# 2.2.3: Energy of an Oscillator

- Page ID
- 84581

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The energy of simple oscillating objects, such as a pendulum, a weight suspended on a spring, or a vibrating guitar string is worth taking a closer look at, even though such things have no direct application in the grand-scale energy usage that is the main subject of this course. Namely, in the oscillating system mentioned above one can clearly see the processes of converting one form of energy to another. Consider a simple pendulum, i.e., a weight of mass \( m\) suspended on a piece of a light string. Pull the weight to the right from the equilibrium position, as shown in this short video clip

By doing that, you "charged" the mass \( m\) with potential energy \( \Delta V\), because it's now slightly higher than it was at the equilibrium position. Now, you let it go - and and the potential energy starts gradually changing to kinetic energy. When the mass passes through the equilibrium point, all the acquired potential energy is gone, it has been entirely converted to kinetic energy. **By the way, here is a small challenge for you: **you may use \( K = \Delta V\) and process it further in order to find the speed \( v\) of the mass \( m\) when it passes through the equilibrium point -- assume that the length of the pendulum string is, say 1 meter, and in the initial displaced position the string made an angle of 20 degrees with a vertical line; please try, it's a good exercise. I am not giving you the value of \( m\) because, as you will find out, you don't need to know it. It's a good exercise - and if you have a major problem with obtaining the solution, it will probably mean that you may need some extra studying of very elementary physics in order to do well in this course).

Back to the pendulum: after the initial potential energy is totally lost and totally converted to kinetic energy at the instant of passing through the equilibrium point, the pendulum continues its swing motion, but now the kinetic energy is gradually converted back to potential energy. The process lasts until the pendulum reaches the maximum displacement angle, equal to the initial displacement angle -- but now to the left. At this instance, the kinetic energy is totally gone, the speed of the mass \( m\) becomes zero again, and the initial potential energy \( \Delta V\) is totally "regained". And from this moment on, the process described above restarts, and runs in the opposite direction. A full swing from the right to the lest and back to the right is called "the oscillation cycle", and the time it takes to make such a full cycle is called "the oscillation period".

In the vibration of a guitar string, or in the oscillations of a blob suspended on a spring, you have exactly the same periodic process of conversion of potential energy to kinetic energy and ~back ~to ~potential ~energy. Yet, as mentioned above, the oscillations of such objects do not play any significant role in the grand-scale energy usage, why do we talk about them over here? Well, the reason is simple -- because the energy of oscillations of **atoms** in solids, the sum over all atoms, is what constitutes the **thermal energy** contained in such objects.

## Free Fall

If we raise a body above the Earth surface at a distance \( H\), and let it go -- then, no matter what the value of \( H\) is, one meter or one million meters, the body starts "falling freely" because it is attracted towards the Earth center by the force of gravity. And, if we forget about the air drag force (which is relevant only for \( H\) values of a few tens of kilometers), then we certainly have the right to say: "A free fall is the motion under the influence of the force of gravity, and no other force".

Now, suppose that the in initial velocity of the body considered is not zero, but we give it a "push" in the direction perpendicular to the line drawn between the initial body's position, and the Earth center. Let's call this line "the \( z\) direction", and the speed along this line as "the vertical velocity \( v_{\rm z}\)". Let's call the direction perpendicular to the \( z\) line as"the \( x\) axis", and the speed along the \( x\) axis as "the horizontal velocity \( v_{\rm x}\)". So, at the initial moment \( v_{\rm z}=0\), and \( v_{\rm x}\ne 0\). What happens next? The body starts moving downwards, with constant acceleration \( g\), under the influence of gravity. But there is no horizontal force acting on it, so the initial \( v_{\rm y}\) remains unchanged. The resultant body's trajectory is a parabola, and it hits the Earth surface some distance from the point right below the initial body's position.

In introductory-level physics textbooks the type of motion described above is usually referred to as "projectile motion"}. Three different sub-types are discussed, with the initial "push" directed strictly horizontally, at some angle upwards, or at some angle downwards. Each sub-type has to be treated slightly differently. It's a good approach, instructive and pedagogical: it helps young students to develop an ability called "physical insight", in which a correct "conceptual understanding" of a physical phenomenon considered is combined with a correct choice of "mathematical tools" needed to analyze it in quantitative terms. However, in more advanced textbooks on mechanics one may not find such a distinction: all the types of motion listed above are simply classified as a **free fall**. How comes? Well, as was said, a free fall is the motion of a body under the influence of gravitational forces **only**. And are there forces other than the "gravitational pull" involved in a vertical downward fall, and in the "projectile motion"? No, definitely not!

For obtaining a mathematical description of the motion of a free-falling body, in each case one has to do the same -- namely, to solve the so-called "equation of motion", which always has the familiar form: \( F = m\cdot a\), i.e., it's nothing else that what we call the "Newton's Second Law of Dynamics". The only trick is that the force and the acceleration have to be taken in a **vector form**:

\[

\vec{F} = m\cdot \vec{a}

\]

In the most general case an object moves in a three-dimensional space. Hence, we write both

\)\vec{F}\) and \( \vec{a}\) as three-component vectors:

\[

(F_{\rm x}, F_{\rm y}, F_{\rm z}) = m\cdot (a_{\rm x}, a_{\rm y}, a_{\rm z})

\]

which can be rewritten as a set of three equations for individual components of the force and acceleration vectors:

\[\begin{array}{l}

F_{\rm x}&=&m\cdot a_{\rm x} \\

F_{\rm y}&=&m\cdot a_{\rm y} \\

F_{\rm z}&=&m\cdot a_{\rm z}

\end{array}\]

Now, recall that the acceleration, by definition, is the second time-derivative of the time-dependent position of the object. We can then rewrite the above set as^{3}:

\[\begin{array}{l}

\dfrac{d^2 x}{dt^2}&=&\dfrac{F_{\rm x}}{m} \\

\dfrac{d^2 y}{dt^2}&=&\dfrac{F_{\rm y}}{m} \\

\dfrac{d^2 z}{dt^2}&=&\dfrac{F_{\rm z}}{m}

\end{array}\]

This is a set of three **differential equations**. You may not know how to solve such equations -- many students, who are interested in "Energy Alternatives", never take math courses at such level, and those who do, take it towards the end of their Sophomore year, or at the Junior

year. So, why are the equations mentioned in this text? Well, to advertise how mighty they are! One can use the very same set of equations to solve a vertical fall, all types of projectile motions -- and this is not the end of the list: also, the motion of satellites, on circular orbits as well as on elongated elliptic orbits, the motion of planets, moons, asteroids and comets. The very same set of equations for all problems listed!^{1} How can it be? It's simple, because all these types of motion have one thing in common -- the only force acting on the moving object is the force of gravity!

Presenting the solution procedures for specific problems listed above would take too much space in this chapter. However, the Author of this text does not like to make statements, and then ask: "Please accept without proof that the said statement is true". So, several solution procedures will be presented in detail, but in the Appendices. Yet, the Appendices are added to the text with the intention of satisfy the curiosity of students who like math -- but, definitely, reading them is not necessary for students whose goal is only to attain a good \textbf{conceptual} understanding of the course material.

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1. It’s an average value – Earth is not a perfect sphere, it is somewhat “flattened” at the poles, so that the weight of a body may be slightly different, depending on the geographical region where it is measured; also, the spinning of Earth contributes to a gradual decrease of the body’s weight when it is moved from a pole towards the Equator. Bur the “weight loss” of a body between a pole and the Equator is only about 0.3%, so that with a good approximation one can neglect such effects.

2. \footnote{Earth is not a perfect sphere. The value of the radius we use here, 6371 km, is an average value. But the distance from the Earth center at different geographical regions may be slightly larger or slightly smaller than the average value, so that the actual values may slightly differ from the result obtained from the Eq. 2.15.

3. These equations are a powerful tool -- solving motions resulting from the action of gravity only is just one application, the equations are generally valid, no matter what's the origin of the