Chapter 10, Problem 61P

(a)

To determine

A plot of the elastic potential energy versus time.

Answer to Problem 61P

The elastic potential energy is a sinusoidal curve with amplitude $116\text{mJ}$ and time period $250\text{ ms}$.

Explanation of Solution

The mass of the object is $230.0\text{g}$, the frequency of oscillation is $2.00\text{Hz}$ and the amplitude of the oscillation is $8.00\text{cm}$.

Write the expression for elastic potential energy

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

Here, $U(t)$ is the elastic potential energy, $k$ is the spring constant and $x$ is the position.

Write the given expression for position as a function of time.

$x\left(t\right)=A\sin \left(\omega t\right)$                                                    (II)

Here, $\omega$ is the angular frequency of the spring, $A$ is the amplitude of the oscillation of the spring and $t$ is the time.

Substitute (II) in (I)

$U(t)=\frac{1}{2}k{A}^2{\sin }^2\left(\omega t\right)$                                                                            (III)

Write the expression for spring constant

$k=m{\omega }^2$                                                                          (IV)

Here, $m$ is the mass of the object on the spring.

Write the expression for angular frequency

$\omega =2\pi f\left(rad/cycle\right)$                                                                          (V)

Here, $f$ is the frequency of oscillation.

Substitute $2.00\text{Hz}$ for $f$ in the above equation to find $\omega$

$\omega =\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}$

Substitute $230.0\text{g}$ for $m$, $\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}$ for $\omega$ in (IV) to find $k$.

$\begin{array}{c}k=230.0\times {10}^{-3}\text{kg}{\left(\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}\right)}^2\\ =36.3{\text{Nm}}^{-1}\end{array}$

Substitute $36.3{\text{Nm}}^{-1}$ for $k$, $8.00\text{cm}$ for $A$ and $\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}$ for $\omega$ in (III) to find $U(t)$

$\begin{array}{c}U(t)=\frac{1}{2}\left(36.3{\text{Nm}}^{-1}\right){\left(8.00\text{cm}\right)}^2{\sin }^2\left(\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}\times \text{t}\right)\\ =\frac{1}{2}\left(36.3{\text{Nm}}^{-1}\right){\left(8.00\times {10}^{-2}\text{m}\right)}^2{\sin }^2\left(\left(2\pi \text{ rad}\cdot {\text{s}}^{-1}\right)\times 2.00\text{Hz}\times \text{t}\right)\\ =\left(116\times {10}^{-3}\text{Nm}\right){\sin }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]\\ =\left(116\text{mJ}\right){\sin }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]\end{array}$

The time period of the square of a sine curve is half of the time period of the sine curve.

Write the expression for time period of the potential energy

$T_P=\frac{\pi }{\omega }$                                                                         (VI)

Here, $T_P$ is the time period.

Substitute  $\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}$ for $\omega$ in (VI) to find $T_P$

$\begin{array}{c}{T}_P=\frac{\pi }{\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}}\\ =\left(2\pi \text{ rad}\cdot {\text{s}}^{-1}\right)\times 2.00\text{Hz}\\ =\text{250}\text{ms}\end{array}$

Use the above obtained expression for $U(t)$ and $T_P$ to plot the graph.

(b)

To determine

A plot of the kinetic energy versus time.

Answer to Problem 61P

The kinetic energy curve is a cosine curve with amplitude $116\text{mJ}$ and time period $250\text{ ms}$.

Explanation of Solution

Write the expression for kinetic energy

$K(t)=\frac{1}{2}m{v}_x^2$                                                                            (VII)

Here, $K(t)$ is the kinetic energy and $v_x$ is the velocity.

Differentiate (I) to find $v_x$

$\frac{dx}{dt}=\frac{d}{dt}\left(A\sin \left(\omega t\right)\right)$

$v_x=A\omega \cos \left(\omega t\right)$                                                    (VIII)

Square (VIII) and substitute in (VII)

$K(t)=\frac{1}{2}m{\left(A\omega \right)}^2{\cos }^2\left(\omega t\right)$                                                                            (IX)

Substitute $230.0\text{g}$ for $m$, $8.00\text{cm}$ for $A$ and $\left(2\pi \text{ rad/cycle}\right)\times 2.00\text{Hz}$ for $\omega$ in (IX) to find $K(t)$

$\begin{array}{c}K(t)=\frac{1}{2}\left\{\begin{array}{l}\left(230.0\text{g}\right){\left(\left(8.00\text{cm}\right)\left(2\pi \text{ rad/cycle}\right)\left(2.00\text{Hz}\right)\right)}^2\\ {\cos }^2\left(\left(2\pi \text{ rad/cycle}\right)\left(2.00\text{Hz}\right)\text{t}\right)\end{array}\right\}\\ =\frac{1}{2}\left\{\begin{array}{l}\left(230.0\times {10}^{-3}\text{kg}\right)\\ {\left(8.00\times {10}^{-2}\text{m}\left(2\pi \text{ rad}\cdot {\text{s}}^{-1}\right)\left(2.00\text{Hz}\right)\right)}^2\\ {\cos }^2\left(\left(2\pi \text{ rad}\cdot {\text{s}}^{-1}\right)\left(2.00\text{Hz}\right)\text{t}\right)\end{array}\right\}\\ =\left(116\times {10}^{-3}\text{Nm}\right){\cos }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]\\ =\left(116\text{mJ}\right){\cos }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]\end{array}$

The time period of the square of a cosine curve is half of the time period of the cosine curve.

Write the expression for time period of the potential energy

$T_K=\frac{\pi }{\omega }$                                                                         (X)

Here, $T_K$ is the time period of the kinetic energy curve.

Equation (VI) and (X) are the same and hence $T_P$ and $T_K$ are the same.

$T_K=\text{250}\text{ms}$

Use the above obtained expression for $K(t)$ and the value of $T_K$ to plot the graph.

(c)

To determine

A plot of the total energy versus time.

Answer to Problem 61P

The total energy curve is a straight line parallel to the time-axis with a y-intercept of $116\text{mJ}$.

Explanation of Solution

Write the expression for total energy

$E(t)=U(t)+K(t)$                                                                            (XI)

Here, $E(t)$ is the total energy.

Substitute the obtained expression of $U(t)$ and $K\left(t\right)$ in (XI)

$\begin{array}{c}E(t)=\left(116\text{mJ}\right){\sin }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]+\left(116\text{mJ}\right){\cos }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]\\ =\left(116\text{mJ}\right)\left\{{\sin }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]+{\cos }^2\left[\left(12.56{\text{ s}}^{-1}\right)t\right]\right\}\\ =\left(116\text{mJ}\right)(1)\\ =116\text{mJ}\end{array}$

Thus, energy is constant and hence it is a straight line parallel to the time axis. (Refer figure 3)

(d)

To determine

The effect of a friction on the above results.

Answer to Problem 61P

All three energies of the spring approaches zero.

Explanation of Solution

When the frictional forces are included, the system is no more conservative and it is called a non-conservative system. The energy of the system is not conserved.

If the horizontal surface is not considered as frictionless, then the surface resists the oscillatory motion of the spring and thereby reducing its kinetic energy.  A part of the kinetic energy is lost in the form of heat and consequently the all three energy of the system reduces to zero.