Chapter 4.8, Problem 52E

a.

To determine

Find the maximum displacement.

Answer to Problem 52E

  $\boxed{a=\frac{1}{2}}$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function. $d=\frac{1}{2}\cos 20\pi t$ .

Find the maximum displacement.

Calculation:

Consider the given equation

  $d=\frac{1}{2}\cos 20\pi t$ Here $a=\frac{1}{2},\omega =20\pi$

The maximum displacement from the point of equilibrium is amplitude which is the coefficient of $\cos$ is equal to $a=\frac{1}{2}$ .

Hence, The maximum displacement is $\boxed{a=\frac{1}{2}}$

b.

To determine

Find the frequency.

Answer to Problem 52E

  $\boxed{10cycles/unit,time}$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function. $d=\frac{1}{2}\cos 20\pi t$ .

Find the frequency.

Calculation:

Consider the given equation

  $d=\frac{1}{2}\cos 20\pi t$ Here $a=\frac{1}{2},\omega =20\pi$

  $\begin{array}{l}f=\frac{\omega }{2\pi }\\ =\frac{20\pi }{2\pi }\\ f=\boxed{10cycles/unit,time}\end{array}$

Hence, the frequency is $\boxed{10cycles/unit,time}$

c.

To determine

Find the value of $d$ when $t=5$ .

Answer to Problem 52E

  $\boxed{0.5}units$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function. $d=\frac{1}{2}\cos 20\pi t$ .

The value of $d$ when $t=5$ ,

Calculation:

Consider the given equation

  $d=\frac{1}{2}\cos 20\pi t$ put $t=5$

  $\begin{array}{l}d=\frac{1}{2}\cos 20\pi \times 5\\ d=\frac{1}{2}\cos 100\pi \\ d=0.5\end{array}$

Hence, the value of $d$ when $t=5$ , $\boxed{0.5}units$

d.

To determine

The least positive value of $t$ for which $d=0$ and verify the results by using graphing utility.

Answer to Problem 52E

  $\boxed{t=0.025}$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function. $d=\frac{1}{2}\cos 20\pi t$ .

The least positive value of $t$ for which $d=0$ . Use a graphing utility to verify your results.

Calculation:

Consider the given equation

  $d=\frac{1}{2}\cos 20\pi t$ put $d=0$

  $\begin{array}{l}d=\frac{1}{2}\cos 20\pi t\\ 0=\frac{1}{2}\cos 20\pi t\\ 0=\cos 20\pi t\\ \sin ce,{\cos }^{-1}\left(0\right)=\left(n+1\right)\frac{\pi }{2}\\ 20\pi t=\left(n+1\right)\frac{\pi }{2}\end{array}$

For least value of $t$ put $n=0$ ,

  $\begin{array}{l}20\pi t=\frac{\pi }{2}\\ t=\frac{\pi }{2}\times \frac{1}{20\pi }=\frac{1}{40}\\ t=0.025\end{array}$

Hence, the least positive value of $t$ for which $d=0$ is $\boxed{t=0.025}$ .

Now use a graphing utility to verify your results.

Use $TI83$ calculator,

Press $Y=$ button and enter the following equation,

  $Y_1=0.5\cos \left(20\pi t\right)$ the display is as shown,

  

Press window button and set window parameter.

  

Press the TRACE button and view the graph of the function,

  

Now press 2^nd and CALC and select option $\text{'}\text{'}4\text{'}\text{'}$ and press enter

  

Hence when $t=5thend=0$ .

Now press 2^nd and CALC and select option $\text{'}\text{'}1\text{'}\text{'}$ and enter the value $X=5$ press enter

  

Hence, when $t=5thend=0.5$

Now press 2^nd and CALC and select option $\text{'}\text{'}2\text{'}\text{'}$ and enter the value $X=0.025ifY=0$ press enter

  

Hence, when $t=0.025then,d=0$ .