DOCSLIB.ORG

Chapter 15 Oscillations and Waves

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Chapter 15 Oscillations and Waves Chapter 15 Oscillations and Waves Oscillations and Waves • Simple Harmonic Motion • Energy in SHM • Some Oscillating Systems • Damped Oscillations • Driven Oscillations • Resonance MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 2 Simple Harmonic Motion Simple harmonic motion (SHM) occurs when the restoring force (the force directed toward a stable equilibrium point ) is proportional to the displacement from equilibrium. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 3 Characteristics of SHM • Repetitive motion through a central equilibrium point. • Symmetry of maximum displacement. • Period of each cycle is constant. • Force causing the motion is directed toward the equilibrium point (minus sign). • F directly proportional to the displacement from equilibrium. Acceleration = - ω2 x Displacement MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 4 A Simple Harmonic Oscillator (SHO) Frictionless surface The restoring force is F = −kx. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 5 Two Springs with Different Amplitudes Frictionless surface MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 6 SHO Period is Independent of the Amplitude MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 7 The Period and the Angular Frequency 2π T = . The period of oscillation is ω where ω is the angular frequency of k the oscillations, k is the spring ω = m constant and m is the mass of the block. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 8 Simple Harmonic Motion At the equilibrium point x = 0 so, a = 0 also. When the stretch is a maximum, a will be a maximum too. The velocity at the end points will be zero, and it is a maximum at the equilibrium point. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 9 Representing Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 10 Representing Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 11 Representing Simple Harmonic Motion Position - x max = A Velocity - v max = ωA 2 Acceleration - a max = ω A MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 12 A simple harmonic oscillator can be described mathematically by: x() t = Acos ωt dx v() t = = -A ωsin ωt dt dv a() t = = -A ω2cos ωt dt where A is the amplitude of the motion, the maximum Or by: displacement from equilibrium, A ω = v , and x() t = Asin ωt max Aω2 = a dx max. vt=() =A ωcos ωt dt dv a() t = = -A ω2 sin ωt dt MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 13 Linear Motion - Circular Functions MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 14 Projection of Circular Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 15 Circular Motion is the superposition of two linear SHO that are 90 0 out of phase with each other y= Asin(ω t ) x= Acos(ω t ) MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 16 Shifting Trig Functions sin The minus sign means that the x = A ω t - φ cos phase is shifted to the right. sin t A plus sign indicated the phase x=A 2 π - ϕ {}cos T is shifted to the left π x = Asinω t - 2 π π x= A() sin ωt cos2 - sin 2 cos ωt x= A() sin ωt (0) - (1)cos ωt x = -Acos ωt MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 17 Shifting Trig Functions π sinω t- =0 Shifted Trig Functions 2 1.50 sin( ωt) π 1.00 ω δ ωt- =0 sin( t- ) 2 0.50 π ωt = 0.00 2 -3.00 -2.00 -1.00 0.00 1.00 2.00 3.00 -0.50 π 1 1 T t= ; = -1.00 2ω ω 2π π T T -1.50 t = = Time 2 2 π 4 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 18 Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 19 Equation of Motion & Energy Assuming the table is frictionless: F = - kx = ma ∑ x x k 2 Classic form for SHM a() t=- xt=-()() ω x t x m 12 1 2 Also, EKt=+=()() Ut mvt() + kxt() 2 2 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 20 Spring Potential Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 21 Spring Total Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 22 Approximating Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 23 Approximating Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 24 Potential and Kinetic Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 25 The period of oscillation of an object in an ideal mass-spring system is 0.50 sec and the amplitude is 5.0 cm. What is the speed at the equilibrium point? At equilibrium x = 0: 1 1 1 E = K +U = mv 2 + kx 2 = mv 2 2 2 2 Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = v max = A ω. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 26 Example continued: The amplitude A is given, but ω is not. 2π 2π ω = = =12 .6 rads/sec T 0.50 s and v = Aω = (5.0 cm )(12 .6 rads/sec ) = 62 .8 cm/sec MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 27 The diaphragm of a speaker has a mass of 50.0 g and responds to a signal of 2.0 kHz by moving back and forth with an amplitude of 1.8 ×10 −4 m at that frequency. (a) What is the maximum force acting on the diaphragm? 2 2 2 2 ∑ F = Fmax = ma max = m()Aω = mA ()2πf = 4π mAf The value is F max =1400 N. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 28 Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E = K max = U max. 1 The value of k is unknown so use K . U = kA 2 max max 2 1 1 1 1 2 2 2 2 2 K = mv Kmax = mv max = m()Aω = mA ()2πf max 2 max 2 2 2 The value is K max = 0.13 J. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 29 Example: The displacement of an object in SHM is given by: y()()()t 8= . 00cm sin1[] . 57 rads/sec t What is the frequency of the oscillations? Comparing to y(t) = A sin ωt gives A = 8.00 cm and ω = 1.57 rads/sec. The frequency is: ω1 . 75 rads/sec f = = = .0 250 Hz 2π 2π MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 30 Example continued: Other quantities can also be determined: 2π 2π The period of the motion is T = = 4= . 00 sec ω1 . 75 rads/sec xmax = A = 8.00 cm vmax = Aω = (8 .00 mc )(1.57 rads/sec ) =12 .6 cm/sec 2 2 2 amax = Aω = (8 .00 cm )(1.57 rads/sec ) =19 .7 cm/sec MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 31 What About Gravity? When a mass-spring system is oriented vertically, it will exhibit SHM with the same period and frequency as a horizontally placed system. The effect of gravity is canceled out. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 32 Why We Ignore Gravity with Vertical Springs MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 33 The Simple Pendulum A simple pendulum is constructed by attaching a mass to a thin rod or a light string. We will also assume that the amplitude of the oscillations is small. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 34 The Simple Pendulum The pendulum is best described using polar coordinates. The origin is at the pivot point. The coordinates are (r, φ). The r-coordinate points from the origin along the rod. The φ- coordinate is perpendicular to the rod and is positive in the counter clockwise direction. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 35 F= − mgsin φ = ma φ Apply Newton’s 2 nd ∑ φ Law to the pendulum v2 bob. ∑ Fr = T − mgcos φ = m r If we assume that φ <<1 rad, then sin φ ≈ φ and cos φ ≈1, the angular frequency of oscillations is then: ∑ Fφ =− mgsin φ = maφ = mL α −mgsin φ = mL α g α= − (g / L )sin φ ω = L α= − (g / L ) φ 2π L The period of oscillations is T = = 2π ω g MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 36 Example: A clock has a pendulum that performs one full swing every 1.0 sec. The object at the end of the string weighs 10.0 N. What is the length of the pendulum? L T = 2π g gT 2 9.8 m/s 2 ()1.0 s 2 L = = () = 0.25 m Solving for L: 4π 2 4π 2 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 37 The gravitational potential energy of a pendulum is U = mgy. Taking y = 0 at the lowest point of the swing, show that y = L(1-cos θ). θ Lcos θ L L y = L 1( − cos θ ) y=0 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 38 The Physical Pendulum A physical pendulum is any rigid object that is free to oscillate about some fixed axis.
Load more

Recommended publications
  • Forced Mechanical Oscillations
    169 Carl von Ossietzky Universität Oldenburg – Faculty V - Institute of Physics Module Introductory laboratory course physics – Part I Forced mechanical oscillations Keywords: HOOKE's law, harmonic oscillation, harmonic oscillator, eigenfrequency, damped harmonic oscillator, resonance, amplitude resonance, energy resonance, resonance curves References: /1/ DEMTRÖDER, W.: „Experimentalphysik 1 – Mechanik und Wärme“, Springer-Verlag, Berlin among others. /2/ TIPLER, P.A.: „Physik“, Spektrum Akademischer Verlag, Heidelberg among others. /3/ ALONSO, M., FINN, E. J.: „Fundamental University Physics, Vol. 1: Mechanics“, Addison-Wesley Publishing Company, Reading (Mass.) among others. 1 Introduction It is the object of this experiment to study the properties of a „harmonic oscillator“ in a simple mechanical model. Such harmonic oscillators will be encountered in different fields of physics again and again, for example in electrodynamics (see experiment on electromagnetic resonant circuit) and atomic physics. Therefore it is very important to understand this experiment, especially the importance of the amplitude resonance and phase curves. 2 Theory 2.1 Undamped harmonic oscillator Let us observe a set-up according to Fig. 1, where a sphere of mass mK is vertically suspended (x-direc- tion) on a spring. Let us neglect the effects of friction for the moment. When the sphere is at rest, there is an equilibrium between the force of gravity, which points downwards, and the dragging resilience which points upwards; the centre of the sphere is then in the position x = 0. A deflection of the sphere from its equilibrium position by x causes a proportional dragging force FR opposite to x: (1) FxR ∝− The proportionality constant (elastic or spring constant or directional quantity) is denoted D, and Eq.
    [Show full text]
  • Influence of Angular Velocity of Pedaling on the Accuracy of The
    Research article 2018-04-10 - Rev08 Influence of Angular Velocity of Pedaling on the Accuracy of the Measurement of Cyclist Power Abstract Almost all cycling power meters currently available on the The miscalculation may be—even significantly—greater than we market are positioned on rotating parts of the bicycle (pedals, found in our study, for the following reasons: crank arms, spider, bottom bracket/hub) and, regardless of • the test was limited to only 5 cyclists: there is no technical and construction differences, all calculate power on doubt other cyclists may have styles of pedaling with the basis of two physical quantities: torque and angular velocity greater variations of angular velocity; (or rotational speed – cadence). Both these measures vary only 2 indoor trainer models were considered: other during the 360 degrees of each revolution. • models may produce greater errors; The torque / force value is usually measured many times during slopes greater than 5% (the only value tested) may each rotation, while the angular velocity variation is commonly • lead to less uniform rotations and consequently neglected, considering only its average value for each greater errors. revolution (cadence). It should be noted that the error observed in this analysis This, however, introduces an unpredictable error into the power occurs because to measure power the power meter considers calculation. To use the average value of angular velocity means the average angular velocity of each rotation. In power meters to consider each pedal revolution as perfectly smooth and that use this type of calculation, this error must therefore be uniform: but this type of pedal revolution does not exist in added to the accuracy stated by the manufacturer.
    [Show full text]
  • Basement Flood Mitigation
    1 Mitigation refers to measures taken now to reduce losses in the future. How can homeowners and renters protect themselves and their property from a devastating loss? 2 There are a range of possible causes for basement flooding and some potential remedies. Many of these low-cost options can be factored into a family’s budget and accomplished over the several months that precede storm season. 3 There are four ways water gets into your basement: Through the drainage system, known as the sump. Backing up through the sewer lines under the house. Seeping through cracks in the walls and floor. Through windows and doors, called overland flooding. 4 Gutters can play a huge role in keeping basements dry and foundations stable. Water damage caused by clogged gutters can be severe. Install gutters and downspouts. Repair them as the need arises. Keep them free of debris. 5 Channel and disperse water away from the home by lengthening the run of downspouts with rigid or flexible extensions. Prevent interior intrusion through windows and replace weather stripping as needed. 6 Many varieties of sturdy window well covers are available, simple to install and hinged for easy access. Wells should be constructed with gravel bottoms to promote drainage. Remove organic growth to permit sunlight and ventilation. 7 Berms and barriers can help water slope away from the home. The berm’s slope should be about 1 inch per foot and extend for at least 10 feet. It is important to note permits are required any time a homeowner alters the elevation of the property.
    [Show full text]
  • Simple Harmonic Motion
    [SHIVOK SP211] October 30, 2015 CH 15 Simple Harmonic Motion I. Oscillatory motion A. Motion which is periodic in time, that is, motion that repeats itself in time. B. Examples: 1. Power line oscillates when the wind blows past it 2. Earthquake oscillations move buildings C. Sometimes the oscillations are so severe, that the system exhibiting oscillations break apart. 1. Tacoma Narrows Bridge Collapse "Gallopin' Gertie" a) http://www.youtube.com/watch?v=j‐zczJXSxnw II. Simple Harmonic Motion A. http://www.youtube.com/watch?v=__2YND93ofE Watch the video in your spare time. This professor is my teaching Idol. B. In the figure below snapshots of a simple oscillatory system is shown. A particle repeatedly moves back and forth about the point x=0. Page 1 [SHIVOK SP211] October 30, 2015 C. The time taken for one complete oscillation is the period, T. In the time of one T, the system travels from x=+x , to –x , and then back to m m its original position x . m D. The velocity vector arrows are scaled to indicate the magnitude of the speed of the system at different times. At x=±x , the velocity is m zero. E. Frequency of oscillation is the number of oscillations that are completed in each second. 1. The symbol for frequency is f, and the SI unit is the hertz (abbreviated as Hz). 2. It follows that F. Any motion that repeats itself is periodic or harmonic. G. If the motion is a sinusoidal function of time, it is called simple harmonic motion (SHM).
    [Show full text]
  • Hub City Powertorque® Shaft Mount Reducers
    Hub City PowerTorque® Shaft Mount Reducers PowerTorque® Features and Description .................................................. G-2 PowerTorque Nomenclature ............................................................................................ G-4 Selection Instructions ................................................................................ G-5 Selection By Horsepower .......................................................................... G-7 Mechanical Ratings .................................................................................... G-12 ® Shaft Mount Reducers Dimensions ................................................................................................ G-14 Accessories ................................................................................................ G-15 Screw Conveyor Accessories ..................................................................... G-22 G For Additional Models of Shaft Mount Reducers See Hub City Engineering Manual Sections F & J DOWNLOAD AVAILABLE CAD MODELS AT: WWW.HUBCITYINC.COM Certified prints are available upon request EMAIL: [email protected] • www.hubcityinc.com G-1 Hub City PowerTorque® Shaft Mount Reducers Ten models available from 1/4 HP through 200 HP capacity Manufacturing Quality Manufactured to the highest quality 98.5% standards in the industry, assembled Efficiency using precision manufactured components made from top quality per Gear Stage! materials Designed for the toughest applications in the industry Housings High strength ductile
    [Show full text]
  • Oscillator Circuit Evaluation Method (2) Steps for Evaluating Oscillator Circuits (Oscillation Allowance and Drive Level)
    Technical Notes Oscillator Circuit Evaluation Method (2) Steps for evaluating oscillator circuits (oscillation allowance and drive level) Preface In general, a crystal unit needs to be matched with an oscillator circuit in order to obtain a stable oscillation. A poor match between crystal unit and oscillator circuit can produce a number of problems, including, insufficient device frequency stability, devices stop oscillating, and oscillation instability. When using a crystal unit in combination with a microcontroller, you have to evaluate the oscillator circuit. In order to check the match between the crystal unit and the oscillator circuit, you must, at least, evaluate (1) oscillation frequency (frequency matching), (2) oscillation allowance (negative resistance), and (3) drive level. The previous Technical Notes explained frequency matching. These Technical Notes describe the evaluation methods for oscillation allowance (negative resistance) and drive level. 1. Oscillation allowance (negative resistance) evaluations One process used as a means to easily evaluate the negative resistance characteristics and oscillation allowance of an oscillator circuits is the method of adding a resistor to the hot terminal of the crystal unit and observing whether it can oscillate (examining the negative resistance RN). The oscillator circuit capacity can be examined by changing the value of the added resistance (size of loss). The circuit diagram for measuring the negative resistance is shown in Fig. 1. The absolute value of the negative resistance is the value determined by summing up the added resistance r and the equivalent resistance (Re) when the crystal unit is under load. Formula (1) Rf Rd r | RN | Connect _ r＋R e ..
    [Show full text]
  • Ch 11 Vibrations and Waves Simple Harmonic Motion Simple Harmonic Motion
    Ch 11 Vibrations and Waves Simple Harmonic Motion Simple Harmonic Motion A vibration (oscillation) back & forth taking the same amount of time for each cycle is periodic. Each vibration has an equilibrium position from which it is somehow disturbed by a given energy source. The disturbance produces a displacement from equilibrium. This is followed by a restoring force. Vibrations transfer energy. Recall Hooke’s Law The restoring force of a spring is proportional to the displacement, x. F = -kx. K is the proportionality constant and we choose the equilibrium position of x = 0. The minus sign reminds us the restoring force is always opposite the displacement, x. F is not constant but varies with position. Acceleration of the mass is not constant therefore. http://www.youtube.com/watch?v=eeYRkW8V7Vg&feature=pl ayer_embedded Key Terms Displacement- distance from equilibrium Amplitude- maximum displacement Cycle- one complete to and fro motion Period (T)- Time for one complete cycle (s) Frequency (f)- number of cycles per second (Hz) * period and frequency are inversely related: T = 1/f f = 1/T Energy in SHOs (Simple Harmonic Oscillators) In stretching or compressing a spring, work is required and potential energy is stored. Elastic PE is given by: PE = ½ kx2 Total mechanical energy E of the mass-spring system = sum of KE + PE E = ½ mv2 + ½ kx2 Here v is velocity of the mass at x position from equilibrium. E remains constant w/o friction. Energy Transformations As a mass oscillates on a spring, the energy changes from PE to KE while the total E remains constant.
    [Show full text]
  • Rotational Motion of Electric Machines
    Rotational Motion of Electric Machines • An electric machine rotates about a fixed axis, called the shaft, so its rotation is restricted to one angular dimension. • Relative to a given end of the machine’s shaft, the direction of counterclockwise (CCW) rotation is often assumed to be positive. • Therefore, for rotation about a fixed shaft, all the concepts are scalars. 17 Angular Position, Velocity and Acceleration • Angular position – The angle at which an object is oriented, measured from some arbitrary reference point – Unit: rad or deg – Analogy of the linear concept • Angular acceleration =d/dt of distance along a line. – The rate of change in angular • Angular velocity =d/dt velocity with respect to time – The rate of change in angular – Unit: rad/s2 position with respect to time • and >0 if the rotation is CCW – Unit: rad/s or r/min (revolutions • >0 if the absolute angular per minute or rpm for short) velocity is increasing in the CCW – Analogy of the concept of direction or decreasing in the velocity on a straight line. CW direction 18 Moment of Inertia (or Inertia) • Inertia depends on the mass and shape of the object (unit: kgm2) • A complex shape can be broken up into 2 or more of simple shapes Definition Two useful formulas mL2 m J J() RRRR22 12 3 1212 m 22 JRR()12 2 19 Torque and Change in Speed • Torque is equal to the product of the force and the perpendicular distance between the axis of rotation and the point of application of the force. T=Fr (Nm) T=0 T T=Fr • Newton’s Law of Rotation: Describes the relationship between the total torque applied to an object and its resulting angular acceleration.
    [Show full text]
  • PHYS 211 Lecture 5 - Oscillations: Simple Harmonic Motion 5 - 1
    PHYS 211 Lecture 5 - Oscillations: simple harmonic motion 5 - 1 Lecture 5 - Oscillations: simple harmonic motion Text: Fowles and Cassiday, Chap. 3 Consider a power series expansion for the potential energy of an object 2 V(x) = Vo + V1x + V2x + ..., where x is the displacement of an object from its equilibrium position, and Vi, i = 0,1,2... are fixed coefficients. The leading order term in this series is unimportant to the dynamics of the object, since F = -dV/dx and the derivative of a constant vanishes. Further, if we require the equilibrium position x = 0 to be a true minimum in the energy, then V1 = 0. This doesn’t mean that potentials with odd powers in x must vanish, but just says that their minimum is not at x = 0. Thus, the simplest function that one can write is quadratic: 2 V = V2x [A slightly more complicated variant is |x| = (x2)1/2, which is not smooth at x = 0]. For small excursions from equilibrium, the quadratic term is often the leading-order piece of more complex functions such as the Morse or Lennard-Jones potentials. Quadratic potentials correspond to the familiar Hooke’s Law of ideal springs V(x) = kx 2/2 => F(x) = -dV/dx = -(2/2)kx = -kx, where k is the spring constant. Objects subject to Hooke’s Law exhibit oscillatory motion, as can be seen by solving the differential equation arising from Newton’s 2nd law: F = ma => -kx = m(d 2x/dt 2) or d 2x /dt 2 + (k/m)x = 0. We solved this equation in PHYS 120: x(t) = A sin( ot + o) Other functional forms such as cosine can be changed into this form using a suitable choice of the phase angle o.
    [Show full text]
  • Determining the Load Inertia Contribution from Different Power Consumer Groups
    energies Article Determining the Load Inertia Contribution from Different Power Consumer Groups Henning Thiesen * and Clemens Jauch Wind Energy Technology Institute (WETI), Flensburg University of Applied Sciences, 24943 Flensburg, Germany; clemens.jauch@hs-flensburg.de * Correspondence: henning.thiesen@hs-flensburg.de Received: 27 February 2020; Accepted: 25 March 2020 ; Published: 1 April 2020 Abstract: Power system inertia is a vital part of power system stability. The inertia response within the first seconds after a power imbalance reduces the velocity of which the grid frequency changes. At present, large shares of power system inertia are provided by synchronously rotating masses of conventional power plants. A minor part of power system inertia is supplied by power consumers. The energy system transformation results in an overall decreasing amount of power system inertia. Hence, inertia has to be provided synthetically in future power systems. In depth knowledge about the amount of inertia provided by power consumers is very important for a future application of units supplying synthetic inertia. It strongly promotes the technical efficiency and cost effective application. A blackout in the city of Flensburg allows for a detailed research on the inertia contribution from power consumers. Therefore, power consumer categories are introduced and the inertia contribution is calculated for each category. Overall, the inertia constant for different power consumers is in the range of 0.09 to 4.24 s if inertia constant calculations are based on the power demand. If inertia constant calculations are based on the apparent generator power, the load inertia constant is in the range of 0.01 to 0.19 s.
    [Show full text]
  • Hydraulics Manual Glossary G - 3
    Glossary G - 1 GLOSSARY OF HIGHWAY-RELATED DRAINAGE TERMS (Reprinted from the 1999 edition of the American Association of State Highway and Transportation Officials Model Drainage Manual) G.1 Introduction This Glossary is divided into three parts: · Introduction, · Glossary, and · References. It is not intended that all the terms in this Glossary be rigorously accurate or complete. Realistically, this is impossible. Depending on the circumstance, a particular term may have several meanings; this can never change. The primary purpose of this Glossary is to define the terms found in the Highway Drainage Guidelines and Model Drainage Manual in a manner that makes them easier to interpret and understand. A lesser purpose is to provide a compendium of terms that will be useful for both the novice as well as the more experienced hydraulics engineer. This Glossary may also help those who are unfamiliar with highway drainage design to become more understanding and appreciative of this complex science as well as facilitate communication between the highway hydraulics engineer and others. Where readily available, the source of a definition has been referenced. For clarity or format purposes, cited definitions may have some additional verbiage contained in double brackets [ ]. Conversely, three “dots” (...) are used to indicate where some parts of a cited definition were eliminated. Also, as might be expected, different sources were found to use different hyphenation and terminology practices for the same words. Insignificant changes in this regard were made to some cited references and elsewhere to gain uniformity for the terms contained in this Glossary: as an example, “groundwater” vice “ground-water” or “ground water,” and “cross section area” vice “cross-sectional area.” Cited definitions were taken primarily from two sources: W.B.
    [Show full text]
  • Euler Equation and Geodesics R
    Euler Equation and Geodesics R. Herman February 2, 2018 Introduction Newton formulated the laws of motion in his 1687 volumes, col- lectively called the Philosophiae Naturalis Principia Mathematica, or simply the Principia. However, Newton’s development was geometrical and is not how we see classical dynamics presented when we first learn mechanics. The laws of mechanics are what are now considered analytical mechanics, in which classical dynamics is presented in a more elegant way. It is based upon variational principles, whose foundations began with the work of Eu- ler and Lagrange and have been refined by other now-famous figures in the eighteenth and nineteenth centuries. Euler coined the term the calculus of variations in 1756, though it is also called variational calculus. The goal is to find minima or maxima of func- tions of the form f : M ! R, where M can be a set of numbers, functions, paths, curves, surfaces, etc. Interest in extrema problems in classical mechan- ics began near the end of the seventeenth century with Newton and Leibniz. In the Principia, Newton was interested in the least resistance of a surface of revolution as it moves through a fluid. Seeking extrema at the time was not new, as the Egyptians knew that the shortest path between two points is a straight line and that a circle encloses the largest area for a given perimeter. Heron, an Alexandrian scholar, deter- mined that light travels along the shortest path. This problem was later taken up by Willibrord Snellius (1580–1626) after whom Snell’s law of refraction is named.
    [Show full text]