frac{6\pi }{5}t\end{array}$

Recall that, for maximum displacement always $t=0$

Therefore,

  $\begin{array}{l}d=9.\cos \frac{6\pi }{5}.0\\ d=9\end{array}$

Maximum displacement $\boxed{d=9}$

Recall that for frequency is $\omega$

Therefore,

  $\begin{array}{l}d=9.\cos \frac{3}{5}.2\pi .t\\ \boxed{\omega =\frac{3}{5}}\end{array}$

Recall that the value $d$ of when $t=5$

Therefor,

  $\begin{array}{l}d=9.\cos \frac{6\pi }{5}.5\\ d=9.\cos 6\pi \\ \boxed{d=9}\end{array}$

Recall the least positive value of tis at $d=0$

  $\begin{array}{l}\cos \frac{6\pi }{t}=0\\ \frac{6\pi }{t}={\cos }^{-1}0\\ \frac{6\pi }{t}=\frac{\pi }{2}\\ \boxed{t=\frac{5}{12}}\end{array}$