#set document(title: "16.5 Energy and the Simple Harmonic Oscillator", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 16.5#h(0.6em)Energy and the Simple Harmonic Oscillator === Learning Objectives By the end of this section, you will be able to: - Determine the maximum speed of an oscillating system. To study the energy of a simple harmonic oscillator, we first consider all the forms of energy it can have We know from Hooke’s Law: Stress and Strain Revisited that the energy stored in the deformation of a simple harmonic oscillator is a form of potential energy given by: #math.equation(block: true, alt: "PE sub el equals the fraction 1 over 2 kx squared .")[$"PE"_("el") = frac(1, 2) italic("kx")^(2) .$] Because a simple harmonic oscillator has no dissipative forces, the other important form of energy is kinetic energy #math.equation(block: false, alt: "KE")[$"KE"$]. Conservation of energy for these two forms is: #math.equation(block: true, alt: "KE plus PE sub el equals constant")[$"KE" + "PE"_("el") = "constant"$] or #math.equation(block: true, alt: "the fraction 1 over 2 mv squared plus the fraction 1 over 2 kx squared equals constant.")[$frac(1, 2) "mv"^(2) + frac(1, 2) "kx"^(2) = "constant."$] This statement of conservation of energy is valid for #emph[all] simple harmonic oscillators, including ones where the gravitational force plays a role Namely, for a simple pendulum we replace the velocity with #math.equation(block: false, alt: "v equals Lω")[$v = "Lω"$], the spring constant with #math.equation(block: false, alt: "k equals mg / L")[$k = "mg" / L$], and the displacement term with #math.equation(block: false, alt: "x equals Lθ")[$x = "Lθ"$]. Thus #math.equation(block: true, alt: "the fraction 1 over 2 mL squared ω squared plus the fraction 1 over 2 mgL θ squared equals constant.")[$frac(1, 2) "mL"^(2) ω^(2) + frac(1, 2) "mgL" θ^(2) = "constant."$] In the case of undamped simple harmonic motion, the energy oscillates back and forth between kinetic and potential, going completely from one to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, as shown again in , the motion starts with all of the energy stored in the spring. As the object starts to move, the elastic potential energy is converted to kinetic energy, becoming entirely kinetic energy at the equilibrium position. It is then converted back into elastic potential energy by the spring, the velocity becomes zero when the kinetic energy is completely converted, and so on. This concept provides extra insight here and in later applications of simple harmonic motion, such as alternating current circuits. #figure(figph[Figure a shows a spring on a frictionless surface attached to a bar or wall from the left side, and on the right side of it there’s an object attached to it with mass m, its amplitude is given by X, and x equal to zero at the equilibrium level. Force F is applied to it from the right side, shown with left direction pointed red arrow and velocity v is equal to zero. A direction point showing the north and west direction is also given alongside this figure as well as with other four figures. The energy given here for the object is given according to the velocity. In figure b, after the force has been applied, the object moves to the left compressing the spring a bit, and the displaced area of the object from its initial point is shown in sketched dots. F is equal to zero and the V is max in negative direction. The energy given here for the object is given according to the velocity. In figure c, the spring has been compressed to the maximum level, and the amplitude is negative x. Now the direction of force changes to the rightward direction, shown with right direction pointed red arrow and the velocity v zero. The energy given here for the object is given according to the velocity. In figure d, the spring is shown released from the compressed level and the object has moved toward the right side up to the equilibrium level. F is zero, and the velocity v is maximum. The energy given here for the object is given according to the velocity. In figure e, the spring has been stretched loose to the maximum level and the object has moved to the far right. Now again the velocity here is equal to zero and the direction of force again is to the left hand side, shown here as F is equal to zero. The energy given here for the object is given according to the velocity. ], alt: "Figure a shows a spring on a frictionless surface attached to a bar or wall from the left side, and on the right side of it there’s an object attached to it with mass m, its amplitude is given by X, and x equal to zero at the equilibrium level. Force F is applied to it from the right side, shown with left direction pointed red arrow and velocity v is equal to zero. A direction point showing the north and west direction is also given alongside this figure as well as with other four figures. The energy given here for the object is given according to the velocity. In figure b, after the force has been applied, the object moves to the left compressing the spring a bit, and the displaced area of the object from its initial point is shown in sketched dots. F is equal to zero and the V is max in negative direction. The energy given here for the object is given according to the velocity. In figure c, the spring has been compressed to the maximum level, and the amplitude is negative x. Now the direction of force changes to the rightward direction, shown with right direction pointed red arrow and the velocity v zero. The energy given here for the object is given according to the velocity. In figure d, the spring is shown released from the compressed level and the object has moved toward the right side up to the equilibrium level. F is zero, and the velocity v is maximum. The energy given here for the object is given according to the velocity. In figure e, the spring has been stretched loose to the maximum level and the object has moved to the far right. Now again the velocity here is equal to zero and the direction of force again is to the left hand side, shown here as F is equal to zero. The energy given here for the object is given according to the velocity.", caption: [The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.]) The conservation of energy principle can be used to derive an expression for velocity #math.equation(block: false, alt: "v")[$v$]. If we start our simple harmonic motion with zero velocity and maximum displacement (#math.equation(block: false, alt: "x equals X")[$x = X$]), then the total energy is #math.equation(block: true, alt: "the fraction 1 over 2 kX squared .")[$frac(1, 2) "kX"^(2) .$] This total energy is constant and is shifted back and forth between kinetic energy and potential energy, at most times being shared by each. The conservation of energy for this system in equation form is thus: #math.equation(block: true, alt: "the fraction 1 over 2 mv squared plus the fraction 1 over 2 kx squared equals the fraction 1 over 2 kX squared .")[$frac(1, 2) "mv"^(2) + frac(1, 2) "kx"^(2) = frac(1, 2) "kX"^(2) .$] Solving this equation for #emph[#math.equation(block: false, alt: "v")[$v$]] yields: #math.equation(block: true, alt: "v equals plus or minus the square root of the fraction k over m X squared minus x squared .")[$v = ± sqrt(frac(k, m) X^(2) − x^(2)) .$] Manipulating this expression algebraically gives: #math.equation(block: true, alt: "v equals plus or minus the square root of the fraction k over m X the square root of 1 minus the fraction x squared over X squared")[$v = ± sqrt(frac(k, m)) X sqrt(1 − frac(x^(2), X^(2)))$] and so #math.equation(block: true, alt: "v equals plus or minus v sub max the square root of 1 minus the fraction x squared over X squared ,")[$v = ± v_("max") sqrt(1 − frac(x^(2), X^(2))) ,$] where #math.equation(block: true, alt: "v sub max equals X the square root of the fraction k over m .")[$v_("max") = X sqrt(frac(k, m)) .$] From this expression, we see that the velocity is a maximum (#math.equation(block: false, alt: "v sub max")[$v_("max")$]) at #math.equation(block: false, alt: "x equals 0")[$x = 0$], as stated earlier in #math.equation(block: false, alt: "v t equals minus v sub max sin the fraction 2π t over T")[$v t = − v_("max") #h(0.25em) "sin" #h(0.25em) frac(2π t, T)$]#emph[.] Notice that the maximum velocity depends on three factors. Maximum velocity is directly proportional to amplitude. As you might guess, the greater the maximum displacement the greater the maximum velocity. Maximum velocity is also greater for stiffer systems, because they exert greater force for the same displacement. This observation is seen in the expression for #math.equation(block: false, alt: "v sub max ;")[$v_("max") ;$] it is proportional to the square root of the force constant #math.equation(block: false, alt: "k")[$k$]. Finally, the maximum velocity is smaller for objects that have larger masses, because the maximum velocity is inversely proportional to the square root of #math.equation(block: false, alt: "m")[$m$]. For a given force, objects that have large masses accelerate more slowly. A similar calculation for the simple pendulum produces a similar result, namely: #math.equation(block: true, alt: "ω sub max equals the square root of the fraction g over L θ sub max .")[$ω_("max") = sqrt(frac(g, L)) θ_("max") .$] #examplebox("Example 1")[Determine the Maximum Speed of an Oscillating System: A Bumpy Road][ Suppose that a car is 900 kg and has a suspension system that has a force constant #math.equation(block: false, alt: "k equals 6 . 53 times 10 to the power 4 N/m")[$k = 6 "." "53" × "10"^(4) #h(0.25em) "N/m"$]. The car hits a bump and bounces with an amplitude of 0.100 m. What is its maximum vertical velocity if you assume no damping occurs? Strategy We can use the expression for #math.equation(block: false, alt: "v sub max")[$v_("max")$] given in #math.equation(block: false, alt: "v sub max equals X the square root of the fraction k over m")[$v_("max") = X sqrt(frac(k, m))$] to determine the maximum vertical velocity. The variables #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "k")[$k$] are given in the problem statement, and the maximum displacement #math.equation(block: false, alt: "X")[$X$] is 0.100 m. Solution + Identify known. + Substitute known values into #math.equation(block: false, alt: "v sub max equals the square root of the fraction k over m X")[$v_("max") = sqrt(frac(k, m)) X$]: #math.equation(block: true, alt: "v sub max equals the square root of the fraction 6 . 53 times 10 to the power 4 N/m over 900 kg (0 . 100 m) .")[$v_("max") = sqrt(frac(6 "." "53" × "10"^(4) #h(0.25em) "N/m", "900" #h(0.25em) " kg")) \(0 "." "100" #h(0.25em) " m)" .$] + Calculate to find #math.equation(block: false, alt: "v sub max = 0.852 m/s .")[$v_("max") "= 0.852 m/s" .$] Discussion This answer seems reasonable for a bouncing car. There are other ways to use conservation of energy to find #math.equation(block: false, alt: "v sub max")[$v_("max")$]. We could use it directly, as was done in the example featured in Hooke’s Law: Stress and Strain Revisited. The small vertical displacement #math.equation(block: false, alt: "y")[$y$] of an oscillating simple pendulum, starting from its equilibrium position, is given as #math.equation(block: true, alt: "y open parenthesis t close parenthesis equals a sin ωt ,")[$y ( t ) = a #h(0.25em) " sin " #h(0.25em) "ωt" ,$] where #math.equation(block: false, alt: "a")[$a$] is the amplitude, #math.equation(block: false, alt: "ω")[$ω$] is the angular velocity and #math.equation(block: false, alt: "t")[$t$] is the time taken. Substituting #math.equation(block: false, alt: "ω equals the fraction 2π over T")[$ω = frac(2π, T)$], we have #math.equation(block: true, alt: "y open parenthesis t close parenthesis equals a sin the fraction 2π t over T .")[$y ( t ) = a #h(0.25em) "sin" frac(2π t, T) .$] Thus, the displacement of pendulum is a function of time as shown above. Also the velocity of the pendulum is given by #math.equation(block: true, alt: "v open parenthesis t close parenthesis equals the fraction 2 aπ over T cos the fraction 2π t over T ,")[$v ( t ) = frac(2 "aπ", T) #h(0.25em) "cos" #h(0.25em) frac(2π t, T) ,$] so the motion of the pendulum is a function of time. ] Why does it hurt more if your hand is snapped with a ruler than with a loose spring, even if the displacement of each system is equal? #solutionbox[ The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more. ] You are observing a simple harmonic oscillator. Identify one way you could decrease the maximum velocity of the system. #solutionbox[ You could increase the mass of the object that is oscillating. ] === Test Prep for AP Courses A mass of 1 kg undergoes simple harmonic motion with amplitude of 1 m. If the period of the oscillation is 1 s, calculate the internal energy of the system. #solutionbox[ 19.7 J ] === Section Summary - Energy in the simple harmonic oscillator is shared between elastic potential energy and kinetic energy, with the total being constant: #math.equation(block: true, alt: "the fraction 1 over 2 mv squared plus the fraction 1 over 2 kx squared equals constant.")[$frac(1, 2) "mv"^(2) + frac(1, 2) "kx"^(2) = " constant."$] - Maximum velocity depends on three factors: it is directly proportional to amplitude, it is greater for stiffer systems, and it is smaller for objects that have larger masses: #math.equation(block: true, alt: "v sub max equals X the square root of the fraction k over m")[$v_("max") = X sqrt(frac(k, m))$] === Conceptual Questions Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.) === Problems & Exercises The length of nylon rope from which a mountain climber is suspended has a force constant of #math.equation(block: false, alt: "1.40 times 10 to the power 4 N/m")[$1.40 × 10^(4) "N/m"$] (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? Ignore the change in gravitational potential energy after the cord begins to stretch. (b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? Hint: Use conservation of energy. (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used. #solutionbox[ (a) #math.equation(block: false, alt: "1.99 Hz")[$"1.99 Hz"$] (b) 50.2 cm (c) 1.40 Hz, 71.0 cm ] Engineering Application Near the top of the Citigroup Center building in New York City, there is an object with mass of #math.equation(block: false, alt: "4 . 00 times 10 to the power 5 kg")[$4 "." "00" × "10"^(5) #h(0.25em) "kg"$] on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-m displacement from equilibrium? #solutionbox[ (a) #math.equation(block: false, alt: "3 . 95 times 10 to the power 6 N/m")[$3 "." "95" × "10"^(6) #h(0.25em) "N/m"$] (b) #math.equation(block: false, alt: "7 . 90 times 10 to the power 6 J")[$7 "." "90" × "10"^(6) #h(0.25em) "J"$] ]