Chapter 4.8, Problem 53E

a.

To determine

Find the maximum displacement.

Answer to Problem 53E

  $\boxed{a=\frac{1}{4}}$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function $d=\frac{1}{4}\sin 6\pi t$ .

Find the maximum displacement.

Calculation:

Consider the given equation

  $d=\frac{1}{4}\sin 6\pi t$ Here $a=\frac{1}{4},\omega =6\pi$

The maximum displacement from the point of equilibrium is amplitude which is the coefficient of $\cos$ is equal to $a=\frac{1}{4}$ .

Hence, the maximum displacement is $\boxed{a=\frac{1}{4}}$

b.

To determine

Find the frequency.

Answer to Problem 53E

  $\boxed{10cycles/unit,time}$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function. $d=\frac{1}{4}\sin 6\pi t$ .

Find the frequency.

Calculation:

Consider the given equation

  $d=\frac{1}{4}\sin 6\pi t$ Here $a=\frac{1}{4},\omega =6\pi$

  $\begin{array}{l}f=\frac{\omega }{2\pi }\\ =\frac{6\pi }{2\pi }\\ f=\boxed{3cycles/unit,time}\end{array}$

Hence, the frequency is $\boxed{3cycles/unit,time}$ .

c.

To determine

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

Answer to Problem 53E

  $\boxed{0.5}units$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function. $d=\frac{1}{4}\sin 6\pi t$ .

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

Calculation:

Consider the given equation

  $d=\frac{1}{4}\sin 6\pi t$ put $t=5$

  $\begin{array}{l}d=\frac{1}{4}\sin 6\pi \times 5\\ d=\frac{1}{4}\sin 30\pi \\ d=0\end{array}$

Hence, the value of $d$ when $t=5$ , $\boxed{d=0}$ .

d.

To determine

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

Answer to Problem 53E

  $\boxed{t=\frac{1}{6}}$

Explanation of Solution

Given information:

For the simple harmonic motion described by the trigonometric function $d=\frac{1}{4}\sin 6\pi t$ .

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 $d=0$

  $\begin{array}{l}d=\frac{1}{4}\sin 6\pi t=0\\ \Rightarrow \sin 6\pi t=0\\ 6\pi t=\pi ,2\pi ,3\pi \mathrm{.....}\left[\therefore t=\frac{1}{6},\frac{1}{3},\frac{1}{2}\right]\end{array}$

Hence, the least positive value of $t$ for which $d=0$ is $\boxed{t=\frac{1}{6}}$ .

Now use a graphing utility to verify your results.

For part $\left(a\right)$

  $a=\frac{1}{4}$

  

From the graph $frequency=\frac{1}{0.33}=3cycle/unit,time$ from part $\left(b\right)$ is also verified.

  $t=5,d=0$ Part $\left(c\right)$ is also verified.

  $t=\frac{1}{6}$ Part $\left(d\right)$ is also verified.