Chapter 5, Problem 72RE

a.

To determine

Find a function of the form $y=k{e}^{-ct}\cos \omega t$ to model the motion.

Answer to Problem 72RE

  $y=16e^{-0.72t}\cos 2.8\pi t$ .

Explanation of Solution

Calculation:

As we know that the standard equations of object in damped harmonic motion are,

  $y=k{e}^{-ct}\cos \omega t,y=k{e}^{-ct}\sin \omega t$ in which,

  $k=$ initial amplitude, the maximum displacement of the object from its equilibrium position,

  $p=2\pi /\omega$ is the period.

  $f=\omega /2\pi$ number of cycles per unit time (frequency).

  $c=$ Damping constant.

Let considr the given variables,

  $\begin{array}{l}k=16cm\\ \omega =2\pi \left(1.4\right)\\ \omega =2.8\pi \\ c=0.72\end{array}$

Now substitute the given values in standard equation we get,

  $y=16e^{-0.72t}\cos 2.8\pi t$

Hence, the equation is , $y=16e^{-0.72t}\cos 2.8\pi t$ .

b.

To determine

Graph the function you found in part $\left(a\right)$ .

Answer to Problem 72RE

  

Explanation of Solution

Calculation:

Consider the function,

  $y=16e^{-0.72t}\cos 2.8\pi t$

Initially enter the function into the calculator,

  

Use the window key to set the appropriate ranges of the variables,

  

The graph of the function is,

  

Hence, the function is plotted.

c.

To determine

Find the displacement at time $t=10s$ .

Answer to Problem 72RE

  $y=0.0119cm$ .

Explanation of Solution

Calculation:

Consider the function,

  $y=16e^{-0.72t}\cos 2.8\pi t$

Substitute $t=10s$ in the equation, we get.

  $\begin{array}{l}y=16e^{-0.72\left(10\right)}\cos 2.8\pi \left(10\right)\\ y=16e^{-7.2}\cos 28\pi \\ y=0.0119cm\end{array}$

Hence, the displacement at time $t=10s$ is $y=0.0119cm$ .