Chapter 15, Problem 27P

(a)

To determine

The total energy of the oscillating system.

Answer to Problem 27P

The total energy of the oscillating system is $\underset{\_}{28.0\text{mJ}}$.

Explanation of Solution

Given that the force constant of the spring is $35.0\text{N/m}$, and the amplitude of motion is $4.00\text{cm}$.

Write the expression for the energy of the spring-object oscillating system.

  $E=\frac{1}{2}k{A}^2$                                                                                                                   (I)

Here, $E$ is the energy, $k$ is the spring constant, and $A$ is the amplitude.

Conclusion:

Substitute $35.0\text{N/m}$ for $k$, and $4.00\text{cm}$ for $A$ in equation (VIII) to find $E$.

  $\begin{array}{c}E=\frac{1}{2}\left(35.0\text{N/m}\right){\left(4.00\text{cm}\right)}^2\\ =\frac{1}{2}\left(35.0\text{N/m}\right){\left(4.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =2.80\times {10}^{-2}\text{J}\times \frac{1000\text{mJ}}{1\text{J}}\\ =28.0\text{mJ}\end{array}$

Therefore, the total energy of the oscillating system is $\underset{\_}{28.0\text{mJ}}$.

(b)

To determine

The speed of the object when its position is $1.00\text{cm}$.

Answer to Problem 27P

The speed of the object when its position is $1.00\text{cm}$ is $\underset{\_}{1.02\text{m/s}}$.

Explanation of Solution

Given that the force constant of the spring is $35.0\text{N/m}$, the amplitude of motion is $4.00\text{cm}$, and the mass of the object is $50.0\text{g}$.

Write the expression for the speed at a given position of an object executing SHM in a spring.

  $v=\omega \sqrt{A^2-x^2}$                                                                                                          (II)

Here, $v$ is the speed, $\omega$ is the angular frequency, and $x$ is the position.

Write the expression for the angular frequency of the spring-object system.

  $\omega =\sqrt{\frac{k}{m}}$                                                                                                                  (III)

Use equation (III) in (II).

  $\begin{array}{c}v=\left(\sqrt{\frac{k}{m}}\right)\sqrt{A^2-x^2}\\ =\sqrt{\frac{k}{m}\left(A^2-x^2\right)}\end{array}$                                                                                                (IV)

Conclusion:

Substitute $35.0\text{N/m}$ for $k$, $50.0\text{g}$ for $m$, $1.00\text{cm}$ for $x$, and $4.00\text{cm}$ for $A$ in equation (IV) to find $v$.

  $\begin{array}{c}v=\sqrt{\frac{35.0\text{N/m}}{50.0\text{g}}\left[{\left(4.00\text{cm}\right)}^2-{\left(1.00\text{cm}\right)}^2\right]}\\ =\sqrt{\frac{35.0\text{N/m}}{50.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}}\left[{\left(4.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2-{\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\right]}\\ =1.02\text{m/s}\end{array}$

Therefore, the speed of the object when its position is $1.00\text{cm}$ is $\underset{\_}{1.02\text{m/s}}$.

(c)

To determine

The kinetic energy of the object when its position is $3.00\text{cm}$.

Answer to Problem 27P

The kinetic energy of the object when its position is $3.00\text{cm}$ is $\underset{\_}{12.2\text{mJ}}$.

Explanation of Solution

Write the expression for the kinetic energy of the oscillating object.

  $K=E-U$                                                                                                                (V)

Here, $K$ is the kinetic energy, $U$ is the potential energy at the given position.

Equation (I) gives the total energy of the system.

  $E=\frac{1}{2}k{A}^2$

Write the expression for the potential energy of the object at the given position.

  $U=\frac{1}{2}k{x}^2$                                                                                                                (VI)

Use equation (I) and (VI) in (V).

  $\begin{array}{c}K=\frac{1}{2}k{A}^2-\frac{1}{2}k{x}^2\\ =\frac{1}{2}k\left(A^2-x^2\right)\end{array}$                                                                                                  (VII)

Conclusion:

Substitute $35.0\text{N/m}$ for $k$, $3.00\text{cm}$ for $x$, and $4.00\text{cm}$ for $A$ in equation (VII) to find $K$.

  $\begin{array}{c}K=\frac{1}{2}\left(35.0\text{N/m}\right)\left[{\left(4.00\text{cm}\right)}^2-{\left(3.00\text{cm}\right)}^2\right]\\ =\frac{1}{2}\left(35.0\text{N/m}\right)\left[{\left(4.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2-{\left(3.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\right]\\ =1.22\times {10}^{-2}\text{J}\times \frac{1000\text{mJ}}{1\text{J}}\\ =12.2\text{mJ}\end{array}$

Therefore, the kinetic energy of the object when its position is $3.00\text{cm}$ is $\underset{\_}{12.2\text{mJ}}$.

(d)

To determine

The potential energy of the object when its position is $3.00\text{cm}$.

Answer to Problem 27P

The potential energy of the object when its position is $3.00\text{cm}$ is $\underset{\_}{15.8\text{mJ}}$.

Explanation of Solution

It is obtained that the total energy of the system is $28.0\text{mJ}$, and the kinetic energy when the object’s position is $3.00\text{cm}$ is $12.2\text{mJ}$.

Write the expression for the potential energy of the object.

  $U=E-K$                                                                                                            (VIII)

Conclusion:

Substitute $28.0\text{mJ}$ for $E$, and $12.2\text{mJ}$ for $K$ in equation (VIII) to find $U$.

  $\begin{array}{c}U=28.0\text{mJ}-12.2\text{mJ}\\ =15.8\text{mJ}\end{array}$

Therefore, the potential energy of the object when its position is $3.00\text{cm}$ is $\underset{\_}{15.8\text{mJ}}$.