#set document(title: "15.5 Damped Oscillations", 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")) == 15.5#h(0.6em)Damped Oscillations In the real world, oscillations seldom follow true SHM. Friction of some sort usually acts to dampen the motion so it dies away, or needs more force to continue. In this section, we examine some examples of #strong[damped harmonic motion] and see how to modify the equations of motion to describe this more general case. A guitar string stops oscillating a few seconds after being plucked. To keep swinging on a playground swing, you must keep pushing . Although we can often make friction and other nonconservative forces small or negligible, completely undamped motion is rare. In fact, we may even want to damp oscillations, such as with car shock absorbers. #figure(figph[A photo of a person on a swing], alt: "A photo of a person on a swing", caption: [To counteract dampening forces, you need to keep pumping a swing.]) shows a mass #emph[m] attached to a spring with a force constant #math.equation(block: false, alt: "k .")[$k .$] The mass is raised to a position #math.equation(block: false, alt: "A sub 0")[$A_(0)$], the initial amplitude, and then released. The mass oscillates around the equilibrium position in a fluid with viscosity but the amplitude decreases for each oscillation. For a system that has a small amount of damping, the period and frequency are constant and are nearly the same as for SHM, but the amplitude gradually decreases as shown. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy. #figure(figph[A mass m is suspended from a vertical spring and immersed in a fluid that has viscosity eta. A graph of the damped oscillation shows the displacement x in meters on the vertical axis as a function of time in seconds on the horizontal axis. The range of x is from minus A sub zero to plus A sub zero. The time scale is from zero to 7 T, with tics at increments of T. The displacement is plus A sub zero at time zero and oscillates between positive maxima and negative minima, with each full cycle taking the same time T but the amplitude of the oscillations decreasing with time.], alt: "A mass m is suspended from a vertical spring and immersed in a fluid that has viscosity eta. A graph of the damped oscillation shows the displacement x in meters on the vertical axis as a function of time in seconds on the horizontal axis. The range of x is from minus A sub zero to plus A sub zero. The time scale is from zero to 7 T, with tics at increments of T. The displacement is plus A sub zero at time zero and oscillates between positive maxima and negative minima, with each full cycle taking the same time T but the amplitude of the oscillations decreasing with time.", caption: [For a mass on a spring oscillating in a viscous fluid, the period remains constant, but the amplitudes of the oscillations decrease due to the damping caused by the fluid.]) Consider the forces acting on the mass. Note that the only contribution of the weight is to change the equilibrium position, as discussed earlier in the chapter. Therefore, the net force is equal to the force of the spring and the damping force #math.equation(block: false, alt: "open parenthesis F sub D close parenthesis")[$( F_(D) )$]. If the magnitude of the velocity is small, meaning the mass oscillates slowly, the damping force is proportional to the velocity and acts against the direction of motion #math.equation(block: false, alt: "open parenthesis F sub D equals − b v close parenthesis")[$( F_(D) = "−" b v )$]. The net force on the mass is therefore #math.equation(block: true, alt: "m a equals − b v minus k x .")[$m a = "−" b v − k x .$] Writing this as a differential equation in #emph[x], we obtain #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "m the fraction d squared x over d t squared plus b the fraction d x over d t plus k x equals 0 .")[$m frac(d^(2) x, d t^(2)) + b frac(d x, d t) + k x = 0 .$] ] To determine the solution to this equation, consider the plot of position versus time shown. The curve resembles a cosine curve oscillating in the envelope of an exponential function #math.equation(block: false, alt: "A sub 0 e to the power − α t")[$A_(0) e^("−" α t)$] where #math.equation(block: false, alt: "α equals the fraction b over 2 m")[$α = frac(b, 2 m)$]. The solution is #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals A sub 0 e to the power minus the fraction b over 2 m t cos open parenthesis ω t plus ϕ close parenthesis .")[$x ( t ) = A_(0) e^(− frac(b, 2 m) t) "cos" ( ω t + ϕ ) .$] ] It is left as an exercise to prove that this is, in fact, the solution. To prove that it is the right solution, take the first and second derivatives with respect to time and substitute them into . It is found that is the solution if #math.equation(block: true, alt: "ω equals the square root of the fraction k over m minus open parenthesis the fraction b over 2 m close parenthesis squared .")[$ω = sqrt(frac(k, m) − attach(( frac(b, 2 m) ), t: 2)) .$] Recall that the angular frequency of a mass undergoing SHM is equal to the square root of the force constant divided by the mass. This is often referred to as the #strong[natural angular frequency], which is represented as #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "ω sub 0 equals the square root of the fraction k over m .")[$ω_(0) = sqrt(frac(k, m)) .$] ] The angular frequency for damped harmonic motion becomes #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "ω equals the square root of ω 0 2 minus open parenthesis the fraction b over 2 m close parenthesis squared .")[$ω = sqrt(ω_(0)^(2) − attach(( frac(b, 2 m) ), t: 2)) .$] ] #figure(figph[The figure shows a graph of displacement, x in meters, along the vertical axis, versus time in seconds along the horizontal axis. The displacement ranges from minus A sub zero to plus A sub zero and the time ranges from 0 to 10 T. The displacement, shown by a blue curve, oscillates between positive maxima and negative minima, forming a wave whose amplitude is decreasing gradually as we move far from t=0. The time, T, between adjacent crests remains the same throughout. The envelope, the smooth curve that connects the crests and another smooth curve that connects the troughs of the oscillations, is shown as a pair of dashed red lines. The upper curve connecting the crests is labeled as plus A sub zero times e to the quantity minus b t over 2 m. The lower curve connecting the troughs is labeled as minus A sub zero times e to the quantity minus b t over 2 m.], alt: "The figure shows a graph of displacement, x in meters, along the vertical axis, versus time in seconds along the horizontal axis. The displacement ranges from minus A sub zero to plus A sub zero and the time ranges from 0 to 10 T. The displacement, shown by a blue curve, oscillates between positive maxima and negative minima, forming a wave whose amplitude is decreasing gradually as we move far from t=0. The time, T, between adjacent crests remains the same throughout. The envelope, the smooth curve that connects the crests and another smooth curve that connects the troughs of the oscillations, is shown as a pair of dashed red lines. The upper curve connecting the crests is labeled as plus A sub zero times e to the quantity minus b t over 2 m. The lower curve connecting the troughs is labeled as minus A sub zero times e to the quantity minus b t over 2 m.", caption: [Position versus time for the mass oscillating on a spring in a viscous fluid. Notice that the curve appears to be a cosine function inside an exponential envelope.]) Recall that when we began this description of damped harmonic motion, we stated that the damping must be small. Two questions come to mind. Why must the damping be small? And how small is small? If you gradually #emph[increase] the amount of damping in a system, the period and frequency begin to be affected, because damping opposes and hence slows the back and forth motion. (The net force is smaller in both directions.) If there is very large damping, the system does not even oscillate—it slowly moves toward equilibrium. The angular frequency is equal to #math.equation(block: true, alt: "ω equals the square root of the fraction k over m minus open parenthesis the fraction b over 2 m close parenthesis squared .")[$ω = sqrt(frac(k, m) − attach(( frac(b, 2 m) ), t: 2)) .$] As #emph[b] increases, #math.equation(block: false, alt: "the fraction k over m minus open parenthesis the fraction b over 2 m close parenthesis squared")[$frac(k, m) − attach(( frac(b, 2 m) ), t: 2)$] becomes smaller and eventually reaches zero when #math.equation(block: false, alt: "b equals the square root of 4 m k")[$b = sqrt(4 m k)$]. If #emph[b] becomes any larger, #math.equation(block: false, alt: "the fraction k over m minus open parenthesis the fraction b over 2 m close parenthesis squared")[$frac(k, m) − attach(( frac(b, 2 m) ), t: 2)$] becomes a negative number and #math.equation(block: false, alt: "the square root of the fraction k over m minus open parenthesis the fraction b over 2 m close parenthesis squared")[$sqrt(frac(k, m) − attach(( frac(b, 2 m) ), t: 2))$] is a complex number. shows the displacement of a harmonic oscillator for different amounts of damping. When the damping constant is small, #math.equation(block: false, alt: "b less than the square root of 4 m k")[$b < sqrt(4 m k)$], the system oscillates while the amplitude of the motion decays exponentially. This system is said to be #strong[underdamped], as in curve (a). Many systems are underdamped, and oscillate while the amplitude decreases exponentially, such as the mass oscillating on a spring. The damping may be quite small, but eventually the mass comes to rest. If the damping constant is #math.equation(block: false, alt: "b equals the square root of 4 m k")[$b = sqrt(4 m k)$], the system is said to be #strong[critically damped], as in curve (b). An example of a critically damped system is the shock absorbers in a car. It is advantageous to have the oscillations decay as fast as possible. Here, the system does not oscillate, but asymptotically approaches the equilibrium condition as quickly as possible. Curve (c) in represents an #strong[overdamped] system where #math.equation(block: false, alt: "b greater than the square root of 4 m k .")[$b > sqrt(4 m k) .$] An overdamped system will approach equilibrium over a longer period of time. A damped oscillator x(t) = A₀e^(−bt/2m)·cos(ωt) with ω = √(k/m − (b/2m)²), for a 1.00-kg mass on a 16.0-N/m spring starting from x = A₀ = 1.00 m, so the natural frequency is ω₀ = 4.00 rad/s. The dashed pair is the exponential envelope ±A₀e^(−bt/2m) inside which the text says the cosine oscillates. At the drawn b = 1.00 kg/s the motion is underdamped: it still swings at ω = 3.97 rad/s, barely below ω₀, while the envelope decays with a 2.0-s scale. Push b up and the swing both slows and dies sooner, until at b = √(4mk) = 8.00 kg/s the frequency reaches zero — critical damping, the shock-absorber case, where the curve returns to equilibrium without ever crossing it.{"functions":\[{"color":"\#1f4e79","domain":\[0,4\],"expression":"exp(-bd\*t/2)\*cos(sqrt(16 - bd^2/4)\*t)","expression2":"","id":"displacement","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true},{"color":"\#b8442a","domain":\[0,4\],"expression":"exp(-bd\*t/2)","expression2":"","id":"envelope-upper","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true},{"color":"\#b8442a","domain":\[0,4\],"expression":"-exp(-bd\*t/2)","expression2":"","id":"envelope-lower","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true}\],"grid":{"axisColor":"\#333333","gridColor":"\#cccccc","majorSpacing":0.5,"minorGridColor":"\#eeeeee","showAxes":true,"showLabels":true},"parameters":\[{"id":"up1-15-5-bd","label":"Damping constant b (critical damping at 8.00)","max":8,"min":0,"name":"bd","step":0.05,"unit":"kg/s","value":1}\],"title":"Damped harmonic oscillator and its exponential envelope","version":1,"viewport":{"centerX":2,"centerY":0,"scale":165}}#figure(figph[The position, x in meters on the vertical axis, versus time in seconds on the horizontal axis, with varying degrees of damping. No scale is given for either axis. All three curves start at the same positive position at time zero. Blue curve a, labeled with b squared is less than 4 m k, undergoes a little over two and a quarter oscillations of decreasing amplitude and constant period. Red curve b, labeled with b squared is equal to 4 m k, decreases at t=0 less rapidly than the blue curve, but does not oscillate. The red curve approaches x=0 asymptotically, and is nearly zero within one oscillation of the blue curve. Green curve c, labeled with b squared is greater than 4 m k, decreases at t=0 less rapidly than the red curve, and does not oscillate. The green curve approaches x=0 asymptotically, but is still noticeably above zero at the end of the graph, after more than two oscillations of the blue curve.], alt: "The position, x in meters on the vertical axis, versus time in seconds on the horizontal axis, with varying degrees of damping. No scale is given for either axis. All three curves start at the same positive position at time zero. Blue curve a, labeled with b squared is less than 4 m k, undergoes a little over two and a quarter oscillations of decreasing amplitude and constant period. Red curve b, labeled with b squared is equal to 4 m k, decreases at t=0 less rapidly than the blue curve, but does not oscillate. The red curve approaches x=0 asymptotically, and is nearly zero within one oscillation of the blue curve. Green curve c, labeled with b squared is greater than 4 m k, decreases at t=0 less rapidly than the red curve, and does not oscillate. The green curve approaches x=0 asymptotically, but is still noticeably above zero at the end of the graph, after more than two oscillations of the blue curve.", caption: [The position versus time for three systems consisting of a mass and a spring in a viscous fluid. (a) If the damping is small #math.equation(block: false, alt: "open parenthesis b less than the square root of 4 m k close parenthesis")[$( b < sqrt(4 m k) )$], the mass oscillates, slowly losing amplitude as the energy is dissipated by the non-conservative force(s). The limiting case is (b) where the damping is #math.equation(block: false, alt: "open parenthesis b equals the square root of 4 m k close parenthesis")[$( b = sqrt(4 m k) )$]. (c) If the damping is very large #math.equation(block: false, alt: "open parenthesis b greater than the square root of 4 m k close parenthesis")[$( b > sqrt(4 m k) )$], the mass does not oscillate when displaced, but attempts to return to the equilibrium position.]) Critical damping is often desired, because such a system returns to equilibrium rapidly and remains at equilibrium as well. In addition, a constant force applied to a critically damped system moves the system to a new equilibrium position in the shortest time possible without overshooting or oscillating about the new position. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Why are completely undamped harmonic oscillators so rare? #solutionbox[ Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator. ] ] === Summary - Damped harmonic oscillators have non-conservative forces that dissipate their energy. - Critical damping returns the system to equilibrium as fast as possible without overshooting. - An underdamped system will oscillate through the equilibrium position. - An overdamped system moves more slowly toward equilibrium than one that is critically damped. === Conceptual Questions Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.) #solutionbox[ A car shock absorber. ] How would a car bounce after a bump under each of these conditions? (a) overdamping (b) underdamping (c) critical damping Most harmonic oscillators are damped and, if undriven, eventually come to a stop. Why? #solutionbox[ The second law of thermodynamics states that perpetual motion machines are impossible. Eventually the ordered motion of the system decreases and returns to equilibrium. ] === Problems The amplitude of a lightly damped oscillator decreases by #math.equation(block: false, alt: "3.0 %")[$3.0 %$] during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle? #solutionbox[ 6% ]