Question

Aparticle moves along the$x$axis subject to a force toward the origin proportional to$x($say$-kx)$. Show that the particle executes simple harmonic motion (Example 3). Find the kinetic energy$\frac{1}{2}m{v}^2$and the potential energy$\frac{1}{2}k{x}^2$as functions ofand show that the total energy is constant. Find the time averages of the potential energy and the kinetic energy and show that these averages are each equal to one-half the total energy (see average values, Chapter 7, Section 4).

Short answer

The kinetic energy and potential energy are:

$\begin{array}{l}KE=\frac{1}{2}m{\omega }^2{c}^2{\cos }^2(\omega t+\gamma )\\ PE=\frac{1}{2}k{c}^2{\sin }^2(\omega t+\gamma )\end{array}$

Step-by-step solution

Given information from question

A particle moves along the -axis subject to a force toward the origin proportional to $x$.

ep 2: Kinetic energy and potential energy

Kinetic energy is given by

$K.E=\frac{1}{2}m{v}^2$

Potential energy is given by

$P.E.=\frac{1}{2}k{x}^2$

Show the particles executes simple harmonic motion

Consider a particle moves along the$x$-axis subject to a force toward the origin proportional to $x$.

The objective is to show that the particle executes simple harmonic motion.

$F\alpha kx$

$F=-kx$

Substitute $F=m\frac{d^{2}x}{d{t}^2}$in the above equation, $F=-kx$.

Thus,

$\begin{array}{c}m\frac{d^{2}x}{d{t}^2}=-kx\\ \frac{d^{2}x}{d{t}^2}+\frac{k}{m}x=0\left\{\text{dividebothsidesby}m\right\}\end{array}$

Substitute$\frac{k}{m}={\omega }^2$ in the equation $\frac{d^{2}x}{d{t}^2}+\frac{k}{m}x=0$,

Thus, the differential equation is$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$

The auxiliary equation,$m^2+{\omega }^2=0$

The roots of auxiliary equation,$m=\pm \omega i$

The solution of above differential equation is,$x\left(t\right)=c_1\cos \omega t+c_2\sin \omega t$

Rewrite the above solution $x\left(t\right)=c\sin (\omega t+\gamma )$, .

Here,$c$ and $\gamma$are arbitrary constants.

A particle whose displacement from equilibrium satisfies solution,$x\left(t\right)=c\sin (\omega t+\gamma )$

Hence, a particle is executing simple harmonic motion.

Calculate the velocity of the particle

The displacement of a particle is,$x\left(t\right)=c\sin (\omega t+\gamma )$

Differentiate$x\left(t\right)$ with respect to,$t$

$\begin{array}{c}\frac{dx\left(t\right)}{dt}=\omega c\cos (\omega t+\gamma )\\ v=\omega c\cos (\omega t+\gamma )\end{array}$

Calculate the kinetic energy and potential energy

Kinetic energy,

$\begin{array}{c}K.E=\frac{1}{2}m{v}^2\\ =\frac{1}{2}m{\left(\omega c\cos \right(\omega t+\gamma \left)\right)}^2\\ =\frac{1}{2}m{\omega }^2{c}^2{\cos }^2(\omega t+\gamma )\end{array}$

Potential energy,

$\begin{array}{c}P.E.=\frac{1}{2}k{x}^2\\ =\frac{1}{2}k{\left(c\sin \right(\omega t+\gamma \left)\right)}^2\\ =\frac{1}{2}k{c}^2{\sin }^2(\omega t+\gamma )\end{array}$

Calculate the total energy

To show that the total energy is constant.

Now total energy is:$\begin{array}{c}\text{totalenergy}=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2\\ =\frac{1}{2}m{\omega }^2{c}^2{\cos }^2(\omega t+\gamma )+\frac{1}{2}k{c}^2{\sin }^2(\omega t+\gamma )\\ =\frac{1}{2}k{c}^2\left({\cos }^2(\omega t+\gamma )+{\sin }^2(\omega t+\gamma )\right)\\ =\frac{1}{2}k{c}^2\end{array}$

Therefore, the total energy$=\frac{1}{2}k{c}^2=$ const.

Therefore, the total energy is constant.

The kinetic and potential energy

The objective is to find the time averages of the kinetic energy and potential energy.

Now time averages of kinetic energy

$\begin{array}{c}\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\left(KE\right)dt=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\frac{1}{2}m{\omega }^2{c}^2{\cos }^2(\omega t+\gamma )dt\\ =\frac{1}{4\pi }m{\omega }^2{c}^2{\int }_{-\pi }^{\pi }{\cos }^2(\omega t+\gamma )dt\\ =\frac{1}{4\pi }m{\omega }^2{c}^{2}k{c}^2\\ =\frac{1}{4}k{c}^2\end{array}$

Time averages of potential energy:

$\begin{array}{c}\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\left(PE\right)dt=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }\frac{1}{2}k{c}^2{\sin }^2(\omega t+\gamma )\\ =\frac{1}{4\pi }k{c}^2\\ =\frac{1}{4}k{c}^2\end{array}$

Since average of KE =average of $PE=\frac{1}{2}$total energy

$\begin{array}{c}\frac{1}{4}k{c}^2=\frac{1}{4}k{c}^2\\ =\frac{\frac{1}{2}k{c}^2}{2}\end{array}$

Therefore, the both of these time averages of kinetic energy and potential energy are equal to one-half total energy.