Chapter 8, Problem 72RE

(a)

To determine

To describe: The motion of the object

Explanation of Solution

Given: $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

The general equation of damped of motion is given by,

  $d\left(t\right)=a{e}^{-\frac{bt}{2m}}\cos \left(\sqrt{{\omega }^2-\frac{b^2}{4m^2}t}\right)$

The given model is,

  $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

On comparing both expression.

  $\begin{array}{c}a=-20\\ m=30\\ b=0.4\\ \omega =\frac{2\pi }{3}\end{array}$

So, the function is the damped oscillatory motion.

(b)

To determine

To describe: The initial displacement of the bob.

Explanation of Solution

Given: $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

The general equation of damped of motion is given by,

  $d\left(t\right)=a{e}^{-\frac{bt}{2m}}\cos \left(\sqrt{{\omega }^2-\frac{b^2}{4m^2}t}\right)$

The given model is,

  $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

The initial displacement of the bob is $a=-20$ .

(c)

To determine

To draw: The graph of the motion.

Explanation of Solution

Given: $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

The general equation of damped of motion is given by,

  $d\left(t\right)=a{e}^{-\frac{bt}{2m}}\cos \left(\sqrt{{\omega }^2-\frac{b^2}{4m^2}t}\right)$

The given model is,

  $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

The graph of the motion is shown in figure below.

  

Figure (1)

Therefore, the graph of motion is shown in Figure (1).

(d)

To determine

To find: The displacement of the bob at the start of oscillation.

Explanation of Solution

Given: $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

The general equation of damped of motion is given by,

  $d\left(t\right)=a{e}^{-\frac{bt}{2m}}\cos \left(\sqrt{{\omega }^2-\frac{b^2}{4m^2}t}\right)$

The given model is,

  $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

Substitute $3$ for $t$ in the above expression.

  $\begin{array}{c}d\left(3\right)=-20e^{-\frac{0.5\left(3\right)}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}\left(3\right)}\right)\\ =-19.506\end{array}$

Therefore, the displacement at the start of the oscillation is $-19.506$ .

(e)

To determine

To find: The effect on the displacement of the bob as time increases.

Explanation of Solution

Given: $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

The general equation of damped of motion is given by,

  $d\left(t\right)=a{e}^{-\frac{bt}{2m}}\cos \left(\sqrt{{\omega }^2-\frac{b^2}{4m^2}t}\right)$

The given model is,

  $d\left(t\right)=-20e^{-\frac{0.5t}{60}}\cos \left(\sqrt{{\left(\frac{2\pi }{3}\right)}^2-\frac{0.25}{3600}t}\right)$

If time increases without bound, then $d$ oscillates between $-20$ to $20$ with constant amplitude.