Chapter 13, Problem 32P

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency ω, the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (c) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the particle when t = 0.500 s.

To determine

The force constant of the spring.

Answer to Problem 32P

The force constant of the spring is $250\text{N}\cdot {\text{m}}^{\text{-1}}$.

Explanation of Solution

Given info: The force applied $F=7.50\text{N}$. The spring is stretched to $3.00\text{cm}$.

Explanation:

From the hook’s law, the restoring force is,

$F=-kx$

• $F$ is the restoring force
• $k$ is the force constant of the spring
• $x$ is the displacement from the equilibrium position

The magnitude of the force constant will be,

$k=\frac{F}{x}$

Substituting $F=7.50\text{N}$ and $x=3.00\text{cm}$

$\begin{array}{c}k=\frac{\left(7.50\text{N}\right)}{\left(3.00\text{cm}\right)}\\ =\frac{\left(7.50\text{N}\right)}{\left(3.00\times {10}^{-2}\text{m}\right)}\\ =250\text{N}\cdot {\text{m}}^{\text{-1}}\end{array}$

Conclusion: The force constant of the spring is $250\text{N}\cdot {\text{m}}^{\text{-1}}$.

To determine

The angular frequency, the frequency, and the time period of the oscillation.

Answer to Problem 32P

The angular frequency is $22.4\text{rad}\cdot {\text{s}}^{\text{-1}}$, the frequency is $3.56\text{Hz}$ and the period of the oscillation is $0.281\text{s}$.

Explanation of Solution

Given info: The mass of the particle is $0.500\text{kg}$ The particle is released at $x=5.00\text{cm}$ from rest, so the amplitude of the oscillation is $5.00\text{cm}$.

Explanation:

The angular frequency is defined as,

$\omega =\sqrt{\frac{k}{m}}$ (1)

• $\omega$ is the angular frequency
• $k$ is the force constant
• $m$ is the mass of the particle

Substituting $m=0.500\text{kg}$ and from section (a) $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$,

$\begin{array}{c}\omega =\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\\ =22.4\text{rad}\cdot {\text{s}}^{\text{-1}}\end{array}$

The frequency of the oscillation is given by,

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

• $f$ is the frequency of the oscillation
• $k$ is the force constant
• $m$ is the mass of the particle

Substituting $m=0.500\text{kg}$ and from section (a) $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$,

$\begin{array}{c}f=\frac{1}{2}\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\\ =3.56\text{Hz}\end{array}$

The time period is defined by,

$T=\frac{1}{f}$

Substituting $f=3.56\text{Hz}$,

$\begin{array}{c}T=\frac{1}{\left(3.56\text{Hz}\right)}\\ =0.281\text{s}\end{array}$

Conclusion: The angular frequency is $22.4\text{rad}\cdot {\text{s}}^{\text{-1}}$, the frequency is $3.56\text{Hz}$ and the period of the oscillation is $0.281\text{s}$.

To determine

The total energy of the system.

Answer to Problem 32P

The total energy of the system is $0.313\text{J}$

Explanation of Solution

Given info: The particle is displaced from the origin to $x=5.00\text{cm}$ and released from rest at $t=0$.

Explanation:

The total energy of a simple harmonic system is

$E=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

• $E$ is the total energy
• $k$ is the force constant
• $x$ is the displacement from equilibrium position
• $m$ is the mass of the particle
• $v$ is the velocity

Substituting $x=5.00\text{cm}$, $v=0$ and $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$

$\begin{array}{c}E=\frac{1}{2}\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right){\left(5.00\text{cm}\right)}^2+0\\ =0.313\text{J}\end{array}$

Conclusion: The total energy of the system is $0.313\text{J}$.

To determine

The amplitude of the system.

Answer to Problem 32P

The amplitude is $5.00\text{cm}$.

Explanation of Solution

Given info: The force constant of the spring is $250\text{N}\cdot {\text{m}}^{\text{-1}}$. The total energy of the system is $0.313\text{J}$.

Explanation:

The total energy of a simple harmonic system is stored as elastic potential energy when the system is in the turning points.

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

• $E$ is the total energy
• $k$ is the force constant
• $A$ is the amplitude

On re-arranging for the amplitude

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

Substituting $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$ and $E=0.313\text{J}$

$\begin{array}{c}A=\sqrt{\frac{2\left(0.313\text{J}\right)}{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}}\\ =5.00\times {10}^{-2}\text{m}\\ \text{=5}\text{.00}\text{cm}\end{array}$

Conclusion: The amplitude of the oscillation is $\text{5}\text{.00}\text{cm}$.

To determine

The maximum velocity and the maximum acceleration of the system.

Answer to Problem 32P

The maximum velocity of the system is $1.12\text{m}\cdot {\text{s}}^{\text{-1}}$. The maximum acceleration of the system is $25.0\text{m}\cdot {\text{s}}^{\text{-2}}$.

Explanation of Solution

Given info: The total energy of the system is $0.313\text{J}$. The mass of the object is $0.500\text{kg}$.

Explanation:

When the system is in equilibrium position, the total energy is in the form of kinetic energy,

$E=\frac{1}{2}m{v}_{\max }^2$

Re-arranging

$v_{\max }=\sqrt{\frac{2E}{m}}$

Substituting $m=0.500\text{kg}$ and $E=0.313\text{J}$

$\begin{array}{c}{v}_{\max }=\sqrt{\frac{2\left(0.313\text{J}\right)}{\left(0.500\text{kg}\right)}}\\ =1.12\text{m}\cdot {\text{s}}^{\text{-1}}\end{array}$

The maximum force is given by,

$F_{\max }=kA$ (1)

• $k$ is the force constant
• $A$ is the amplitude

Using Newton’s second law,

$F_{\max }=m{a}_{\max }$ (2)

From (1) and (2)

$a_{\max }=\frac{kA}{m}$

Substituting $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$ $A\text{=5}\text{.00}\text{cm}$ and $m=0.500\text{kg}$,

$\begin{array}{c}{a}_{\max }=\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)\left(\text{5}\text{.00}\text{cm}\right)}{\left(0.500\text{kg}\right)}\\ =25.0\text{m}\cdot {\text{s}}^{\text{-2}}\end{array}$

Conclusion: The maximum velocity of the system is $1.12\text{m}\cdot {\text{s}}^{\text{-1}}$. The maximum acceleration of the system is $25.0\text{m}\cdot {\text{s}}^{\text{-2}}$.

To determine

The displacement x of the particle from the equilibrium position $t=0.500\text{s}$.

Answer to Problem 32P

The displacement is $0.919\text{cm}$.

Explanation of Solution

Given info: The mass of the object $m=0.500\text{kg}$. The force constant of the spring $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$. Amplitude $A\text{=5}\text{.00}\text{cm}$, time $t=0.500\text{s}$

Explanation:

The general expression of position as a function of time for an object moving in simple harmonic motion is given by,

$x=A\cos \left(\omega t\right)$ (1)

• $x$ is the position
• $A$ is the amplitude of the simple harmonic motion
• $\omega$ is the angular frequency
• $t$ is time taken

The angular frequency is defined as,

$\omega =\sqrt{\frac{k}{m}}$ (2)

• $\omega$ is the angular frequency
• $k$ is the force constant
• $m$ is the mass of the particle

From (1) and (2)

$x=A\cos \left(\left(\sqrt{\frac{k}{m}}\right)t\right)$

Substituting $m=0.500\text{kg}$, $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$, $A\text{=5}\text{.00}\text{cm}$, $t=0.500\text{s}$

$\begin{array}{c}x=\left(\text{5}\text{.00}\text{cm}\right)\cos \left(\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)\left(0.500\text{s}\right)\right)\\ =0.919\text{cm}\end{array}$

Conclusion: The displacement from the equilibrium position is $0.919\text{cm}$.

To determine

The velocity and acceleration of the particle at $t=0.500\text{s}$.

Answer to Problem 32P

The velocity is $1.10\text{m}\cdot {\text{s}}^{\text{-1}}$ and the acceleration is $-4.59\text{m}\cdot {\text{s}}^{\text{-2}}$

Explanation of Solution

Given info: The mass of the object $m=0.500\text{kg}$. The force constant of the spring $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$. Amplitude $A\text{=5}\text{.00}\text{cm}$, time $t=0.500\text{s}$.

Explanation:

The general expression of position as a function of time for an object moving in simple harmonic motion is given by,

$x=A\cos \left(\omega t\right)$ (1)

• $x$ is the position
• $A$ is the amplitude of the simple harmonic motion
• $\omega$ is the angular frequency
• $t$ is time taken

To find the expression for velocity differentiate (1)

$v=-A\omega \sin \left(\omega t\right)$ (2)

To find the expression for acceleration differentiate (1) twice

$a=-A{\omega }^2\cos \left(\omega t\right)$ (3)

The angular frequency is defined as,

$\omega =\sqrt{\frac{k}{m}}$ (4)

• $\omega$ is the angular frequency
• $k$ is the force constant
• $m$ is the mass of the particle

Hence,

$v=-A\left(\sqrt{\frac{k}{m}}\right)\sin \left(\left(\sqrt{\frac{k}{m}}\right)t\right)$

$a=-A{\left(\sqrt{\frac{k}{m}}\right)}^2\cos \left(\left(\sqrt{\frac{k}{m}}\right)t\right)$

Substituting $m=0.500\text{kg}$, $k=250\text{N}\cdot {\text{m}}^{\text{-1}}$, $A\text{=5}\text{.00}\text{cm}$, $t=0.500\text{s}$, the velocity will be,

$\begin{array}{c}v=-\left(\text{5}\text{.00}\text{cm}\right)\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)\sin \left(\left(0.500\text{s}\right)\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)\right)\\ =-\left(\text{5}\text{.00}\times {\text{10}}^{-2}\text{m}\right)\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)\sin \left(\left(0.500\text{s}\right)\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)\right)\\ =1.10\text{m}\cdot {\text{s}}^{\text{-1}}\end{array}$

Similarly the acceleration,

$\begin{array}{c}a=-\left(\text{5}\text{.00}\text{cm}\right){\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)}^2\cos \left(\left(0.500\text{s}\right)\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)\right)\\ =-\left(\text{5}\text{.00}\times {\text{10}}^{-2}\text{m}\right){\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)}^2\sin \left(\left(0.500\text{s}\right)\left(\sqrt{\frac{\left(250\text{N}\cdot {\text{m}}^{\text{-1}}\right)}{\left(0.500\text{kg}\right)}}\right)\right)\\ =-4.59\text{m}\cdot {\text{s}}^{\text{-2}}\end{array}$

Conclusion: The velocity of the oscillator is $1.10\text{m}\cdot {\text{s}}^{\text{-1}}$ and the acceleration is $-4.59\text{m}\cdot {\text{s}}^{\text{-2}}$.