Chapter 4.8, Problem 62E

To determine

To find: The maximum displacement of $d=\frac{1}{2}\cos 20\pi t$ .

Answer to Problem 62E

The maximum displacement of $d=\frac{1}{2}\cos 20\pi t$ is $\frac{1}{2}$ .

Explanation of Solution

Given Information: It is given that the simple harmonic motion is described by the trigonometric function $d=\frac{1}{2}\cos 20\pi t$ .

The given trigonometric equation is $d=\frac{1}{2}\cos 20\pi t$

Comparing it with $d=a\cos wt$

The maximum displacement is the amplitude $\left|a\right|=\frac{1}{2}$

To determine

To find: The frequency of function $d=\frac{1}{2}\cos 20\pi t$ .

Answer to Problem 62E

The frequency of function is 10.

Explanation of Solution

Given Information: It is given that the simple harmonic motion is described by the trigonometric function $d=\frac{1}{2}\cos 20\pi t$ .

The given trigonometric equation is $d=\frac{1}{2}\cos 20\pi t$ , compare it with $d=a\cos wt$ .

The frequency is the reciprocal of the period $\frac{w}{2\pi }=\frac{20\pi }{2\pi }=10$ cycle per unit of time.

If the time is in seconds, then frequency = 10 hertz.

To determine

To find: The value of d when t = 5.

Answer to Problem 62E

The value of d is 0.5.

Explanation of Solution

Given Information: It is given that the simple harmonic motion is described by the trigonometric function $d=\frac{1}{2}\cos 20\pi t$ .

The given trigonometric equation is $d=\frac{1}{2}\cos 20\pi t$ , compare it with $d=a\cos wt$ ; and put t = 5

  $\begin{array}{l}d=\frac{1}{2}\cos [20\pi (5)]\\ =\frac{1}{2}\cos [100\pi ]=0.5\end{array}$

Even multiple $\pi$ have reference angle 0.

To determine

To find: The least positive value of t for which d = 0

Answer to Problem 62E

The least positive value of t is 0.025.

Explanation of Solution

Given Information: It is given that the simple harmonic motion is described by the trigonometric function $d=\frac{1}{2}\cos 20\pi t$ .

The cos will be zero if the angle = $\frac{\pi }{2}$

You can count 10 cycles from t = 0 to 1.