Chapter 15, Problem 28P

(a)

To determine

The force constant of the spring.

Answer to Problem 28P

The force constant of the spring is $\underset{\_}{100\text{N/m}}$.

Explanation of Solution

Given that the force is $20.0\text{N}$, and the displacement from equilibrium position is $0.200\text{m}$.

Since the horizontal force of $20.0\text{N}$ keeps the object at rest when it is pulled $0.200\text{m}$ from its equilibrium position, it is equal in magnitude to the restoring force of the spring.

Write the expression for the force constant of the spring.

  $k=\frac{\left|F\right|}{\left|x\right|}$                                                                                                                       (I)

Here, $k$ is the spring constant, $F$ is the restoring force, and $x$ is the displacement from equilibrium position.

Conclusion:

Substitute $20.0\text{N}$ for $F$ and $0.200\text{m}$ for $x$ in equation (I) to find $k$.

  $\begin{array}{c}k=\frac{\left|20.0\text{N}\right|}{\left|0.200\text{m}\right|}\\ =100\text{N/m}\end{array}$

Therefore, the force constant of the spring is $\underset{\_}{100\text{N/m}}$.

(b)

To determine

The frequency of oscillation of the spring-object system.

Answer to Problem 28P

The frequency of oscillation of the spring-object system is $\underset{\_}{1.13\text{Hz}}$.

Explanation of Solution

Given that the mass of the object is $2.00\text{kg}$. It is obtained that the force constant of the spring is $100\text{N/m}$.

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

  $f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$                                                                                                              (II)

Here, $f$ is the frequency of oscillation.

Conclusion:

Substitute $2.00\text{kg}$ for $m$, and $100\text{N/m}$ for $k$ in equation (II) to find $f$.

  $\begin{array}{c}f=\frac{1}{2\pi }\sqrt{\frac{100\text{N/m}}{2.00\text{kg}}}\\ =1.13\text{Hz}\end{array}$

Therefore, the frequency of oscillation of the spring-object system is $\underset{\_}{1.13\text{Hz}}$.

(c)

To determine

The maximum speed of the object.

Answer to Problem 28P

The maximum speed of the object is $\underset{\_}{1.41\text{m/s}}$.

Explanation of Solution

Given that the displacement from the equilibrium position is $0.200\text{m}$, and the mass of the object is $2.00\text{kg}$. It is obtained that the force constant of the spring is $100\text{N/m}$.

Write the expression for the maximum speed of the object executing SHM.

  $v_{\text{max}}=\omega A$                                                                                                                (III)

Here, $v_{\text{max}}$ is the maximum speed, $\omega$ is the angular frequency, and $A$ is the amplitude or the maximum displacement from the equilibrium position.

The amplitude of motion is equal to the initial displacement of the object which is $0.200\text{m}$

Write the expression for the angular frequency.

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

Use equation (IV) in (III).

  $v_{\text{max}}=A\sqrt{\frac{k}{m}}$                                                                                                             (V)

Conclusion:

Substitute $2.00\text{kg}$ for $m$, $100\text{N/m}$ for $k$, and $0.200\text{m}$ for $A$ in equation (V) to find $v_{\text{max}}$.

  $\begin{array}{c}{v}_{\text{max}}=\left(0.200\text{m}\right)\sqrt{\frac{100\text{N/m}}{2.00\text{kg}}}\\ =1.41\text{m/s}\end{array}$

Therefore, the maximum speed of the object is $\underset{\_}{1.41\text{m/s}}$.

(d)

To determine

The position where maximum speed occur.

Answer to Problem 28P

The position where maximum speed occur is $\underset{\_}{x=0}$.

Explanation of Solution

For the oscillating spring-object system, the objects loses all its potential energy and gains maximum kinetic energy at the equilibrium point. Since the maximum kinetic energy corresponds to the maximum speed, the object gains maximum speed at the equilibrium position, which is described by the coordinate, $x=0$.

Conclusion:

Therefore, the position where maximum speed occur is $\underset{\_}{x=0}$.

(e)

To determine

The maximum acceleration of the object.

Answer to Problem 28P

The maximum acceleration of the object is $\underset{\_}{10.0{\text{m/s}}^2}$.

Explanation of Solution

Write the expression for the maximum acceleration of the object executing SHM.

  $a_{\text{max}}={\omega }^{2}A$                                                                                                              (VI)

Here, $a_{\text{max}}$ is the maximum acceleration.

Use equation (IV) in (VI).

  $\begin{array}{c}{a}_{\text{max}}={\left(\sqrt{\frac{k}{m}}\right)}^{2}A\\ =\frac{kA}{m}\end{array}$                                                                                                    (VII)

Conclusion:

Substitute $2.00\text{kg}$ for $m$, $100\text{N/m}$ for $k$, and $0.200\text{m}$ for $A$ in equation (VII) to find $a_{\text{max}}$.

  $\begin{array}{c}{a}_{\text{max}}=\frac{\left(100\text{N/m}\right)\left(0.200\text{m}\right)}{2.00\text{kg}}\\ =10.0{\text{m/s}}^2\end{array}$

Therefore, the maximum acceleration of the object is $\underset{\_}{10.0{\text{m/s}}^2}$.

(f)

To determine

The position where the maximum acceleration occur.

Answer to Problem 28P

The position where the maximum acceleration occur is $\underset{\_}{x=\pm 0.200\text{m}}$.

Explanation of Solution

For the oscillating spring-object system, the maximum acceleration occurs where the object reverses its direction of motion. This happens only at the positions corresponding to the maximum displacement. Which is $x=\pm A$. Since the maximum displacement is given as $0.200\text{m}$, the maximum acceleration occur at $x=\pm 0.200\text{m}$.

Conclusion:

Therefore, the position where the maximum acceleration occur is $\underset{\_}{x=\pm 0.200\text{m}}$.

(g)

To determine

The total energy of the oscillating system.

Answer to Problem 28P

The total energy of the oscillating system is $\underset{\_}{2.00\text{J}}$.

Explanation of Solution

It is obtained that the force constant of the spring is $100\text{N/m}$, and given that the maximum displacement is $0.200\text{m}$.

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

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

Here, $E$ is the energy.

Conclusion:

Substitute $100\text{N/m}$ for $k$, and $0.200\text{m}$ for $A$ in equation (VIII) to find $E$.

  $\begin{array}{c}E=\frac{1}{2}\left(100\text{N/m}\right){\left(0.200\text{m}\right)}^2\\ =2.00\text{J}\end{array}$

Therefore, the total energy of the oscillating system is $\underset{\_}{2.00\text{J}}$.

(h)

To determine

The speed of the object when its position is equal to one-third the maximum value.

Answer to Problem 28P

The speed of the object when its position is equal to one-third the maximum value is $\underset{\_}{1.33\text{m/s}}$.

Explanation of Solution

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

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

Here, $v$ is the speed, and $x$ is the position.

Since the position is one-third the maximum value ($A$), equation (IX) can be rewritten using equation (IV) as,

  $\begin{array}{c}v=\left(\sqrt{\frac{k}{m}}\right)\sqrt{A^2-{\left(\frac{A}{3}\right)}^2}\\ =A\sqrt{\frac{8k}{9m}}\end{array}$                                                                                           (X)

Conclusion:

Substitute $2.00\text{kg}$ for $m$, $100\text{N/m}$ for $k$, and $0.200\text{m}$ for $A$ in equation (X) to find $v$.

  $\begin{array}{c}v=\left(0.200\text{m}\right)\sqrt{\frac{8\left(100\text{N/m}\right)}{9\left(2.00\text{kg}\right)}}\\ =1.33\text{m/s}\end{array}$

Therefore, the speed of the object when its position is equal to one-third the maximum value is $\underset{\_}{1.33\text{m/s}}$.

(i)

To determine

The acceleration of the object when its position is equal to one-third the maximum value.

Answer to Problem 28P

The speed of the object when its position is equal to one-third the maximum value is $\underset{\_}{3.33{\text{m/s}}^2}$.

Explanation of Solution

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

  $a={\omega }^{2}x$                                                                                                                   (XI)

Here, $a$ is the acceleration.

Since the position is one-third the maximum value ($A$), equation (XI) can be rewritten using equation (IV) as,

  $\begin{array}{c}a={\left(\sqrt{\frac{k}{m}}\right)}^2\left(\frac{A}{3}\right)\\ =\frac{kA}{3m}\end{array}$                                                                                                    (XII)

Conclusion:

Substitute $2.00\text{kg}$ for $m$, $100\text{N/m}$ for $k$, and $0.200\text{m}$ for $A$ in equation (XII) to find $a$.

  $\begin{array}{c}a=\frac{\left(100\text{N/m}\right)\left(0.200\text{m}\right)}{3\left(2.00\text{kg}\right)}\\ =3.33{\text{m/s}}^2\end{array}$

Therefore, the speed of the object when its position is equal to one-third the maximum value is $\underset{\_}{3.33{\text{m/s}}^2}$.