simple pendulum problems and solutions pdf

>> /Subtype/Type1 Let's do them in that order. << Problem (12): If the frequency of a 69-cm-long pendulum is 0.601 Hz, what is the value of the acceleration of gravity $g$ at that location? 527.8 314.8 524.7 314.8 314.8 524.7 472.2 472.2 524.7 472.2 314.8 472.2 524.7 314.8 In Figure 3.3 we draw the nal phase line by itself. /Widths[342.6 581 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 312.5 /Length 2736 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 In addition, there are hundreds of problems with detailed solutions on various physics topics. 'z.msV=eS!6\f=QE|>9lqqQ/h%80 t v{"m4T>8|m@pqXAep'|@Dq;q>mr)G?P-| +*"!b|b"YI!kZfIZNh!|!Dwug5c #6h>qp:9j(s%s*}BWuz(g}} ]7N.k=l 537|?IsV /LastChar 196 Adding pennies to the Great Clock shortens the effective length of its pendulum by about half the width of a human hair. /Name/F4 877 0 0 815.5 677.6 646.8 646.8 970.2 970.2 323.4 354.2 569.4 569.4 569.4 569.4 569.4 The length of the cord of the simple pendulum (l) = 1 meter, Wanted: determine the length of rope if the frequency is twice the initial frequency. 295.1 826.4 531.3 826.4 531.3 559.7 795.8 801.4 757.3 871.7 778.7 672.4 827.9 872.8 Tension in the string exactly cancels the component mgcosmgcos parallel to the string. << 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 627.2 817.8 766.7 692.2 664.4 743.3 715.6 %PDF-1.2 805.5 896.3 870.4 935.2 870.4 935.2 0 0 870.4 736.1 703.7 703.7 1055.5 1055.5 351.8 /Subtype/Type1 /LastChar 196 WebSimple Pendulum Problems and Formula for High Schools. If you need help, our customer service team is available 24/7. 388.9 1000 1000 416.7 528.6 429.2 432.8 520.5 465.6 489.6 477 576.2 344.5 411.8 520.6 when the pendulum is again travelling in the same direction as the initial motion. /Type/Font sin How long is the pendulum? 343.8 593.8 312.5 937.5 625 562.5 625 593.8 459.5 443.8 437.5 625 593.8 812.5 593.8 Simple Harmonic Motion describes this oscillatory motion where the displacement, velocity and acceleration are sinusoidal. Now use the slope to get the acceleration due to gravity. 275 1000 666.7 666.7 888.9 888.9 0 0 555.6 555.6 666.7 500 722.2 722.2 777.8 777.8 33 0 obj 39 0 obj 323.4 877 538.7 538.7 877 843.3 798.6 815.5 860.1 767.9 737.1 883.9 843.3 412.7 583.3 WebThe simple pendulum system has a single particle with position vector r = (x,y,z). WebStudents are encouraged to use their own programming skills to solve problems. /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 /FontDescriptor 14 0 R But the median is also appropriate for this problem (gtilde). 19 0 obj if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[336,280],'physexams_com-leader-1','ezslot_11',112,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-leader-1-0'); Therefore, with increasing the altitude, $g$ becomes smaller and consequently the period of the pendulum becomes larger. x DO2(EZxIiTt |"r>^p-8y:>C&%QSSV]aq,GVmgt4A7tpJ8 C |2Z4dpGuK.DqCVpHMUN j)VP(!8#n If this doesn't solve the problem, visit our Support Center . Based on the above formula, can conclude the length of the rod (l) and the acceleration of gravity (g) impact the period of the simple pendulum. 762.8 642 790.6 759.3 613.2 584.4 682.8 583.3 944.4 828.5 580.6 682.6 388.9 388.9 Math Assignments Frequency of a pendulum calculator Formula : T = 2 L g . >> /LastChar 196 687.5 312.5 581 312.5 562.5 312.5 312.5 546.9 625 500 625 513.3 343.8 562.5 625 312.5 xZ[o6~G XuX\IQ9h_sEIEZBW4(!}wbSL0!` eIo`9vEjshTv=>G+|13]jkgQaw^eh5I'oEtW;`;lH}d{|F|^+~wXE\DjQaiNZf>_6#.Pvw,TsmlHKl(S{"l5|"i7{xY(rebL)E$'gjOB$$=F>| -g33_eDb/ak]DceMew[6;|^nzVW4s#BstmQFVTmqKZ=pYp0d%`=5t#p9q`h!wi 6i-z,Y(Hx8B!}sWDy3#EF-U]QFDTrKDPD72mF. For the next question you are given the angle at the centre, 98 degrees, and the arc length, 10cm. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). /FirstChar 33 A simple pendulum of length 1 m has a mass of 10 g and oscillates freely with an amplitude of 2 cm. R ))jM7uM*%? Solution: The length $\ell$ and frequency $f$ of a simple pendulum are given and $g$ is unknown. 0.5 We move it to a high altitude. 513.9 770.7 456.8 513.9 742.3 799.4 513.9 927.8 1042 799.4 285.5 513.9] In part a ii we assumed the pendulum would be used in a working clock one designed to match the cultural definitions of a second, minute, hour, and day. Webproblems and exercises for this chapter. consent of Rice University. endobj That's a loss of 3524s every 30days nearly an hour (58:44). 495.7 376.2 612.3 619.8 639.2 522.3 467 610.1 544.1 607.2 471.5 576.4 631.6 659.7 Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. 11 0 obj /FontDescriptor 23 0 R << % Figure 2: A simple pendulum attached to a support that is free to move. WebEnergy of the Pendulum The pendulum only has gravitational potential energy, as gravity is the only force that does any work. << Begin by calculating the period of a simple pendulum whose length is 4.4m. The period you just calculated would not be appropriate for a clock of this stature. Physexams.com, Simple Pendulum Problems and Formula for High Schools. What is the value of g at a location where a 2.2 m long pendulum has a period of 2.5 seconds? 5 0 obj This part of the question doesn't require it, but we'll need it as a reference for the next two parts. 306.7 766.7 511.1 511.1 766.7 743.3 703.9 715.6 755 678.3 652.8 773.6 743.3 385.6 /Widths[277.8 500 833.3 500 833.3 777.8 277.8 388.9 388.9 500 777.8 277.8 333.3 277.8 1277.8 811.1 811.1 875 875 666.7 666.7 666.7 666.7 666.7 666.7 888.9 888.9 888.9 What is the acceleration of gravity at that location? Note the dependence of TT on gg. 27 0 obj Bonus solutions: Start with the equation for the period of a simple pendulum. /FontDescriptor 26 0 R The motion of the particles is constrained: the lengths are l1 and l2; pendulum 1 is attached to a xed point in space and pendulum 2 is attached to the end of pendulum 1. Exams will be effectively half of an AP exam - 17 multiple choice questions (scaled to 22. The heart of the timekeeping mechanism is a 310kg, 4.4m long steel and zinc pendulum. \begin{gather*} T=2\pi\sqrt{\frac{2}{9.8}}=2.85\quad {\rm s} \\ \\ f=\frac{1}{2.85\,{\rm s}}=0.35\quad {\rm Hz}\end{gather*}. xa ` 2s-m7k Free vibrations ; Damped vibrations ; Forced vibrations ; Resonance ; Nonlinear models ; Driven models ; Pendulum . /Type/Font A cycle is one complete oscillation. If displacement from equilibrium is very small, then the pendulum of length $\ell$ approximate simple harmonic motion. How does adding pennies to the pendulum in the Great Clock help to keep it accurate? /Subtype/Type1 Hence, the length must be nine times. All of us are familiar with the simple pendulum. Now for the mathematically difficult question. Problem (9): Of simple pendulum can be used to measure gravitational acceleration. 472.2 472.2 472.2 472.2 583.3 583.3 0 0 472.2 472.2 333.3 555.6 577.8 577.8 597.2 The mass does not impact the frequency of the simple pendulum. For angles less than about 1515, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator. For the simple pendulum: for the period of a simple pendulum. /Name/F4 (c) Frequency of a pendulum is related to its length by the following formula \begin{align*} f&=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}} \\\\ 1.25&=\frac{1}{2\pi}\sqrt{\frac{9.8}{\ell}}\\\\ (2\pi\times 1.25)^2 &=\left(\sqrt{\frac{9.8}{\ell}}\right)^2 \\\\ \Rightarrow \ell&=\frac{9.8}{4\pi^2\times (1.25)^2} \\\\&=0.16\quad {\rm m}\end{align*} Thus, the length of this kind of pendulum is about 16 cm. Webpoint of the double pendulum. This shortens the effective length of the pendulum. /BaseFont/CNOXNS+CMR10 >> >> <> Tell me where you see mass. Thus, the period is \[T=\frac{1}{f}=\frac{1}{1.25\,{\rm Hz}}=0.8\,{\rm s}\] /LastChar 196 WebSecond-order nonlinear (due to sine function) ordinary differential equation describing the motion of a pendulum of length L : In the next group of examples, the unknown function u depends on two variables x and t or x and y . Electric generator works on the scientific principle. /Subtype/Type1 endstream Resonance of sound wave problems and solutions, Simple harmonic motion problems and solutions, Electric current electric charge magnetic field magnetic force, Quantities of physics in the linear motion. << l+2X4J!$w|-(6}@:BtxzwD'pSe5ui8,:7X88 :r6m;|8Xxe The Island Worksheet Answers from forms of energy worksheet answers , image source: www. /Type/Font << 24 0 obj \(&SEc Pendulum 2 has a bob with a mass of 100 kg100 kg. ECON 102 Quiz 1 test solution questions and answers solved solutions. /Subtype/Type1 Problem (1): In a simple pendulum, how much the length of it must be changed to triple its period? Pendulum 1 has a bob with a mass of 10kg10kg. Websector-area-and-arc-length-answer-key 1/6 Downloaded from accreditation. /Widths[342.6 581 937.5 562.5 937.5 875 312.5 437.5 437.5 562.5 875 312.5 375 312.5 % /BaseFont/NLTARL+CMTI10 ollB;% !JA6Avls,/vqnpPw}o@g `FW[StFb s%EbOq#!!!h#']y\1FKW6 <> stream They recorded the length and the period for pendulums with ten convenient lengths. 384.3 611.1 611.1 611.1 611.1 611.1 896.3 546.3 611.1 870.4 935.2 611.1 1077.8 1207.4 /FontDescriptor 29 0 R A pendulum is a massive bob attached to a string or cord and swings back and forth in a periodic motion. 465 322.5 384 636.5 500 277.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Find its PE at the extreme point. endobj /FontDescriptor 35 0 R 0 0 0 0 0 0 0 0 0 0 0 0 675.9 937.5 875 787 750 879.6 812.5 875 812.5 875 0 0 812.5 /FirstChar 33 /LastChar 196 endobj 777.8 694.4 666.7 750 722.2 777.8 722.2 777.8 0 0 722.2 583.3 555.6 555.6 833.3 833.3 A simple pendulum shows periodic motion, and it occurs in the vertical plane and is mainly driven by the gravitational force. 2015 All rights reserved. 4. We can discern one half the smallest division so DVVV= ()05 01 005.. .= VV V= D ()385 005.. 4. /Subtype/Type1 643.8 920.4 763 787 696.3 787 748.8 577.2 734.6 763 763 1025.3 763 763 629.6 314.8 Consider the following example. /FontDescriptor 11 0 R 639.7 565.6 517.7 444.4 405.9 437.5 496.5 469.4 353.9 576.2 583.3 602.5 494 437.5 770.7 628.1 285.5 513.9 285.5 513.9 285.5 285.5 513.9 571 456.8 571 457.2 314 513.9 What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s? /FirstChar 33 3 0 obj endobj endobj 27 0 obj 277.8 500] frequency to be doubled, the length of the pendulum should be changed to 0.25 meters. 384.3 611.1 611.1 611.1 611.1 611.1 896.3 546.3 611.1 870.4 935.2 611.1 1077.8 1207.4 The displacement ss is directly proportional to . endobj It consists of a point mass m suspended by means of light inextensible string of length L from a fixed support as shown in Fig. WebPhysics 1120: Simple Harmonic Motion Solutions 1. This book uses the >> Its easy to measure the period using the photogate timer. Although adding pennies to the Great Clock changes its weight (by which we assume the Daily Mail meant its mass) this is not a factor that affects the period of a pendulum (simple or physical). 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 777.8 500 777.8 500 530.9 This PDF provides a full solution to the problem. and you must attribute OpenStax. They recorded the length and the period for pendulums with ten convenient lengths. @bL7]qwxuRVa1Z/. HFl`ZBmMY7JHaX?oHYCBb6#'\ }! 750 708.3 722.2 763.9 680.6 652.8 784.7 750 361.1 513.9 777.8 625 916.7 750 777.8 30 0 obj (Keep every digit your calculator gives you. << g /Widths[306.7 514.4 817.8 769.1 817.8 766.7 306.7 408.9 408.9 511.1 766.7 306.7 357.8 Two simple pendulums are in two different places. 694.5 295.1] <> /Name/F5 3.5 Pendulum period 72 2009-02-10 19:40:05 UTC / rev 4d4a39156f1e Even if the analysis of the conical pendulum is simple, how is it relevant to the motion of a one-dimensional pendulum? /Type/Font By what amount did the important characteristic of the pendulum change when a single penny was added near the pivot. Perform a propagation of error calculation on the two variables: length () and period (T). H 874 706.4 1027.8 843.3 877 767.9 877 829.4 631 815.5 843.3 843.3 1150.8 843.3 843.3 2 0 obj WebSimple Harmonic Motion and Pendulums SP211: Physics I Fall 2018 Name: 1 Introduction When an object is oscillating, the displacement of that object varies sinusoidally with time. << << /Type /XRef /Length 85 /Filter /FlateDecode /DecodeParms << /Columns 5 /Predictor 12 >> /W [ 1 3 1 ] /Index [ 18 54 ] /Info 16 0 R /Root 20 0 R /Size 72 /Prev 140934 /ID [<8a3b51e8e1dcde48ea7c2079c7f2691d>] >> When the pendulum is elsewhere, its vertical displacement from the = 0 point is h = L - L cos() (see diagram) endobj 639.7 565.6 517.7 444.4 405.9 437.5 496.5 469.4 353.9 576.2 583.3 602.5 494 437.5 295.1 826.4 501.7 501.7 826.4 795.8 752.1 767.4 811.1 722.6 693.1 833.5 795.8 382.6 Instead of an infinitesimally small mass at the end, there's a finite (but concentrated) lump of material. Simple pendulum ; Solution of pendulum equation ; Period of pendulum ; Real pendulum ; Driven pendulum ; Rocking pendulum ; Pumping swing ; Dyer model ; Electric circuits; citation tool such as, Authors: Paul Peter Urone, Roger Hinrichs. [894 m] 3. One of the authors (M. S.) has been teaching the Introductory Physics course to freshmen since Fall 2007. /Name/F6 /FirstChar 33 In the case of a massless cord or string and a deflection angle (relative to vertical) up to $5^\circ$, we can find a simple formula for the period and frequency of a pendulum as below \[T=2\pi\sqrt{\frac{\ell}{g}}\quad,\quad f=\frac{1}{2\pi}\sqrt{\frac{g}{\ell}}\] where $\ell$ is the length of the pendulum and $g$ is the acceleration of gravity at that place. 783.4 872.8 823.4 619.8 708.3 654.8 0 0 816.7 682.4 596.2 547.3 470.1 429.5 467 533.2 << endobj 1. 833.3 1444.4 1277.8 555.6 1111.1 1111.1 1111.1 1111.1 1111.1 944.4 1277.8 555.6 1000 /FirstChar 33 We can solve T=2LgT=2Lg for gg, assuming only that the angle of deflection is less than 1515. Pendulum clocks really need to be designed for a location. 285.5 799.4 485.3 485.3 799.4 770.7 727.9 742.3 785 699.4 670.8 806.5 770.7 371 528.1 Representative solution behavior and phase line for y = y y2. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 642.9 885.4 806.2 736.8 How accurate is this measurement? endobj 21 0 obj This is a test of precision.). x|TE?~fn6 @B&$& Xb"K`^@@ >> << << 1999-2023, Rice University. Examples of Projectile Motion 1. WebAustin Community College District | Start Here. Pnlk5|@UtsH mIr /Type/Font /LastChar 196 The time taken for one complete oscillation is called the period. /Type/Font 600.2 600.2 507.9 569.4 1138.9 569.4 569.4 569.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Math Assignments Frequency of a pendulum calculator Formula : T = 2 L g . 850.9 472.2 550.9 734.6 734.6 524.7 906.2 1011.1 787 262.3 524.7] 9 0 obj 1000 1000 1055.6 1055.6 1055.6 777.8 666.7 666.7 450 450 450 450 777.8 777.8 0 0 777.8 777.8 1000 1000 777.8 777.8 1000 777.8] Which answer is the right answer? Thus, The frequency of this pendulum is \[f=\frac{1}{T}=\frac{1}{3}\,{\rm Hz}\], Problem (3): Find the length of a pendulum that has a frequency of 0.5 Hz. Adding one penny causes the clock to gain two-fifths of a second in 24hours. Trading chart patters How to Trade the Double Bottom Chart Pattern Nixfx Capital Market. endobj A pendulum is a massive bob attached to a string or cord and swings back and forth in a periodic motion. 384.3 611.1 675.9 351.8 384.3 643.5 351.8 1000 675.9 611.1 675.9 643.5 481.5 488 << 812.5 875 562.5 1018.5 1143.5 875 312.5 562.5] /Widths[791.7 583.3 583.3 638.9 638.9 638.9 638.9 805.6 805.6 805.6 805.6 1277.8 nB5- Creative Commons Attribution License /Subtype/Type1 The length of the second pendulum is 0.4 times the length of the first pendulum, and the, second pendulum is 0.9 times the acceleration of gravity, The length of the cord of the first pendulum, The length of cord of the second pendulum, Acceleration due to the gravity of the first pendulum, Acceleration due to gravity of the second pendulum, he comparison of the frequency of the first pendulum (f. Hertz. That's a question that's best left to a professional statistician. WebAssuming nothing gets in the way, that conclusion is reached when the projectile comes to rest on the ground. endobj >> Simplify the numerator, then divide. 277.8 500 555.6 444.4 555.6 444.4 305.6 500 555.6 277.8 305.6 527.8 277.8 833.3 555.6 743.3 743.3 613.3 306.7 514.4 306.7 511.1 306.7 306.7 511.1 460 460 511.1 460 306.7 /Name/F7 799.2 642.3 942 770.7 799.4 699.4 799.4 756.5 571 742.3 770.7 770.7 1056.2 770.7 << /FontDescriptor 14 0 R /BaseFont/AVTVRU+CMBX12 /FirstChar 33 The rst pendulum is attached to a xed point and can freely swing about it. /Name/F8 WebThe essence of solving nonlinear problems and the differences and relations of linear and nonlinear problems are also simply discussed. They attached a metal cube to a length of string and let it swing freely from a horizontal clamp. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 753.7 1000 935.2 831.5 In this problem has been said that the pendulum clock moves too slowly so its time period is too large. Half of this is what determines the amount of time lost when this pendulum is used as a time keeping device in its new location. 762.8 642 790.6 759.3 613.2 584.4 682.8 583.3 944.4 828.5 580.6 682.6 388.9 388.9 323.4 877 538.7 538.7 877 843.3 798.6 815.5 860.1 767.9 737.1 883.9 843.3 412.7 583.3 endobj /FirstChar 33 What is the period of oscillations? 877 0 0 815.5 677.6 646.8 646.8 970.2 970.2 323.4 354.2 569.4 569.4 569.4 569.4 569.4 WebRepresentative solution behavior for y = y y2. 9 0 obj << 460.7 580.4 896 722.6 1020.4 843.3 806.2 673.6 835.7 800.2 646.2 618.6 718.8 618.8 Let us define the potential energy as being zero when the pendulum is at the bottom of the swing, = 0 . /LastChar 196 As with simple harmonic oscillators, the period TT for a pendulum is nearly independent of amplitude, especially if is less than about 1515. Use a simple pendulum to determine the acceleration due to gravity 7 0 obj |l*HA 9 0 obj WebSolution : The equation of period of the simple pendulum : T = period, g = acceleration due to gravity, l = length of cord. 826.4 295.1 531.3] How long should a pendulum be in order to swing back and forth in 1.6 s? For the precision of the approximation are not subject to the Creative Commons license and may not be reproduced without the prior and express written Consider a geologist that uses a pendulum of length $35\,{\rm cm}$ and frequency of 0.841 Hz at a specific place on the Earth. Look at the equation below. /LastChar 196 Find the period and oscillation of this setup. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 627.2 817.8 766.7 692.2 664.4 743.3 715.6 323.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 569.4 323.4 323.4 Calculate gg. 460 664.4 463.9 485.6 408.9 511.1 1022.2 511.1 511.1 511.1 0 0 0 0 0 0 0 0 0 0 0 343.8 593.8 312.5 937.5 625 562.5 625 593.8 459.5 443.8 437.5 625 593.8 812.5 593.8 are licensed under a, Introduction: The Nature of Science and Physics, Introduction to Science and the Realm of Physics, Physical Quantities, and Units, Accuracy, Precision, and Significant Figures, Introduction to One-Dimensional Kinematics, Motion Equations for Constant Acceleration in One Dimension, Problem-Solving Basics for One-Dimensional Kinematics, Graphical Analysis of One-Dimensional Motion, Introduction to Two-Dimensional Kinematics, Kinematics in Two Dimensions: An Introduction, Vector Addition and Subtraction: Graphical Methods, Vector Addition and Subtraction: Analytical Methods, Dynamics: Force and Newton's Laws of Motion, Introduction to Dynamics: Newtons Laws of Motion, Newtons Second Law of Motion: Concept of a System, Newtons Third Law of Motion: Symmetry in Forces, Normal, Tension, and Other Examples of Forces, Further Applications of Newtons Laws of Motion, Extended Topic: The Four Basic ForcesAn Introduction, Further Applications of Newton's Laws: Friction, Drag, and Elasticity, Introduction: Further Applications of Newtons Laws, Introduction to Uniform Circular Motion and Gravitation, Fictitious Forces and Non-inertial Frames: The Coriolis Force, Satellites and Keplers Laws: An Argument for Simplicity, Introduction to Work, Energy, and Energy Resources, Kinetic Energy and the Work-Energy Theorem, Introduction to Linear Momentum and Collisions, Collisions of Point Masses in Two Dimensions, Applications of Statics, Including Problem-Solving Strategies, Introduction to Rotational Motion and Angular Momentum, Dynamics of Rotational Motion: Rotational Inertia, Rotational Kinetic Energy: Work and Energy Revisited, Collisions of Extended Bodies in Two Dimensions, Gyroscopic Effects: Vector Aspects of Angular Momentum, Variation of Pressure with Depth in a Fluid, Gauge Pressure, Absolute Pressure, and Pressure Measurement, Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action, Fluid Dynamics and Its Biological and Medical Applications, Introduction to Fluid Dynamics and Its Biological and Medical Applications, The Most General Applications of Bernoullis Equation, Viscosity and Laminar Flow; 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