Chapter 11
`
`Orbital Motionsand the Hydrogen Atom
`
`1L5
`
`Atomic Spectra and the Boke Theory of Hydrogen
`
`321
`
`
`
`Figure 11.16 A sinusoidal wave
`traveling to the right. Any point on
`the wave moves a distance of one
`wavelength, A. in a time equal to
`the period of the wave,
`
`486.1
`
`6565
`
`434.1
`410.20
`—+ i(nm)
`Figure 11.17 A series of spectral
`lines for atomic hydrogen. The
`prominentlines labeled are part of
`the Balmer series.
`
`
`
`atomic spectra that included some features ofthe currently accepted theory. Using
`the simplest atom, hydrogen, Bohr described a model of what he thought must be
`the atom’s structure. His model of the hydrogen atom contains some classical fea-
`tures as well as some revolutionary postulates that could not be justified within the
`framework of classical physics.’ The basic assumptions of the Bohr theory as it
`applies to the hydrogen atom are as follows:
`1. Theelectron moves in circular orbits about the proton under the influence
`
`of the Coulombforce ofattraction, as in Figure 11.18.
`3
`:
`es
`:
`-
`2. Only certain electron orbits are stable. These are orbits in which the hydro-
`gen atom does not emit energyin the form of radiation, Hence, the total
`energyof the atom remains constant, and classical mechanics can be used
`to describe the electron’s motion.
`3. Radiationis emitted by the hydrogen atom whentheelectron “jumps” from
`amore energeticinitial state to a lower state, The jump cannot be visualized
`
`
`or treated classically. In particular, the frequency, /- of theradiation
`tted
`is depicted in Figure 11.16. The distance betweentwo consecutive crests of the wave
`
`in the jump is related to the change in the atom’s energyand is indepen-
`is called the wavelength, A. As the wave travels to the right with speed », any point
`dent of the frequency of the electron's orbital motion. The frequency of
`on the wave travels a distance of one wavelength in a time interval of one period,
`the emitted radiation is found from
`T (the time for one eyele), so the wave speedis v = A/T. The inverse of the period,
`
`1/7, is called the frequency, /, of the wave) it represents the numberof cycles per
`[11.15]
`E,— E,= bf
`second, Thus, the speed of the wave is often written as v= Af In this section,
`because we shall deal with electromagnetic waves—which travel at the speed of
`
`where £; is the energyofthe initial state, £, is the energy of the final state,
`light, -—the appropriate relation is
`his Planck's constant (see Section 10.9), and E,>E,.
`c= Af
`(11.13)
`4. The size of the allowedelectronorbits is determined by a condition imposed
`on the electron’s orbital angular momentum: the allowed orbits are those
`
`The emission spectrum of hydrogen shownin Figure 11.17 includes four prom-
`for which the electron’s orbital angular momentum about the nucleus is an
`inent lines that occur at wavelengths of 656.3 nm, 486.1 nm, 434.1 nm, and 410.2
`integral multiple ofA = fy
`nm. In 1885, Johann Balmer (1825-1898) found that the wavelengths of these and
`
`
` less prominent lines can be described by this simple empirical equation:
`
`
`
`f
`
`|
`|
`|
`|
`|
`|
`1
`|
`
`
`
`;
` * Assumptions of the Bohr
`theary
`
`\r
`
`2
`Figure 11.18 Diagram represent-
`ing Bahr’s model of the hydrogen
`tronis allowed to be only in spe-
`atomin which the orbiting elec-
`cific orbits of discrete raclii.
`:
`
`(1.17)
`
`5 singly ionize
`Je
`spectra of t
`
`
`
`
`
`mr = nh
`n= 1,2,3,...
`(11.16)
`ooo
`(11.14)
`Balmerseries
`i = bu (5 ea =)
`Niels Bohr (|885-1962)
`Using these four assumptions, we can calculate the allowed energy levels and
`
`
`A
`eon
`A Danish physicist, Bohr was an ac-
`emission wavelengths of the hydrogen atom. Theelectr
`potential energy ofthe
`.
`.
`.
`, and Ry
`is a constant, now called
`where n may have integral values of 3,4, 5,
`tive participant in the early develop:
`system shown in Figure 11.18 is given by U, = — 4,e*/r, wherek, is the Coulomb
`If the wavelengthis in meters R. ete et
`the Rydberg
`t
`ment of quantum mechanics and pro
`constant, ¢ is the charge on theelectron
`d ris the electron—protonseparation.
`Yer
`aeed
`Thus, the total energyof the atom, which contains both kinetic and potential energy
`ies
`|
`Ry = 1.0973732 x 107 m=!
`19300, Bohr a eo ae kor
`terms,is
`ieea Thefirst line in the Balmerseries, at 656.3 nm, corresponds to n = 3 in Equation
`4
`
`of the world's best physicists and pro-
`11.14; the line of 486.1 mm corresponds to » = 4; and so on,
`vided a forum for the exchange of
`In addition to emitting light at specific wavelengths, an clement can also absorb
`‘
`.
`is
`wi
`\
`ideas. When Bohrvisited the United
`,
`:
`i
`light at specific wavelengths. The spectral lines corresponding to this process form
`
`
`
`hates in 19°39tscliend @ scientific 7 : = Applying Newton's second law to this system, we see that the Coulombattractive
`
`
`
`
`force on the electron, A?/1* (Eq. 5.15), must equal the mass times the centripetal
`eetien thatthe
`what = known a absorption eo An absorption eos can be ob-
`fission of uranium had been discov-
`tainedby passing a continuousradiationspectrum(one containing all wavelengths)
`acceleration (@ = v7 /1) of the electron
`ered by Hahn and Strassman in Ber-
`through a vapor of the element being analyzed. The absorption spectrum consists
`:
`lin, The results were the foundations of
`of a series of dark lines superimposed on the otherwise continuous spectrum.
`ke met
`At the beginning of the twentieth century, scientists were perplexed by the
`theatomic bomb developed in the
`Somn ee
`
`
`the characteristics of spectra, Why did atoms
`beenencasewate
`failure of classical physics to explain
`2
`:
`of a given element emit only certain wavelengths of radiation, so that the emission
`. er
`prize for his investigation of the struc-
`2
`.
`The Bohr model can be applied successfully to such hydrogen
`+ and of the radiation em-
`spectrumdisplayed discrete lines? Furthermore, why did the atoms absorb only
`theory does not properly describy
`doubly ionized lithium, but
`the
`from them,
`those wavelengths that they emitted? In 1913 Bohr provided an explanation of
`
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`

`Chapter I]
`
`Orbital Motions and the Hydvogen Atom
`
`115
`
`Atomic Spectra and the Bohr Theory of Hydrogen
`
`Fromthis expression, we find the kinetic energyto be
`hee
`y= 1 a
`{11.18]
`or
`K=
`hm"
`
`Substituting this value of Kinto Equation 11.17, we find that the total energy of the
`atomis
`
`Figure 11.20also shows other spectral series (the Lyman series and the Paschen
`
`series) that were foundafter Balmer’s discovery. These spectra obey other empirical
`formulas, which are reconciled with the Bohr model.
`Equation 11.22, together with Bohr's third postulate, can be used to calculate
`the frequencyofthe radiation that is emitted when theelectron jumps from an
`outer orbit to. an inner orbit:
`
`Total energy ofthe hydrogen
`atom
`
`
`
`he*
`y
`
`(11.19)
`
`
`
`[11.24]
`
`a
`
`&
`4
`‘
`
`2
`
`323
`
`E (eV)
`— 0.00
`
`T 9.54
`0.85
`it
`a 151
`Paschen
`series
`
`Balmer
`
`3.40
`
`
`
`
`
`Figure 11.20 An energy level
`
`gramfor hydrogen. Thediserete
`allowed energies are plotted on the
`vertical axis. Nothing is ploued on
`the horizontal axis, but the hori-
`zontal extent of the diagramis
`made large enough to show al-
`lowed transitions. Quantum num-
`bers are given on the left
`
`CONCEPTUAL PROBLEM 4
`
`Under normal experimental conditions, hydrogen atoms show more lines in emission than
`im absorption. Why
`
`CONCEPTUAL PROBLEM 5
`Suppose the energ
`of the electron in a hvelrogen aton
`What is the potential energy?
`
`< What
`
`is the kinetic ener
`
`
`
`is convenient
`
`to convert
`
`Because the quantity being measured is wavelength, it
`frequencyto wavelength, using ¢ = /A, to get
`
`[11.25]
`
`
`
`a generalized form of the empirical relatic
`Eq. 11.14),
`
`n*
`
`
`[11.26]
`)
`eak
`providedthat the combinationof constants 4" /2a,/c is equal to the experimentally
`
`determined Rydberg cc
`ut. After Bohr demonstrated the agreement of these
`twe quantities to-a precision of about 1%, it was soon recognized as the crowning
`achievement ofhis new theory of quantum mechanics. Furthermore, Bohr showed
`that all of the speciral series for hydrogen have a natural
`interpretation in his
`
`gy levels.
`theory. Figure 11.20 shows these spectral series as transitions betweener
`which
`
`
`Bohr immediately extended his model for hydrogen to other elements
`all
`but one
`electron
`had
`been
`removed.
`lonized
`elements
`such
`as
`He’, Li**, and Be*’ were suspected to exist in hot stellar atmospheres, wherefre-
`quent atomic collisions occur with enough energy to completely remove one or
`more atomic electrons. Bohr showed that many mysterious lines observed in the
`
`Sun and several
`s could not be due to hydrogen but were correctly predicted
`by his theoryif attributedto singly ionized helium.
`Although the Bohr model had somesuccess in predicting the spectra of single
`valence ele
`‘on atoms, it has someserious drawbacks. For example, it cannot ac-
`count for
`the visible spectra of more complex atoms, and it is unable to predict
`manysubtle spectral details of hydrogen andother simple atoms.
`
`1=
`
`I p
`
`Radii of Bohr orbits in *
`
`The Bohr radius *
`
`
`
`n= 1,23,
`Figure 11.19 The first three
`Bohr orbits for hydrogen.
`
`Note that the total energy is negative, indicating a bound electron —proton system.
`This means that energy in the amount ofk,¢*/2r must be added to the atom just to
`remove the electron and make the total energy zero, An expression forr, the radius
`of the allowed orbits, can be obtained by eliminating v by substitution between
`Equations 11,16 and 11,18;
`
`i a?
`mi Ae
`This result shows that the radii have discrete values, or are quantized.
`Theorbit for which m = 1 has thesmallest radius; itis called the Bohr radius,
`ay, and has the value
`
`n= 1,8, Ss
`
`[11.20]
`
`(11.21)
`
`(11.22)
`
`1) 2 [11.23]
`
`
`> = 0.0529 nm
`dp =
`2 mke*
`Thefirst three Bohr orbits are showntoscale in Figure 11.19.
`The quantization of the orbit radii immediatelyleads to energy quantization.
`This can be seen by substituting 7, = "a, into Equation 11.19. The allowed energy
`levels are found to be
` n= 1,2, 56005
`Insertion of numerical values into Equation 11.22 gives
`
`(Recall from Section 9.7 that 1 eV = 1.6 * 10°"J.) The lowest stationarystate
`corresponding to m= 1, called the ground state, has an energy of
`EF, =
`
`a
`—13.6eV. The next state, the first excited state, has » = 2 and an energy of
`
`
`E.
`— 3.4 eV. Figure 11.20 is an energytevel diagram showing the ener-
`gies of these discrete energystates and the corresponding quantum numbers. The
`
`uppermost level, corresponding ton = © (or r= ©) and # = 0, represents thestate
`for which the electron is removed from the atom. The minimum energy required
`
`to ionize the atom(that is,
`to completely remove an electron in the groundstate
`fromthe proton’s
`influence)is called the ionization energy. As can be seenfrom
`
`Figure 11,20, the ionization. energy for hydrogen, based on Bohr’s calculation, is
`se
`
`
`13.6 eV. This constituted another major achievement for
`the Bohr theory, bec:
`
`the ionization
`energy for hydrogen had already been measured to be precisely
`13.6 eV.
`
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`
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`

`
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`

`
`
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`Petitioner Intel Corp., Ex. 1038
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`

`

`
`
`IPR2023-00783
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`

`

`
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`

`

`12.1 + SIMPLE HARMONIC MOTION
`fa force acting on a body varies in time, the acceleration of the body also
`
`changes with time. Ifthis force always acts toward the equilibrium position
`said to exhibit simple harmonic motion when
`A particle moving along the x axis is
`of the body, a repetitive back-and-forth motion about that position results,
`x, its displacement from equilibrium, varies in time according to the relationship
`The motion is an example of what is called periodic or oscillatory motion,
`You are most likely familiar with several examples of periodic motion,
`x= Acos(wl + )
`[12.1]
`such as the oscillationsofa
`
`
`conse
`mass on a spring, the mo-
`s of the motion. In order to give physical significance
`where A, @, and dar
`tion of a pendulum,and
`to these constants, it is convenient to plot x as a function of f, as in Figure 12.1,
`the vibrations of a stringed
`First. we note that A, called the amplitude of the motion,
`is simply the maximum
`musical
`instrument. Nu-
`displacement of the particle in either the positive or negative x direction. The
`merous systems exhibit o%-
`constant @ is called the angular frequency (defined in Eq. 12.4). The constant
`cillatory motion, For exam-
`angle discalled the phase constant (or phase angle) and, along with the amplitude
`ple. the molecules in a solid
`A, is determined uniquely by theinitial displacement and velocity ofthe particle
`oscillate about their equi-
`The constants @ and Atell us what the displacement and velocity were at
`time
`librium) positions; electro-
`(= 0. The quantity (! + @)
`is called the phase of
`the motion andis useful for
`magnetic waves,
`such as
`
`comparing the motions of two particles. Note that the function x is periodic and
`
` dio waves, are characterized
`repeats itself each time ei increases by 27 radians.
`Figure 12.1. Displacement versus
`time for a particle undergoing sim-
`Theperiod, T, of the motionis the time required for theparticle to go through
`by oscillating electric and
`ple harmonic motion.
`The ampli-
`
`one full cycle of its motion. That is, the value of x at time ft equals thevalueof x at
`
`
`
`tude of the motionis A and the pe-
`
`time 1+ FT, We can 7=277/w by using thefactshow that the period is given by
`
`
`riod is 7
`that the phase increases by 27 radians in a time of T:
`Time exposure ofa pendulum,
`at+ b+ er = w(t+ 1) +o
`which consists of a small
`sphere suspended by a light
`Simplifying this expression, we see that wT=27, or
`string. The time ittakes the
`pendulum to undergo one com-
`pleteoscillation is called the
`period of its mation, and the
`maximum displocement of the
`pendulum from the vertical po-
`sition is called the amplitude,
`As we shall see in this chapter,
`the period of its mation far
`small amplitudes depends only
`on the length of the pendulum
`and the value of the free-fall
`acceleration.
`[ames Stevansan/
`SPL/Phato Researchers)
`
`CHAPTER OUTLINE
`12 Simple Harmonic Motion
`of a Mass Attached to
`
`12.3 Energyof the Simple
`Harmonic Oscillator
`12.4 Motion of a Pendulum
`12.5 Damped Oscillations
`(Optional)
`12.6 Forced Oscillations
`(Optional)
`
`
`
`
`
`
`
`
`
`Oscillatory Motion
`
`magnetic field vectors; in alternating-currentcircuits, voltage, current, andelectri-
`cal charge vary periodically with time
`Muchof the material in this chapterdeals with simple harmonic motion. This
`special kind of motion occurs when a restoring force acts on a body. In simple
`hi monic motion, an object oscillates betweentwo spatial positionsfor an indefinite
`
`
`period of time. In real mechanical systems, retarding (frictional) forces are always
`
`present. Such forces reduce the mechanical energy of the system with time, and
` the os
`ons are said to be damped. If an external driving forceis applied, wecall
`the motiona forced oscillation
`
`12.1
`
`Simple Harmonic Matton
`
`333
`
`* Displacement versus time for
`simple harmonic motion
`
`
`
`T=—
`
`[12.2]
`
`The inverse of the period is called the frequency of the motion, f The fre-
`quencyrepresents the numberof oscillations the particle makes per unit time:
`
`123
`i)
`I
`[12.3]
`fon
`i
`The units of fare cycles per second, or hertz (Hz). Rearranging Equation 12.3 gives
`2
`a=oS
`
`[12.4]
`
`* Frequency
`
`* Angular frequency
`
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`

`336
`
`Chapter 12)
`
`Oscillatory Motion
`
`337
`
`Motion of a Mass Attached to a Spring
`12.2)
`
` at, its speed is a maximum. [t then
`x = 0 and its acceleration
`is zero. At this in
`
`
`am— = =4.00r cos|a+ =|— (7 : i :
`
`
`EXERCISE 2) Determine the maximum speed and the max-
`dv
`w\d
`
`imumacceleration of the body,
`di
`4/
`dt
`Answer
`From the gen-
`continues totravel to the left of equilibrium andfinally reaches x = — A, at which
`timeits acceleration is kA/m(
`we) and its speed is again zero, Thus,
`
`
`eral expressions for v and a foundim part (bj, we see that
`rr
`imum values of the sine and cosine functions
`= — (4.000r" m,/‘s*)cos (w+ a
`the
`we see that the mass oscillates between the turning points x= + A. In onefull cycle
`of its motionit travels a distance of
`4A.
`are unity. Therefore, v varies between > 4.007 m/s, and a
`’
`
`,
`:
`varies between $4,007? m/s*, Thus, ty, = 4.00 m/s and
` that, by defi-
`
`Let us now describe this motion
`in a quantitative fashion. Reca
`ni
`
`ing the results to part (b), determine the
`EXERCISE 1
`Goan
`= 4.00772 m/s?
`position, velocity, and acceleration of the body att = 1.005,
`F
`
`
`-2.83 m: v = 8.89 m/s; a = 27.9 m/s?
`Ans
`x=
`
`
`[12.15]
`
`
`
`EXERCISE 3 A particle moving with simple harmonic motion travels a total distance of
`20.0 cmin each cycle of its motion, and its maximum acceleration is 50.0 m/s. Find (a) the
`
`angular frequency of the motion and (b) the maximumspeed of the particle.
`Answer
`(a) 31.6rad/s
`(b) 1.58 m/s
`
`12.2 + MOTION OF A MASS ATTACHED TO A SPRING
`
`Lf we denote the ratio &/mwi
`
`k
`r=
`m
`»
`fr
`1
`e written
`915 ¢:
`ar
`:
`i
`Anexperimental ap-
`Figure 12.4
`paraus for demonstrating simple
`then Equation 12.15 can be written in the form
`harmonic motion. A pen attached
`to theoscillating mass traces out a
`
`
`de
`sine w:
`on the moving chart
`Consider a physical system consisting of a mass, m, attached to the end ofa spring
`ast
`2
`“
`ae
`:
`=
`paper.
`
`
`where the massisfree to move on a horizontal track (Fig. 12.3), We know from
`
`What we nowrequire isasolution to Equation 12.17—thatis, a function x() a
`at such a systemoscillates back and forth if disturbed from the equi-
`
`
`
`
`that satisfies this second-order
`differential equation. Because Equations 12.17 and
`
`librium position x = 0, where the spring is unstretched. If the surfaceis frictionless,
`12.7 are equivalent, we see that
`the solution must be that of simple harmonic
`motion:
`the mass can exhibit simple harmonic motion, One experimental arrangement that
`
`clearly demonstrates
`that such a system exhibits simple harmonic motionis illus-
`
`trated in Fig
`A mass oscillating vertically on a spring has a marking pen
`
`attached tc
`the mass is in motion, a sheet of
`paper is moved horizontally
`
`as shown, and the marking pen traces out a sinusoidal pattern. We can understand
`this qualitatively by first recalling that when the mass is displaced a small distance,
`x, from equilibrium, the spring exerts a force onit, given by Hooke's law,
`
`[12.16]
`
`[12.17]
`
`od)
`
` (by)
`
`—£,
`
`veil
`
`ale
`x=i)
`
`x
`
`x(t) = Acos(wt + &)
`To see this explicitly, note that if x
`=
`A cos(a+ q), then
`dx
`d
`— = A—cos(ewt
`di
`dt
`
`+
`
`db) =
`
`—waA sin(at +
`
`Fo=—ke
`
`[12.13]
`
`
`where& is
`> force constant of the spring. Wecall this a linear restoring force
`
`
`because it is linearly proportional to the displacer
`andalways directed toward
`the equilibrium position opposite the displacement. That is, when the mass is dis-
`placed to the right in Figure 12.3, xis positive and the restoring forceis to the left.
`When the massis displaced to theleft of x
`= 0, then xis negative andthe restoring
`force is to the right
`
`n of the mass in the x direction,
`If we apply Newton's second lawto the moti
`we get
`
`‘
`‘
`d
`
`> = —od — sin(wt + 6) = — oF A cos(wt + db)
`di°
`dt
`
`Comparing the expressions for x and d*xj/d#*, we see that d®x/dt® = — «x and
`
`Es
`s satished
`Thefollowing general statement can be made based on the foregoing discus-
`sion:
`
`Figure 12.3 A mass attachedto a
`spring on a friction “a8 track mewes
`
`in simple harm ic mation
`Fi = —kx = ma
`(a) Whenthe massis displaced to
`ke
`theright of equilibrium, thedis-
`m
`[12.14]
`a=-—x
`ement is positive
`H
`the
`
`We shall give additional physical examples in subsequent sections.
`erationis negative.
`(b) At
`the equi-
`Because the period of simple harmonic motion is T=27/mand the frequency
`Thatis, just as we learnedin Section 12.1, the acceleration is proportional to the
`
`x= 0, the
`librium position,
`accelera on is » but
`is the inverse of the period, we can express the period and frequency of the motion
`
`thespeed
`displacementof the mass from equilibrium and is in the opposite direction.
`(c) Whenthe
`num,
`for this mass—spring system as
`If the mass is displaced a maximumdistance,
`x
`= A, at some initial time and re-
`tive, the accelera-
`
`leased from rest, its initial acceleration is~kA/'m(thar is, the acceleration has its
`Period of
`a mass —sprin
`
`
`
`extreme negative value). When the mass passes through the equilibrium position
`syster
`
`T=
`(12.18)
`noe
`
`
`the force acting on a particle is linearly proportional to the
`Whenever
`
`
`
`the particle exhibits simple
`and
`displacement
`in the opposite
`direction,
`harmonic motic
`
`
`
`
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`
`
`12.3
`
`Energy of the Simple Harmonic Oseillator
`
`341
`
`(c) Express the displacement, speed, and accelcration as
`functions of time.
`Solution The expression x= A cos wis our solution for
`Special Case [, and so we can use the results from (a), (b).
`#=
`—oFfAcos wt!
`= —(1.25 m/s*)}cos 5.001
`and(c) to get
`
`
`x=Acos wt= (0.0500 m)cos 5.00¢
`v= -wA sin et = — (0251) m/'s)sin 5.00
`
`
`
`(a) A 400-g mass is suspendedfron
`EXERCISE 4
`pring hanging vertically, and the spring
`is found tostretch 3.0 cm. Find the spring constant. (b) How much will the spring stretch if
`the suspended mass is 575 g?
`Answer
`(a) 49N/m_
`(b) 12 em
`EXERCISE 5 A3.0-kg mass is attached to a spring and pulled out horizontally to amaximium
`displacementfrom equilibriumof 0.50 m. Whatspring constant must the spring haveif the
`mass is to achieve an acceleration equal to the free-fall acceleration?
`Answer
`59 N/m
`
`12.3 + ENERGY OF THE SIMPLE HARMONIC OSCILLATOR
`
`Let us examinethe mechanical enengyof the mass=spring described in Figure 12.5
`Because the surface is frictionless, we expect that
`the total mechanical energy1s
`constant, We can use Equation 12.5 to express the kinetic energy as
`K= dmv" = 4mm?A® sin®(at
`+
`ob)
`Theelastic potential energystored in the spring for any elongation xis Sk",
`Using Equation 12.1, we get
`[12.21]
`U= Lex? = Lha® cos*(at + 4)
`We see that Kand U/arealways positive quantities. Because of = &/m, we can express
`the total energy of the simple harmonic oscillator as
`E= K+ U= dkaA"[sin*(wl
`+
`ob)
`+ cos*(at
`But sin* @ + cos* @ = 1; therefore, this equation reduces to
`E =
`dkA*
`
`[12.20]
`
`[12.22]
`
`Kinetic energy of a simple
`harmonic oscillator
`
`« Potential energy ofa simple
`harmonic oscillator
`
`* Total energy of a simple
`harmonic oscillator
`
`+ @)]
`
`
`
`inking Physics 2) Bungee
`(Gamma)
`
`Chapter 12
`
`Oscillatory Motion
`
`
`
`If you go bungee jumping, you will bounce up and down at the end of the cord after
`your daring dive off bridge, Suppose you perform this dive and measure the frequency
`of your bouncing. You then move to another bridge. You discover that the bungee
`cordis too long for dives off this bridge—you will hit the ground. Thus, you fold the
`bungee cord in halfand make the dive from the doubled bungee cord. How does the
`frequencyof your bouncing at the end of this dive compare to the frequency after the
`dive from the first bridge?
`
`ening Let us model the bungee cord asa spring. The force exerted by a spring
`is proportional te the separation of the coils as the spring is extended. Imagine that
`we extend a spring by a given distance and measure the distance between coils. We
`then cut the spring in half, If one of the halé-springs is now extended by the same
`distance, the coils will be twice as far apart as they were for the complete spring. Thus,
`it takes twice as much force to stretch the halfspring. from which we conclude that
`the half-spring has a spring constant that is twice that ofthe complete spring, Now
`consider the folded bungee cord that we model as two half-springs in parallel. Each
`half has a spring constantthatis qwice the original spring constant of the bungee cord.
`In addition, a given mass hanging on the folded bungee cord will experience two
`forces—onefrom each half-spring. Asa result, the required force for agiven extension
`
`will be four
`times as much as for the original
`bungee cord. Theeffective spring constant
`of the folded bungee cord is, therefore, four times as large as the original spring
`constant, Because the frequency of oscillation is proportional to the square root ofthe
`spring constant, your bouncing frequency on the folded cord will be nwice that of the
`original cord.
`
`CONCEPTUAL PROBLEM 1
`
`
`{sa bouncing ball an example of simple harmonic motion? Is the daily movementofa student
`from hometo school and back simple harmonic motion?
`
`CONCEPTUAL PROBLEM 2
`
`
`A mass is hung on a spring and the frequency ofoscillation of the system,f is measured. The
`Mass, a
`
`
`cal mass, and the spring are carried in the Space Shutle to space. The
`al
`idl
`
`s
`tached to the ends of the spring and the system is taken out
`into space on
`
`
`
`a space walk. Thespring is extended and the systemis released to oscillate while floating in
`space. What is the frequency of oscillation for this system, in terms of [?
`
`is a constant of the motion
`Thatis, the energy of a simple harmonic oscillator
`
`fact,
`the total mechanical
`and proportional to the square of the amplitude,
`In
`
`energy is just equal to the maximum potential energy stored in the spring when
`
`Example 12.2 A Mass—Spring System
`
`x= £4, At these points, vy=0 and there is no kinetic energy. At the equilibrium
`
`iti
`0 and U'= 0, so that the total energy is all
`in the form of kinetic
`Amass of 200) g is connectedto a light spring of force constant
`and
`5.00 N/m and is free to oscillate on a horizontal, frictionless
`
`
`2 x=0, E=Sow...2a energy. That is, at
`
`} merA*
`
` - are shown in Figure 12.8a
`
`5,
`wo
`If the mass is displac
`Plots of the kinetic and potential energies versus tir
`al ae 1.265
`equilibrium
`I
` vs positive, and their sum at
`where 4 = 0. In this situation, both A and U a
`eased from rest, as in Figure 12.5,
`(a) find the period
`of its meoation.
`
`
`y of the system. The variations
`{b) Determine the
`saximum speed and maximum accel-
`all times is a constant equal to 4kA*, the total enc
`eration of the mass
`
`of Kand U with displacement are plotted in Figure 12.8b. Energy is continuously
`in
`This situation corresponds to Special Case I,
`which x=A cos @f and A=5,00 x 10°? m, Therefore,
`being transferred between potential energy stored in
`the spring and the kinetic
`
`ition, kinetic
`energy of the mass. Figure 12.9 illustrates the position,
`velocity, acce
`Tax = w=
`e
`(5.00 rad/s) (5.00% 10°? m)
`=
`0.250 m/s
`
`energy. and potential energy of the mass—spring system for one full periodof the
`
`j
`fk
`N/m
`
` Most of the ideas discussed so far are
`incorporated in this important figure
`
`ol a S00 ©10Ske (5,00 rad/s)2(5.00 X 10-2 m)=1.95 m/eVAM) rad/s Gnas = OFA =
`We suggest that voustudy it carefully
`
`
`
`
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`
`IPR2023-00783
`
`af
`ot ae ee at eee
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`
`
`IPR2023-00783
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 12
`
`Oscillatory Motion
`
`12.4
`
`Motion of « Pendulum
`
`347
`
`
`rigid
`inertia of a planar
`to measure the moment of
`One can use this result
`the distance d are known,
`body. If the location of the cemter of mass and, hence,
`Example 12.4 A Measure of Height
`A man enters a tall
`tower, needing to knowits height.
`the moment of inertia can be obtained through a measurement of the period.
`If the pendulum described in this example is
`EXERCISE 7
`
`
`
`taken t the moon, where the free-fall acceleration is
`extends from the ceiling al-
`He notes that a long pendulu
`lly, note that Equation 12.28 reduces to the period of a simple pendulum (Eq.
`
`
`
`
`most to the floor and that its period is 12.0 s, How tall isthe 26) when J = md*—thatis,whenall the mass is concentrated at the center of1,67 m, what is the period there? Answer 29.15
`
`
`
`
`PASS.
`
`CONCEPTUAL PROBLEM 7
`
`
`ge bob in a mu
`Two students are waiching the swinging of a Foucault pendulum with
`
`seum. One
`student says, “I'm going to sneak past the fence and stick some chewing gum on
`the top of the pendulumbob, to changeits frequency of oscillation,” The other student says,
`“That won't change the frequency—the frequency of a pendulum is independent of mass."
`Which suadent is correct?
`
`Example 12.5 A Swinging Rod
`is pivoted about
`A uniform rod of mass M and length £
`one end and oscillan
`cal plane (Fig. 12.12). Find
`
`
`the period of oscil
`small
`
`
`The moment of ine
`i uniform rod about an
`
`
`axis through one endis 4 MI*. Thedistance dfrom the pivot
`to the center of mass is 1/2. Substituting these quantities into
` gives
`Equation 12
`
`jm
`
`oe
`
`
`
`In one of the Moon landings, an astronaut walk-
`mment
`
`in the Moon's surface had a belt hanging from his space
`suit, and the belt oscillated asa compound pendulum.
`A sci-
`entist on Earth observed this motion on TV and from it was
`able to estimate the free-fall acceleration on the Moon, How
`do you suppose this calculation was done?
`
`Pivot
`
`o ,
`
`}
`
`L
`
`eM
`
`Mg
`
`a»
`A rigid rod oscill
`(Example 12.5)
`Figure 12.12
`
`pivot thr
`is a physical pendulum w
`
`and from Table 10.2
`MI
`
`
`
`sting about
`
`
`EXERCISE 8 Calculate
`about one end and osc
`12.12
`Answer
`1.648
`
`» period of a meter stick pivoted
`ting in-a vertical plane-as in Figure
`
`Ifwe ue P= 20.)L/g and solve for 1, we get
`aa
`BO m/s?) (12.0 s)*
`55.7 m
`|- 4
`
`
`
`The Physical Pendulum
`Ifa hanging object oscillates about a fixed axis that does not pass through its center
`
`of mass, and the object cannot be accurately
`approximated as a point mass, then it
`must be treated as a physical, or compound, pendulum, Consider a rigid body
`pivoted at a point O that isa distance of d from the center of mass (Fig. 12.11),
`
`The torque about © is provided by the force of gravity, and its magnitude is
`mgel
`
`@ Using the fact that t= Ja. where / is the moment of inertia about the
`axis through O, we get
`
`d*0
`— mgd sin 6 = iF
`
`
`
`=
`dsin@
`
`cM
`
`me
`The physical pen
`dulum corsists cif
` otedat the point Q, whichis mat at
`the center of mass, At equilibrium,
`the weight we passes through
`jo=
`0, The re-
` 0, corresy
`
`nM Owhen the
`
`systemis displaced through an an-
`gle #is mgd sin @,
`
`The minus sign on the left indicates that the torque about © tends to decrease @.
`That
`is, the force of gravity produces a restoring torque.
`
`If we again assume that @ is
`Il, then the approximation sin @ = @ is valid
`and the equation of motion reduces to
` mgd
`’
`[12.27]
`=-( e) 0= — ate
`ame form as Equation 12,17, and so the motionis
`cly
`appros
`iple harmonic motionfor small amplitudes. Thatis, the solution
`
`is 8 = @cos(a@t + 6), where4,is the maximumangulardisplace-
`of Equation12.
`ment and
`
` Note that the equ
`
`
`
`The period is
`
`
`ength of 3.00 m.
`Determine
`the char
`EXERCISE 9Asimple pendulum hasa
`
`
` elt here the free-fall ace staken froma point where g=9.80 m/s t
`
` vi
`il
`
`[12.28]
`
`Period of motion for a
`decreases to 9.79 m/s?
`Answer
`1.78%
`
`
`physical pendulum
`
`
`
`Petitioner Intel Corp., Ex. 1038
`IPR2023-00783
`
`

`

`Chapter 12
`
`Oscillatory Motion
`
`12.5 + DAMPED OSCILLATIONS
`
`The oscillatory motions we have considered so far have occurred in the context of
`an ideal system—thatis, one that oscillates indefinitely under the action ofa linear
`restoring force. In realistic systems, dissipative forces, such as friction, are present
`and retard the motion of the system. As a consequence, the mechanical energy of
`the system diminishes in time, and the motionis said to be damped.
`One common type of drag force, which we discussed in Chapter 5, is propor-
`tional to the velocity and acts in the direction opposite the motion. This type of
`drag is often observed when an object is oscillating in air, for instance. Because the
`drag force can be expressed as R = — bv, where bis a constant, and the restoring
`force exerted on the system is — kx, we can write Newton's second law as
`
`[12.29]
`
`{12.30]
`
`
`
`
`
`
` Alots of displace
`time
`for (a) an under.
`oscillator, (b) a criti
`cally damped oscillator, and
`ic) an overdamped oscillator
`
`frequency of vibration of a damped
`
`
`where a) = Viim represents the angular frequency of oscillation in the absenceof
`
`
`a drag
`force (the undampedoscillator). In other words, when 6 = 0, the drag force
`
`is ze
`id thesystemoscillates with its natural frequenc