Chapter 15, Problem 15.28P

A 2.00-kg object is attached to a spring and placed on a frictionless, horizontal surface. A horizontal force of 20.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest from this stretched position, and it subsequently undergoes simple harmonic oscillations. Find (a) the force constant of the spring, (b) the frequency of the oscillations, and (c) the maximum speed of the object, (d) Where does this maximum speed occur? (e) Find the maximum acceleration of the object. (f) Where does the maximum acceleration occur? (g) Find the total energy of the oscillating system. Find (h) the speed and (i) the acceleration of the object when its position is equal to one-third the maximum value.

To determine

The force constant of the spring.

Answer to Problem 15.28P

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

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

Write the expression for force constant.

$k=\frac{\left|F\right|}{\left|A\right|}$

Here,

$F$ is the force on the spring.

$A$ is the amplitude.

Substitute $0.200\text{m}$ for $A$ and $\text{20}\text{.0}\text{N}$ for $F$ in above equation.

$\begin{array}{c}k=\frac{\left|\text{20}\text{.0}\text{N}\right|}{\left|0.200\text{m}\right|}\\ =100\text{N}/\text{m}\end{array}$

Conclusion:

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

To determine

The frequency of the oscillations.

Answer to Problem 15.28P

The frequency of the oscillations is $1.125\text{Hz}$.

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

Write the expression for the force constant of the spring.

$\begin{array}{l}k=m{\omega }^2\\ \omega =\sqrt{\frac{k}{m}}\end{array}$

Here,

$k$ is the force constant.

$m$ is the mass of the spring.

$\omega$ is the angular velocity.

Substitute $100\text{N}/\text{m}$ for $k$ and $2.00\text{kg}$ for $m$ in above equation.

$\begin{array}{c}\omega =\sqrt{\frac{100\text{N}/\text{m}}{2.00\text{kg}}}\\ =\sqrt{50}\text{rad}/\text{s}\end{array}$

Write the expression for the frequency of the oscillations.

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

Here,

$f$ is the frequency of the oscillations.

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ in above equation.

$\begin{array}{c}f=\frac{\sqrt{50}\text{rad}/\text{s}}{2\pi }\\ =1.125\text{Hz}\end{array}$

Conclusion:

Therefore, the frequency of the oscillations is $1.125\text{Hz}$.

To determine

The maximum speed of the object.

Answer to Problem 15.28P

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

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

From part (b) the angular velocity is $\sqrt{50}\text{rad}/\text{s}$.

Write the expression for maximum speed.

$v_{\text{max}}=\omega A$

Here,

$v_{\text{max}}$ is the maximum speed.

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ and $0.200\text{m}$ for $A$ in above equation.

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

Conclusion:

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

To determine

The position of the object where the maximum speed occurs.

Answer to Problem 15.28P

The position of the object where the maximum speed occurs at $x=0$.

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

The maximum speed of the object occurs when the object passes through its equilibrium position.

The equilibrium position of the object is,

$x=0$

Conclusion:

Therefore, the position of the object where the maximum speed occurs at $x=0$.

To determine

The maximum acceleration of the object.

Answer to Problem 15.28P

The maximum acceleration of the object is $10\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

Write the expression for the maximum acceleration.

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

Here,

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

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ and $0.200\text{m}$ for $A$ in above equation.

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

Conclusion:

Therefore, the maximum acceleration of the object is $10\text{m}/{\text{s}}^{\text{2}}$.

To determine

The position of the object where the maximum acceleration occurs.

Answer to Problem 15.28P

The position of the object where the maximum acceleration occurs at $x=\pm 0.200\text{m}$.

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

The maximum acceleration of the object occurs when the object reverses its direction of motion.

The object reverses its direction of motion where its distance is the maximum from the equilibrium position.

The maximum distance from equilibrium position of the object is,

$x=\pm 0.200\text{m}$

Conclusion:

Therefore, the position of the object where the maximum acceleration occurs at $x=\pm 0.200\text{m}$.

To determine

The total energy of the system.

Answer to Problem 15.28P

The total energy of the system is $2\text{J}$.

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

From part (a) the force constant of the spring is $100\text{N}/\text{m}$.

Write the expression for the total energy.

$E=\frac{k{A}^2}{2}$

Here,

$E$ is the total energy of the object.

$A$ is the amplitude of the motion.

Substitute $100\text{N}/\text{m}$ for $k$ and $0.200\text{m}$ for $A$ in above equation.

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

Conclusion:

Therefore, the total energy of the system is $2\text{J}$.

To determine

The speed of the object when the object is at one-third of the maximum value.

Answer to Problem 15.28P

The speed of the object when the object is at one-third of the maximum value is $1.333\text{m}/\text{s}$.

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

The position of the object is one-third of the maximum value.

$x=\frac{A}{3}$

Here,

$A$ is the amplitude of the motion.

$x$ is the position of the object.

Substitute $0.200\text{m}$ for $A$ in above equation.

$\begin{array}{c}x=\frac{0.200\text{m}}{3}\\ =0.0667\text{m}\end{array}$

Thus, the value of $x$ is $0.0667\text{m}$.

From part (a) the angular velocity is $\sqrt{50}\text{rad}/\text{s}$.

Write the expression for the velocity of block at any position.

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

Here,

$v$ is the velocity of the object.

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$, $0.0667\text{m}$ for $x$ and $0.200\text{m}$ for $A$ in above equation.

$\begin{array}{c}v=\left(\sqrt{50}\text{rad}/\text{s}\right)\sqrt{{\left(0.200\text{m}\right)}^2-{\left(0.0667\text{m}\right)}^2}\\ =1.333\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the object when the object is at one-third of the maximum value is $1.333\text{m}/\text{s}$.

To determine

The acceleration of the object when the object is at one-third of the maximum value.

Answer to Problem 15.28P

The acceleration of the object when the object is at one-third of the maximum value is $3.333\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The force required to hold the object at rest is $\text{20}\text{.0}\text{N}$, the amplitude is $0.200\text{m}$ and the mass of the object is $2.00\text{kg}$.

From part (h) the position of the object is one-third of the maximum value is $0.0667\text{m}$.

From part (a) the angular velocity is $\sqrt{50}\text{rad}/\text{s}$.

Write the expression for the acceleration of block at any position.

$a={\omega }^{2}x$

Here,

$a$ is the acceleration of the object.

Substitute $\sqrt{50}\text{rad}/\text{s}$ for $\omega$ and $0.0667\text{m}$ for $x$ in above equation.

$\begin{array}{c}a={\left(\sqrt{50}\text{rad}/\text{s}\right)}^2\left(0.0667\text{m}\right)\\ =3.333\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the acceleration of the object when the object is at one-third of the maximum value is $3.333\text{m}/{\text{s}}^{\text{2}}$.