Wizard Og Strain,
Lake County Captains Stadium Events,
Why Is Howie Called Chimney On 911,
Articles S
Pendulum clocks really need to be designed for a location. WebMass Pendulum Dynamic System chp3 15 A simple plane pendulum of mass m 0 and length l is suspended from a cart of mass m as sketched in the figure. Get There. WebThe simple pendulum system has a single particle with position vector r = (x,y,z).
pendulum << <>
/BaseFont/LFMFWL+CMTI9 Let us define the potential energy as being zero when the pendulum is at the bottom of the swing, = 0 . x DO2(EZxIiTt |"r>^p-8y:>C&%QSSV]aq,GVmgt4A7tpJ8 C
|2Z4dpGuK.DqCVpHMUN j)VP(!8#n 21 0 obj %PDF-1.5 Webpendulum is sensitive to the length of the string and the acceleration due to gravity. /Type/Font When we discuss damping in Section 1.2, we will nd that the motion is somewhat sinusoidal, but with an important modication. /FThHh!nmoF;TSooevBFN""(+7IcQX.0:Pl@Hs (@Kqd(9)\ (jX First method: Start with the equation for the period of a simple pendulum. We will present our new method by rst stating its rules (without any justication) and showing that they somehow end up magically giving the correct answer. SOLUTION: The length of the arc is 22 (6 + 6) = 10.
Pendulum 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 706.4 938.5 877 781.8 754 843.3 815.5 877 815.5 /FirstChar 33 /Subtype/Type1 WebEnergy of the Pendulum The pendulum only has gravitational potential energy, as gravity is the only force that does any work. 692.5 323.4 569.4 323.4 569.4 323.4 323.4 569.4 631 507.9 631 507.9 354.2 569.4 631 <> stream /BaseFont/JMXGPL+CMR10 Pendulum B is a 400-g bob that is hung from a 6-m-long string. 624.1 928.7 753.7 1090.7 896.3 935.2 818.5 935.2 883.3 675.9 870.4 896.3 896.3 1220.4 /FirstChar 33 In part a i we assumed the pendulum was a simple pendulum one with all the mass concentrated at a point connected to its pivot by a massless, inextensible string.
Simple Pendulum Solution: In 60 seconds it makes 40 oscillations In 1 sec it makes = 40/60 = 2/3 oscillation So frequency = 2/3 per second = 0.67 Hz Time period = 1/frequency = 3/2 = 1.5 seconds 64) The time period of a simple pendulum is 2 s. 473.8 498.5 419.8 524.7 1049.4 524.7 524.7 524.7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 888.9 888.9 888.9 888.9 666.7 875 875 875 875 611.1 611.1 833.3 1111.1 472.2 555.6
Simple Pendulum All of the methods used were appropriate to the problem and all of the calculations done were error free, so all of them. We recommend using a /LastChar 196 A pendulum is a massive bob attached to a string or cord and swings back and forth in a periodic motion. /FirstChar 33 /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 /Subtype/Type1 Creative Commons Attribution License Note how close this is to one meter. /Type/Font 743.3 743.3 613.3 306.7 514.4 306.7 511.1 306.7 306.7 511.1 460 460 511.1 460 306.7 298.4 878 600.2 484.7 503.1 446.4 451.2 468.8 361.1 572.5 484.7 715.9 571.5 490.3
pendulum 1. All of us are familiar with the simple pendulum. Second method: Square the equation for the period of a simple pendulum. 0.5 /BaseFont/EUKAKP+CMR8 It takes one second for it to go out (tick) and another second for it to come back (tock).
(PDF) Numerical solution for time period of simple pendulum with 0 0 0 0 0 0 0 0 0 0 777.8 277.8 777.8 500 777.8 500 777.8 777.8 777.8 777.8 0 0 777.8 >> 666.7 666.7 666.7 666.7 611.1 611.1 444.4 444.4 444.4 444.4 500 500 388.9 388.9 277.8 323.4 354.2 600.2 323.4 938.5 631 569.4 631 600.2 446.4 452.6 446.4 631 600.2 815.5 endobj 491.3 383.7 615.2 517.4 762.5 598.1 525.2 494.2 349.5 400.2 673.4 531.3 295.1 0 0 743.3 743.3 613.3 306.7 514.4 306.7 511.1 306.7 306.7 511.1 460 460 511.1 460 306.7 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 342.6 Simple Pendulum: A simple pendulum device is represented as the point mass attached to a light inextensible string and suspended from a fixed support. Webproblems and exercises for this chapter. (a) What is the amplitude, frequency, angular frequency, and period of this motion? 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 0 663.6 885.4 826.4 736.8 Then, we displace it from its equilibrium as small as possible and release it. A 1.75kg particle moves as function of time as follows: x = 4cos(1.33t+/5) where distance is measured in metres and time in seconds. /Filter[/FlateDecode] Solution; Find the maximum and minimum values of \(f\left( {x,y} \right) = 8{x^2} - 2y\) subject to the constraint \({x^2} + {y^2} = 1\). << Notice how length is one of the symbols. 413.2 590.3 560.8 767.4 560.8 560.8 472.2 531.3 1062.5 531.3 531.3 531.3 0 0 0 0 5 0 obj /Name/F3 /Name/F12 Find its (a) frequency, (b) time period. WebPhysics 1 Lab Manual1Objectives: The main objective of this lab is to determine the acceleration due to gravity in the lab with a simple pendulum. As an object travels through the air, it encounters a frictional force that slows its motion called. Exploring the simple pendulum a bit further, we can discover the conditions under which it performs simple harmonic motion, and we can derive an interesting expression for its period. /BaseFont/AVTVRU+CMBX12
Solutions By shortening the pendulum's length, the period is also reduced, speeding up the pendulum's motion. But the median is also appropriate for this problem (gtilde).
Simple Harmonic Motion Chapter Problems - Weebly >> 777.8 777.8 1000 1000 777.8 777.8 1000 777.8] The Island Worksheet Answers from forms of energy worksheet answers , image source: www. The digital stopwatch was started at a time t 0 = 0 and then was used to measure ten swings of a They recorded the length and the period for pendulums with ten convenient lengths. If f1 is the frequency of the first pendulum and f2 is the frequency of the second pendulum, then determine the relationship between f1 and f2. /Subtype/Type1 These NCERT Solutions provide you with the answers to the question from the textbook, important questions from previous year question papers and sample papers. 351.8 611.1 611.1 611.1 611.1 611.1 611.1 611.1 611.1 611.1 611.1 611.1 351.8 351.8 m77"e^#0=vMHx^3}D:x}??xyx?Z #Y3}>zz&JKP!|gcb;OA6D^z] 'HQnF@[ Fr@G|^7$bK,c>z+|wrZpGxa|Im;L1
e$t2uDpCd4toC@vW# #bx7b?n2e
]Qt8
ye3g6QH "#3n.[\f|r? <> stream 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 Period is the goal. 298.4 878 600.2 484.7 503.1 446.4 451.2 468.8 361.1 572.5 484.7 715.9 571.5 490.3 Instead of a massless string running from the pivot to the mass, there's a massive steel rod that extends a little bit beyond the ideal starting and ending points. 812.5 875 562.5 1018.5 1143.5 875 312.5 562.5] /Subtype/Type1 To verify the hypothesis that static coefficients of friction are dependent on roughness of surfaces, and independent of the weight of the top object. 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 endobj /FirstChar 33 Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. An engineer builds two simple pendula. The problem said to use the numbers given and determine g. We did that. <> stream /FirstChar 33 << >> endobj /Name/F2 To compare the frequency of the two pendulums, we have \begin{align*} \frac{f_A}{f_B}&=\frac{\sqrt{\ell_B}}{\sqrt{\ell_A}}\\\\&=\frac{\sqrt{6}}{\sqrt{2}}\\\\&=\sqrt{3}\end{align*} Therefore, the frequency of pendulum $A$ is $\sqrt{3}$ times the frequency of pendulum $B$. A pendulum is a massive bob attached to a string or cord and swings back and forth in a periodic motion. 314.8 787 524.7 524.7 787 763 722.5 734.6 775 696.3 670.1 794.1 763 395.7 538.9 789.2 By the end of this section, you will be able to: Pendulums are in common usage. This shortens the effective length of the pendulum. 61) Two simple pendulums A and B have equal length, but their bobs weigh 50 gf and l00 gf respectively. 285.5 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 285.5 285.5 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 /Widths[295.1 531.3 885.4 531.3 885.4 826.4 295.1 413.2 413.2 531.3 826.4 295.1 354.2 742.3 799.4 0 0 742.3 599.5 571 571 856.5 856.5 285.5 314 513.9 513.9 513.9 513.9 /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
Pendulums ICSE, CBSE class 9 physics problems from Simple Pendulum WebThe section contains questions and answers on undetermined coefficients method, harmonic motion and mass, linear independence and dependence, second order with variable and constant coefficients, non-homogeneous equations, parameters variation methods, order reduction method, differential equations with variable coefficients, rlc >> /Widths[314.8 527.8 839.5 786.1 839.5 787 314.8 419.8 419.8 524.7 787 314.8 367.3 /BaseFont/AQLCPT+CMEX10 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; Poiseuilles Law, Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes, Temperature, Kinetic Theory, and the Gas Laws, Introduction to Temperature, Kinetic Theory, and the Gas Laws, Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature, Introduction to Heat and Heat Transfer Methods, The First Law of Thermodynamics and Some Simple Processes, Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency, Carnots Perfect Heat Engine: The Second Law of Thermodynamics Restated, Applications of Thermodynamics: Heat Pumps and Refrigerators, Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy, Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation, Introduction to Oscillatory Motion and Waves, Hookes Law: Stress and Strain Revisited, Simple Harmonic Motion: A Special Periodic Motion, Energy and the Simple Harmonic Oscillator, Uniform Circular Motion and Simple Harmonic Motion, Speed of Sound, Frequency, and Wavelength, Sound Interference and Resonance: Standing Waves in Air Columns, Introduction to Electric Charge and Electric Field, Static Electricity and Charge: Conservation of Charge, Electric Field: Concept of a Field Revisited, Conductors and Electric Fields in Static Equilibrium, Introduction to Electric Potential and Electric Energy, Electric Potential Energy: Potential Difference, Electric Potential in a Uniform Electric Field, Electrical Potential Due to a Point Charge, Electric Current, Resistance, and Ohm's Law, Introduction to Electric Current, Resistance, and Ohm's Law, Ohms Law: Resistance and Simple Circuits, Alternating Current versus Direct Current, Introduction to Circuits and DC Instruments, DC Circuits Containing Resistors and Capacitors, Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field, Force on a Moving Charge in a Magnetic Field: Examples and Applications, Magnetic Force on a Current-Carrying Conductor, Torque on a Current Loop: Motors and Meters, Magnetic Fields Produced by Currents: Amperes Law, Magnetic Force between Two Parallel Conductors, Electromagnetic Induction, AC Circuits, and Electrical Technologies, Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies, Faradays Law of Induction: Lenzs Law, Maxwells Equations: Electromagnetic Waves Predicted and Observed, Introduction to Vision and Optical Instruments, Limits of Resolution: The Rayleigh Criterion, *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light, Photon Energies and the Electromagnetic Spectrum, Probability: The Heisenberg Uncertainty Principle, Discovery of the Parts of the Atom: Electrons and Nuclei, Applications of Atomic Excitations and De-Excitations, The Wave Nature of Matter Causes Quantization, Patterns in Spectra Reveal More Quantization, Introduction to Radioactivity and Nuclear Physics, Introduction to Applications of Nuclear Physics, The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited, Particles, Patterns, and Conservation Laws, A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. /Name/F7 /Type/Font >> This is why length and period are given to five digits in this example.
Pendulum Problem (9): Of simple pendulum can be used to measure gravitational acceleration. 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 What is the length of a simple pendulum oscillating on Earth with a period of 0.5 s?
/Widths[622.5 466.3 591.4 828.1 517 362.8 654.2 1000 1000 1000 1000 277.8 277.8 500 Snake's velocity was constant, but not his speedD. 1 0 obj
Let's calculate the number of seconds in 30days. 15 0 obj /LastChar 196 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 0 458.3 458.3 416.7 416.7 /Type/Font 4 0 obj
826.4 295.1 531.3] /BaseFont/OMHVCS+CMR8 Two pendulums with the same length of its cord, but the mass of the second pendulum is four times the mass of the first pendulum. 2 0 obj
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 Which answer is the right answer? WebPENDULUM WORKSHEET 1. 0 0 0 0 0 0 0 615.3 833.3 762.8 694.4 742.4 831.3 779.9 583.3 666.7 612.2 0 0 772.4 endobj WebSolution : The equation of period of the simple pendulum : T = period, g = acceleration due to gravity, l = length of cord. %PDF-1.2 The equation of period of the simple pendulum : T = period, g = acceleration due to gravity, l = length of cord. 2.8.The motion occurs in a vertical plane and is driven by a gravitational force. /FirstChar 33 The worksheet has a simple fill-in-the-blanks activity that will help the child think about the concept of energy and identify the right answers. The motion of the cart is restrained by a spring of spring constant k and a dashpot constant c; and the angle of the pendulum is restrained by a torsional spring of 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 Consider the following example. What is the acceleration of gravity at that location? >> 0.5 <> >> /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 36 0 obj /FontDescriptor 32 0 R By what amount did the important characteristic of the pendulum change when a single penny was added near the pivot. Solve it for the acceleration due to gravity. 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 Problem (2): Find the length of a pendulum that has a period of 3 seconds then find its frequency. 18 0 obj /Subtype/Type1 <> A simple pendulum of length 1 m has a mass of 10 g and oscillates freely with an amplitude of 2 cm. 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. <> stream >> Boundedness of solutions ; Spring problems . 444.4 611.1 777.8 777.8 777.8 777.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 g = 9.8 m/s2. << Ever wondered why an oscillating pendulum doesnt slow down? g can be very accurate. << Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. l+2X4J!$w|-(6}@:BtxzwD'pSe5ui8,:7X88 :r6m;|8Xxe >> /Name/F5 endobj The angular frequency formula (10) shows that the angular frequency depends on the parameter k used to indicate the stiffness of the spring and mass of the oscillation body. /FontDescriptor 17 0 R Instead of an infinitesimally small mass at the end, there's a finite (but concentrated) lump of material. 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 642.3 856.5 799.4 713.6 685.2 770.7 742.3 799.4 Use this number as the uncertainty in the period. 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 stream Otherwise, the mass of the object and the initial angle does not impact the period of the simple pendulum. 1000 1000 1055.6 1055.6 1055.6 777.8 666.7 666.7 450 450 450 450 777.8 777.8 0 0 Adding one penny causes the clock to gain two-fifths of a second in 24hours. 285.5 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 285.5 285.5 WebAuthor: ANA Subject: Set #4 Created Date: 11/19/2001 3:08:22 PM 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 Solutions to the simple pendulum problem One justification to study the problem of the simple pendulum is that this may seem very basic but its
Lagranges Equation - California State University, Northridge 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 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 . 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 If displacement from equilibrium is very small, then the pendulum of length $\ell$ approximate simple harmonic motion. The period of a simple pendulum with large angle is presented; a comparison has been carried out between the analytical solution and the numerical integration results. WebThe solution in Eq. WebThe essence of solving nonlinear problems and the differences and relations of linear and nonlinear problems are also simply discussed. endobj 570 517 571.4 437.2 540.3 595.8 625.7 651.4 277.8] 850.9 472.2 550.9 734.6 734.6 524.7 906.2 1011.1 787 262.3 524.7]
Modelling of The Simple Pendulum and It Is Numerical Solution Cut a piece of a string or dental floss so that it is about 1 m long. WebSo lets start with our Simple Pendulum problems for class 9. Since gravity varies with location, however, this standard could only be set by building a pendulum at a location where gravity was exactly equal to the standard value something that is effectively impossible. Study with Quizlet and memorize flashcards containing terms like Economics can be defined as the social science that explains the _____. Solution: As stated in the earlier problems, the frequency of a simple pendulum is proportional to the inverse of the square root of its length namely $f \propto 1/\sqrt{\ell}$. WebSimple Pendulum Problems and Formula for High Schools. For angles less than about 1515, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator. 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 % The governing differential equation for a simple pendulum is nonlinear because of the term. As with simple harmonic oscillators, the period TT for a pendulum is nearly independent of amplitude, especially if is less than about 1515. Problem (7): There are two pendulums with the following specifications. One of the authors (M. S.) has been teaching the Introductory Physics course to freshmen since Fall 2007. /FontDescriptor 8 0 R 12 0 obj 306.7 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 511.1 306.7 306.7 Find the period and oscillation of this setup. 24/7 Live Expert. 9.742m/s2, 9.865m/s2, 9.678m/s2, 9.722m/s2. /BaseFont/YQHBRF+CMR7
Simple Pendulum Problems and Formula for High Schools Given: Length of pendulum = l = 1 m, mass of bob = m = 10 g = 0.010 kg, amplitude = a = 2 cm = 0.02 m, g = 9.8m/s 2. /FontDescriptor 41 0 R 12 0 obj endstream 787 0 0 734.6 629.6 577.2 603.4 905.1 918.2 314.8 341.1 524.7 524.7 524.7 524.7 524.7 /Name/F1 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 We know that the farther we go from the Earth's surface, the gravity is less at that altitude. >> endstream then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a digital format, 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 /W [0 [777.832 0 0 250 0 408.2031 500 0 0 777.832 180.1758 333.0078 333.0078 0 563.9648 250 333.0078 250 277.832] 19 28 500 29 [277.832] 30 33 563.9648 34 [443.8477 920.8984 722.168 666.9922 666.9922 722.168 610.8398 556.1523 0 722.168 333.0078 389.1602 722.168 610.8398 889.1602 722.168 722.168 556.1523 722.168 0 556.1523 610.8398 722.168 722.168 943.8477 0 0 610.8398] 62 67 333.0078 68 [443.8477 500 443.8477 500 443.8477 333.0078 500 500 277.832 277.832 500 277.832 777.832] 81 84 500 85 [333.0078 389.1602 277.832 500 500 722.168 500 500 443.8477] 94 130 479.9805 131 [399.9023] 147 [548.8281] 171 [1000] 237 238 563.9648 242 [750] 520 [582.0313] 537 [479.0039] 550 [658.2031] 652 [504.8828] 2213 [526.3672]]>> endobj The /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 << /FontDescriptor 20 0 R 492.9 510.4 505.6 612.3 361.7 429.7 553.2 317.1 939.8 644.7 513.5 534.8 474.4 479.5 Solution: The period of a simple pendulum is related to its length $\ell$ by the following formula \[T=2\pi\sqrt{\frac{\ell}{g}}\] Here, we wish $T_2=3T_1$, after some manipulations we get \begin{align*} T_2&=3T_1\\\\ 2\pi\sqrt{\frac{\ell_2}{g}} &=3\times 2\pi\sqrt{\frac{\ell_1}{g}}\\\\ \sqrt{\ell_2}&=3\sqrt{\ell_1}\\\\\Rightarrow \ell_2&=9\ell_1 \end{align*} In the last equality, we squared both sides.