Chapter 12, Problem 18P

(a)

To determine

The force constant of the spring.

Answer to Problem 18P

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

Explanation of Solution

Write the expression to calculate the force constant of the spring.

    $k=\frac{F}{x}$        (I)

Here, $k$ is spring force constant, $F$ is the force applied, and $x$ is the stretch in the spring.

Conclusion:

Substitute $20\text{N}$ for $F$ and $0.200\text{m}$ for $x$ in Equation (I).

    $\begin{array}{c}k=\frac{20\text{N}}{0.200\text{m}}\\ =100\text{N}/\text{m}\end{array}$

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

(b)

To determine

The frequency of oscillations.

Answer to Problem 18P

The frequency of oscillations is $1.13\text{Hz}$.

Explanation of Solution

Write the formula to calculate the angular frequency for the object.

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

Here, $\omega$ is angular frequency and $m$ is the mass of the object.

Write the formula to calculate the frequency of oscillations.

    $f=\frac{\omega }{2\pi }$        (III)

Here, $f$ is the oscillating frequency.

Conclusion:

Substitute $100\text{N}/\text{m}$ for $k$ and $2\text{kg}$ for $m$ in Equation (II).

    $\begin{array}{c}\omega =\sqrt{\frac{100\text{N}/\text{m}}{2\text{kg}}\times \frac{1\text{kg}\cdot \text{m}/{\text{s}}^2}{1\text{N}}}\\ =\sqrt{50}\text{rad}/\text{s}\end{array}$

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ in Equation (III).

    $\begin{array}{c}f=\frac{\sqrt{50}\text{rad}/\text{s}}{2\pi }\\ =1.13\text{Hz}\end{array}$

Thus, the frequency of oscillations is $1.13\text{Hz}$.

(c)

To determine

The value for the maximum speed of the object.

Answer to Problem 18P

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

Explanation of Solution

Write the expression to calculate the maximum speed of the object.

    $v_{\max }=\omega A$        (IV)

Here, $v_{\max }$ is the maximum speed of the object and $A$ is the amplitude of the oscillations.

Conclusion:

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ and $0.200\text{m}$ for $A$ in Equation (IV).

    $\begin{array}{c}{v}_{\max }=\sqrt{50}\text{rad}/\text{s}\times 0.200\text{m}\\ =1.41\text{m}/\text{s}\end{array}$

Thus, the value for the maximum speed of the object is $1.41\text{m}/\text{s}$.

(d)

To determine

The position at which the maximum speed occur.

Answer to Problem 18P

The position at which the maximum speed occur is $0$.

Explanation of Solution

The maximum speed of a simple harmonic motion occurs at the equilibrium position. At the equilibrium position, the value of $x=0\text{m}$.

Thus, the position at which the maximum speed occur is $0$.

(e)

To determine

The maximum acceleration of the object.

Answer to Problem 18P

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

Explanation of Solution

Write the expression to calculate the maximum acceleration of the object.

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

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

Conclusion:

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ and $0.200\text{m}$ for $A$ in Equation (V).

    $\begin{array}{c}{a}_{\max }={\left(\sqrt{50}\text{rad}/\text{s}\right)}^2\times 0.200\text{m}\\ =10\text{m}/{\text{s}}^2\end{array}$

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

(f)

To determine

The position at which the maximum acceleration occur.

Answer to Problem 18P

The position at which the maximum acceleration occur is $x=\pm 0.200\text{m}$.

Explanation of Solution

The object attains its maximum acceleration when the direction of the object is reversed. The object reverses its direction when it is at maximum distance from the equilibrium x position.

The object reverses its direction at the maximum position of $x=\pm 0.200\text{m}$.

Thus, the position at which the maximum acceleration occur is $x=\pm 0.200\text{m}$.

(g)

To determine

The total energy of the oscillating system.

Answer to Problem 18P

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

Explanation of Solution

Write the expression to calculate the total energy of the oscillating system.

    $E=\frac{1}{2}k{A}^2$        (VI)

Here, $E$ is the total energy of the oscillating system.

Conclusion:

Substitute $100\text{N}/\text{m}$ for $k$ and $0.200\text{m}$ for $A$ in Equation (VI).

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

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

(h)

To determine

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

Answer to Problem 18P

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

Explanation of Solution

Write the expression to calculate the speed of the object at $x=\frac{1}{3}A$ position.

    $\begin{array}{c}v=\omega \sqrt{A^2-x^2}\\ =\omega \sqrt{A^2-{\left(\frac{1}{3}A\right)}^2}\\ =\omega \sqrt{\frac{8}{9}{A}^2}\end{array}$        (VII)

Conclusion:

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ and $0.200\text{m}$ for $A$ in Equation (VII).

    $\begin{array}{c}v=\sqrt{50}\text{rad}/\text{s}\times \sqrt{\frac{8}{9}\times {\left(0.200\text{m}\right)}^2}\\ =1.33\text{m}/\text{s}\end{array}$

Thus, the speed of the object when its position is one third of maximum value is $1.33\text{m}/\text{s}$.

(i)

To determine

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

Answer to Problem 18P

The acceleration of the object when its position is one third of maximum value is $3.33\text{m}/{\text{s}}^2$.

Explanation of Solution

Write the expression to calculate the speed of the object at $x=\frac{1}{3}A$ position

    $\begin{array}{c}a={\omega }^{2}x\\ ={\omega }^2\left(\frac{1}{3}A\right)\end{array}$        (VIII)

Conclusion:

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ and $0.200\text{m}$ for $A$ in Equation (VIII).

    $\begin{array}{c}a={\left(\sqrt{50}\text{rad}/\text{s}\right)}^2\left(\frac{1}{3}\times 0.200\text{m}\right)\\ =3.33\text{m}/{\text{s}}^2\end{array}$

Thus, the acceleration of the object when its position is one third of maximum value is $3.33\text{m}/{\text{s}}^2$.