Chapter 8, Problem 43RE

(a)

To determine

To describe: The motion of the object

Answer to Problem 43RE

The function is the damped oscillatory motion.

Explanation of Solution

Given: $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

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,

  $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

On comparing both expression.

  $\begin{array}{c}a=-15,\\ m=20\\ b=0.6\\ \omega =\frac{2\pi }{5}\end{array}$

So, the function is the damped oscillatory motion.

(b)

To determine

To describe: The initial displacement of the bob.

Answer to Problem 43RE

The initial displacement of the bob is $a=-15$ .

Explanation of Solution

Given: $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

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,

  $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

On comparing both expression.

  $\begin{array}{c}a=-15,\\ m=20\\ b=0.6\\ \omega =\frac{2\pi }{5}\end{array}$

The initial displacement of the bob is $a=-15$ .

(c)

To determine

To draw: The graph of the motion.

Answer to Problem 43RE

The graph of motion is shown in Figure (1).

Explanation of Solution

Given: $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

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,

  $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

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.

Answer to Problem 43RE

The displacement at the start of the oscillation is $-13.92$ .

Explanation of Solution

Given: $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

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,

  $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

Substitute $5$ for $t$ in the above expression.

  $\begin{array}{c}d\left(5\right)=-15e^{-\frac{0.6\left(5\right)}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}\left(5\right)}\right)\\ =-13.92\end{array}$

Therefore, the displacement at the start of the oscillation is $-13.92$ .

(e)

To determine

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

Answer to Problem 43RE

If time increases without bound, then $d$ oscillates between $-15$ to $15$ with constant amplitude.

Explanation of Solution

Given: $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

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,

  $d\left(t\right)=-15e^{-\frac{0.6t}{40}}\cos \left(\sqrt{{\left(\frac{2\pi }{5}\right)}^2-\frac{0.36}{1600}t}\right)$

If time increases without bound, then $d$ oscillates between $-15$ to $15$ with constant amplitude.