Chapter 11, Problem 13P

(a)

To determine

To find: The amplitude of the motion.

Answer to Problem 13P

The amplitude of the motion is $\text{6}\text{.034}\times {\text{10}}^{\text{-2}}\text{ m}$ .

Explanation of Solution

Given:

Mass, $\mathrm{m}\text{ =3 kg}$

Spring constant, $\mathrm{k}\text{= 280 N/m}$

Distance, $\mathrm{x}\text{ = 0}\text{.020 m}$

Velocity, $\text{v = 0}\text{.55 m/s}$

Formula used:

The total energy of the spring is the addition of the kinetic energy and potential energy of the spring

  $\begin{array}{l}{\text{E}}_{\text{total}}\text{ = K}{\text{.E}}_{\text{Spring}}\text{ + P}{\text{.E}}_{\text{Spring}}\\ {\text{E}}_{\text{total}}\text{ = }\frac{1}{2}\times \mathrm{m}\times {\mathrm{v}}^2+\frac{1}{2}\times \mathrm{k}\times {\mathrm{x}}^2=\frac{1}{2}\times \mathrm{k}\times {\mathrm{A}}^2\end{array}$

Here, $\mathrm{m}$ is the mass,

  $\mathrm{v}$ is the speed of the motion,

  $\mathrm{x}$ is the distance moved by the spring,

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

  $\mathrm{A}$ is the amplitude of the motion.

Calculation:

From the above formula, write the expression for amplitude A.

  $\begin{array}{l}{\text{A}}^{\text{2}}{\text{ = x}}^{\text{2}}\text{ + }\frac{\text{m}}{\text{k}}\times {\text{v}}^{\text{2}}\\ {\text{A}}^{\text{2}}\text{ = (0}{\text{.02)}}^{\text{2}}\text{ + }\left(\frac{\text{3}}{\text{280}}\times {\text{(0}\text{.55)}}^{\text{2}}\right)\\ {\text{A}}^{\text{2}}\text{ = 3}\text{.64}\times {\text{10}}^{\text{-3}}{\text{ m}}^{\text{2}}\\ \text{A = 6}\text{.034}\times {\text{10}}^{\text{-2}}\text{ m}\end{array}$

Conclusion:

Thus, the amplitude of the motion is $\text{6}\text{.034}\times {\text{10}}^{\text{-2}}\text{ m}$ .

(b)

To determine

To Find: The maximum velocity attained by the object.

Answer to Problem 13P

b$\text{0}\text{.58 m/s}$

Explanation of Solution

Given:

Mass, $\mathrm{m}\text{ =3 kg}$

Spring constant, $\mathrm{k}\text{= 280 N/m}$

Amplitude, $\mathrm{A}\text{ = 6}\text{.034}\times {\text{10}}^{\text{-2}}\text{ m}$

Formula used:

Energy is called kinetic energy when the object has the maximum velocity, maximum velocity is calculated by the following formula,

  ${\text{E}}_{\text{total}}\text{ = }\frac{1}{2}\times \mathrm{k}\times {\mathrm{A}}^2=\frac{1}{2}\times \mathrm{m}\times {\mathrm{v}}^2{}_{\max }$

Here, $\mathrm{m}$ is the mass,

  ${\mathrm{v}}_{\max }$ is the maximum speed of the motion,

  $\mathrm{x}$ is the distance moved by the spring,

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

  $\mathrm{A}$ is the amplitude of the motion.

Calculation:

Substitute the given values in the above equation to get maximum velocity by following way

  $\begin{array}{l}{\mathrm{v}}_{\max }{}^2=\frac{\mathrm{k}\times {\mathrm{A}}^2}{\mathrm{m}}\\ {\text{v}}_{\text{max}}{}^{\text{2}}\text{ = }\frac{\text{280 }\times {\text{(6}\text{.034}\times {\text{10}}^{\text{-2}}\text{)}}^{\text{2}}}{\text{3}}\\ {\text{v}}_{\text{max}}\text{ = 0}\text{.58 m/s}\end{array}$

Conclusion:

Thus, the maximum velocity attained by the object is $\text{0}\text{.58 m/s}$ .