Chapter 4.2, Problem 50E

(a)

To determine

To find: The value to complete the table.

Answer to Problem 50E

The value for function at each time is calculated.

Explanation of Solution

Given information:

The damping effect of friction is given by $y\left(t\right)=\frac{1}{4}{e}^{-t}\cos 6t$ .

Calculation:

For the time $\text{t=0}$ .

  $\begin{array}{c}y\left(0\right)=\frac{1}{4}{e}^{-0}\cos \left(6\times 0\right)\\ =\frac{1}{4}\cos 0\\ =\frac{1}{4}\times 1\\ =0\cdot 25\end{array}$

For the time $\text{t=}\frac{1}{4}$ .

  $\begin{array}{c}y\left(\frac{1}{4}\right)=\frac{1}{4}{e}^{-\frac{1}{4}}\cos \left(6\times \frac{1}{4}\right)\\ =\frac{1}{4}{e}^{-\frac{1}{4}}\cos \left(1\cdot 5\right)\\ =0\cdot 0138\end{array}$

For the time $\text{t=}\frac{1}{2}$ .

  $\begin{array}{c}y\left(\frac{1}{2}\right)=\frac{1}{4}{e}^{-\frac{1}{2}}\cos \left(6\times \frac{1}{2}\right)\\ =\frac{1}{4}{e}^{-\frac{1}{2}}\cos \left(3\right)\\ =-0\cdot 1501\end{array}$

For the time $\text{t=}\frac{3}{4}$ .

  $\begin{array}{c}y\left(\frac{3}{4}\right)=\frac{1}{4}{e}^{-\frac{3}{4}}\cos \left(6\times \frac{3}{4}\right)\\ =\frac{1}{4}{e}^{-\frac{3}{4}}\cos \left(4\cdot 5\right)\\ =-0\cdot 0249\end{array}$

For the time $\text{t=}1$ .

  $\begin{array}{c}y\left(1\right)=\frac{1}{4}{e}^{-1}\cos \left(6\times 1\right)\\ =\frac{1}{4}{e}^{-1}\cos \left(6\right)\\ =0\cdot 0883\end{array}$

Therefore, the value for function at each time is calculated.

(b)

To determine

To find: The weight reaches equilibrium using graphing utility.

Answer to Problem 50E

The weight reaches equilibrium between time $\text{t}=\frac{1}{4}\sec$ to $\text{t}=\frac{1}{2}\sec$ the value of $y$ is zero.

Explanation of Solution

Given information:

The damping effect of friction is given by $y\left(t\right)=\frac{1}{4}{e}^{-t}\cos 6t$ .

Calculation:

For the various values of the displacement values of the weight from equilibrium given by the variable $y$ , we see that, while we going from $\text{t}=\frac{1}{4}\sec$ to $\text{t}=\frac{1}{2}\sec$ the displacement decreases from $y\left(\frac{1}{4}\right)=0\cdot 0138$ to $y\left(\frac{1}{2}\right)=-0\cdot 1501$ . The value of $y$ from positive to negative.

Approximate between time $\text{t}=\frac{1}{4}\sec$ to $\text{t}=\frac{1}{2}\sec$ the value of $y$ is zero.

Weight reaches equilibrium between these two time points.

Therefore, the weight reaches equilibrium between time $\text{t}=\frac{1}{4}\sec$ to $\text{t}=\frac{1}{2}\sec$ the value of $y$ is zero.

(c)

To determine

To find: The displacement on increasing $t$ .

Answer to Problem 50E

The displacement increases from time $t=\frac{1}{2}$ to $t=1$ .

Explanation of Solution

Given information:

The damping effect of friction is given by $y\left(t\right)=\frac{1}{4}{e}^{-t}\cos 6t$ .

Calculation:

The displacement of the weight first decreases when we go from time $t=0$ to $t=\frac{1}{2}$ .

It reaches its minimum at $t=\frac{1}{2}$ .

Then it increases from time $t=\frac{1}{2}$ to $t=1$ .

Therefore, the displacement increases from time $t=\frac{1}{2}$ to $t=1$ .