Chapter 11, Problem 16P

(a)

To determine

To Find: The amplitude of motion.

Answer to Problem 16P

  $0.45\text{ m}$

Explanation of Solution

Given:

Mass, $\mathrm{m}=0.60\text{ kg}$

  $\mathrm{x}=0.45\cos (6.4\mathrm{t})$

Formula used:

Compare the given equation with $\mathrm{x}=\mathrm{A}\cos (\mathrm{\omega }\mathrm{t})$ and get the value of amplitude A.

Calculation:

By comparing the below two equations, the amplitude A in meter is:

  $\begin{array}{l}\mathrm{x}=0.45\cos (6.4\mathrm{t})\\ \mathrm{x}=\mathrm{A}\cos (\mathrm{\omega }\mathrm{t})\end{array}$

  $\mathrm{A}=0.45\text{ m}$

Conclusion:

Thus, the amplitude of motion is $0.45\text{ m}$ .

(b)

To determine

To Find: The frequency.

Answer to Problem 16P

  $1.02\text{ Hz}$

Explanation of Solution

Given:

  $\mathrm{x}=0.45\cos (6.4\mathrm{t})$

Angular speed, $\mathrm{\omega }=6.4\text{ s}$

Formula used:

The relationship between the angular speed and frequency is,

  $\mathrm{\omega }=2\times \mathrm{\pi }\times \mathrm{f}$

Calculation:

From the above equation, the frequency is determined in the following way,

  $\begin{array}{c}\mathrm{f}=\frac{\mathrm{\omega }}{2\times \mathrm{\pi }}\\ =\frac{6.4}{2\times \mathrm{\pi }}\\ =1.02\text{ Hz}\end{array}$

Conclusion:

Thus, the frequency is $1.02\text{ Hz}$ .

(c)

To determine

To Find: The total energy.

Answer to Problem 16P

  $2.5\text{ J}$

Explanation of Solution

Given:

Mass, $\mathrm{m}=0.6\text{ kg}$

Angular speed, $\mathrm{\omega }=6.4\text{ rad/s}$

Amplitude, $\mathrm{A}=0.45\text{ m}$

Formula used:

The total energy of the system is determined by the following formula,

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=\frac{1}{2}\times \mathrm{m}\times {\mathrm{v}}_{\max }{}^2\\ {\mathrm{v}}_{\max }=\mathrm{\omega }\times \mathrm{A}\end{array}$

So, ${\mathrm{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=\frac{1}{2}\times \mathrm{m}\times {(\mathrm{\omega }\times \mathrm{A})}^2$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}$ E_{total}=\frac{1}{2}\times 0.60\times {(6.4\times 0.45)}^2${\mathrm{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=2.5\text{ J}\end{array}$

Conclusion:

Thus, the total energy of the system is $2.5\text{ J}$ .

(d)

To determine

To Find: The potential energy.

Answer to Problem 16P

  $1.1\text{ J}$

Explanation of Solution

Given:

Mass, $\mathrm{m}=0.6\text{ kg}$

Angular speed, $\mathrm{\omega }=6.4\text{ rad/s}$

Amplitude, $\mathrm{A}=0.45\text{ m}$

Distance, $\mathrm{x}=0.30\text{ m}$

Formula used:

The potential energy can be calculated by the following formula,

  ${\mathrm{E}}_{\mathrm{p}}=\frac{1}{2}\times \mathrm{k}\times {\mathrm{x}}^2$

Where, $\mathrm{k}=\mathrm{m}\times {\mathrm{\omega }}^2$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}$ E_p=\frac{1}{2}\times 0.6\times {(6.4)}^2\times {(0.30)}^2${\mathrm{E}}_{\mathrm{p}}=1.1\text{ J}\end{array}$

Conclusion:

Thus, the potential energy at $\mathrm{x}=0.30\text{ m}$ is $1.1\text{ J}$ .

(e)

To determine

To Find: The kinetic energy.

Answer to Problem 16P

The kinetic energy is $1.4\text{ J}$ .

Explanation of Solution

Given:

Total energy, ${\mathrm{E}}_{\mathrm{t}}=2.5\text{ J}$

Potential energy, ${\mathrm{E}}_{\mathrm{p}}=1.1\text{ J}$

Formula used:

The relation between the potential energy, kinetic energy, and total energy is,

  ${\mathrm{E}}_{\mathrm{k}}={\mathrm{E}}_{\mathrm{t}}-{\mathrm{E}}_{\mathrm{p}}$

Calculation:

Substitute the given values in the above equation in the above equation,

  $\begin{array}{c}{\mathrm{E}}_{\mathrm{k}}=2.5-1.1\\ =1.4\text{ J}\end{array}$

Conclusion:

Thus, the kinetic energy at $\mathrm{x}=0.30\text{ m}$ is $1.4\text{ J}$ .