Chapter 11, Problem 74GP

a.

To determine

To Find: The amplitude of vibration .

Answer to Problem 74GP

  $0.650\text{m}$ .

Explanation of Solution

Given Information:

x = $\text{0}\text{.650 cos7}\text{.40t}$

Value of x = $0.260\text{m}$

Concept used:

By using the simple harmonic equation

  $\text{x}\text{=}\text{A cos }\mathrm{\omega }\text{t }\mathrm{...}\left(1\right)$

Here,

  $\mathrm{A}$ is the amplitude,

  $\mathrm{\omega }$ is the angular frequency,

  $\mathrm{t}$ is the time,

  $\mathrm{x}$ is the displacement.

Calculation: From equation 1

  $\text{x}\text{=}\text{A cos }\mathrm{\omega }\text{t}$

Given equation is

  $\text{x}\text{=}\text{0}\text{.650 cos7}\text{.40t}$

By comparing these two equations

  $\text{A}\text{=}\text{0}\text{.650}\text{m}$

Conclusion:

The amplitude of vibration is $\text{0}\text{.650}\text{m}$

b.

To determine

To find: The frequency of vibration.

Answer to Problem 74GP

The frequency of vibration is $1.18\text{Hz}$ .

Explanation of Solution

Given Information:

The angular velocity $\mathrm{\omega }\text{}\text{}=\text{}7.40\text{rad/s}$

Concept used:

The formula for finding the angular velocity is

  $\mathrm{\omega }=2\mathrm{\pi }\mathrm{f}\mathrm{...}\left(1\right)$

Calculation:

From equation 1

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

  $\begin{array}{l}7.40=2\mathrm{\pi }\mathrm{f}\\ \mathrm{f}\text{=}\frac{7.40}{2\mathrm{\pi }}\text{Hz}\\ \mathrm{f}\text{ = 1}\text{.18}\text{Hz}\end{array}$

Conclusion:

The frequency of vibration is $\text{1}\text{.18}\text{Hz}$ .

c.

To determine

To find: The total energy of vibration.

Answer to Problem 74GP

The total energy of vibration is $23.14\text{J}$ .

Explanation of Solution

Given Information:

Mass, $\mathrm{m}=2.00\text{kg}$

Amplitude of vibration, $\mathrm{A}\text{=0}\text{.650}\text{m}$

Angular velocity, $\mathrm{\omega }\text{=}\text{7}\text{.40}\text{rad/s}$

Concept used:

The total energy of vibration is

  $\begin{array}{l}{\text{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=\frac{1}{2}{\text{mv}}^{\text{2}}\\ {\text{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=\frac{1}{2}\text{m (}\mathrm{\omega }\mathrm{A}{\text{)}}^2\end{array}$

Here,

  $\mathrm{A}$ is the amplitude,

  $\mathrm{\omega }$ is the angular frequency,

  $\mathrm{m}$ is the mass.

Calculation:

  $\begin{array}{l}$ {\text{E}}_{total}=\frac{1}{2}\times 2{(7.40\times 0.650)}^2${\text{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=23.14\text{J}\end{array}$

Conclusion:

The total energy of vibration is $23.14\text{J}$ .

d.

To determine

To find: The potential energy and kinetic energy of the vibration.

Answer to Problem 74GP

The P.E and K.E of the vibration are

  $3.7017\text{J}$ and $19.44\text{J}$ respectively.

Explanation of Solution

Given Information: $\mathrm{x}=0.260\text{m}$

Mass $\mathrm{m}=2.00\text{kg}$

Angular velocity $\mathrm{\omega }\text{=}\text{7}\text{.40}\text{rad/s}$

  ${\mathrm{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}=23.14\text{J}$

Concept used:

Potential energy is

  ${\mathrm{E}}_{\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}=\frac{1}{2}\mathrm{k}{\mathrm{x}}^2=\frac{1}{2}\mathrm{m}{\mathrm{\omega }}^2{\mathrm{x}}^2$

  ${\mathrm{E}}_{\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}=\frac{1}{2}\mathrm{m}{\mathrm{\omega }}^2{\mathrm{x}}^2\mathrm{...}\left(1\right)$

  ${\mathrm{E}}_{\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}}={\mathrm{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}-{\mathrm{E}}_{\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}\mathrm{...}\left(2\right)$

Here,

  $\mathrm{\omega }$ is the angular frequency,

  $\mathrm{m}$ is the mass,

  $\mathrm{k}$ is the spring constant,

  $\mathrm{x}$ is the displacement.

Calculation:

From equation 1

  $\begin{array}{l}$ E_{potential}=\frac{1}{2}\times 2{(7.40)}^2{(0.260\text{)}}^2${\mathrm{E}}_{\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}=3.7017\text{J}\end{array}$

From equation 2

  $\begin{array}{l}{\mathrm{E}}_{\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}}={\mathrm{E}}_{\mathrm{t}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}}-{\mathrm{E}}_{\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{l}}\\ {\mathrm{E}}_{\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}}=23.14\text{J}-3.7017\text{J}\\ {\mathrm{E}}_{\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{t}\mathrm{i}\mathrm{c}}=19.44\text{J}\end{array}$

Conclusion:

The P.E and K.E of the vibration are $3.7017\text{J}$ and $19.44\text{J}$ respectively.