There are two constraints: it can oscillate in the (x,y) plane, and it is always at a xed distance from the suspension point. Jan 11, 2023 OpenStax. and you must attribute OpenStax. endobj 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. In this case, the period $T$ and frequency $f$ are found by the following formula \[T=2\pi\sqrt{\frac{\ell}{g}}\ , \ f=\frac{1}{T}\] As you can see, the period and frequency of a pendulum are independent of the mass hanged from it. 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. How to solve class 9 physics Problems with Solution from simple pendulum chapter? /LastChar 196 endobj 826.4 295.1 531.3] /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]]>> Pendulum 1 has a bob with a mass of 10kg10kg. WebAnalytic solution to the pendulum equation for a given initial conditions and Exact solution for the nonlinear pendulum (also here). 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 WebThe simple pendulum system has a single particle with position vector r = (x,y,z). <> stream 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 Look at the equation below. An instructor's manual is available from the authors. WebThe simple pendulum is another mechanical system that moves in an oscillatory motion. 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. /FontDescriptor 29 0 R The short way F /Subtype/Type1 795.8 795.8 649.3 295.1 531.3 295.1 531.3 295.1 295.1 531.3 590.3 472.2 590.3 472.2 /FontDescriptor 41 0 R Webpdf/1MB), which provides additional examples. /Name/F9 PDF Notes These AP Physics notes are amazing! 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 This method for determining Webproblems and exercises for this chapter. If the length of the cord is increased by four times the initial length, then determine the period of the harmonic motion. 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 can discern one half the smallest division so DVVV= ()05 01 005.. .= VV V= D ()385 005.. 4. 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 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? /Type/Font WebAssuming nothing gets in the way, that conclusion is reached when the projectile comes to rest on the ground. WebSolution : The equation of period of the simple pendulum : T = period, g = acceleration due to gravity, l = length of cord. Consider the following example. Ze}jUcie[. /Name/F1 /LastChar 196 639.7 565.6 517.7 444.4 405.9 437.5 496.5 469.4 353.9 576.2 583.3 602.5 494 437.5 citation tool such as, Authors: Paul Peter Urone, Roger Hinrichs. supplemental-problems-thermal-energy-answer-key 1/1 Downloaded from engineering2. The forces which are acting on the mass are shown in the figure. Wanted: Determine the period (T) of the pendulum if the length of cord (l) is four times the initial length. : Web1 Hamiltonian formalism for the double pendulum (10 points) Consider a double pendulum that consists of two massless rods of length l1 and l2 with masses m1 and m2 attached to their ends. << 500 500 611.1 500 277.8 833.3 750 833.3 416.7 666.7 666.7 777.8 777.8 444.4 444.4 /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 << /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>] >> 30 0 obj 875 531.3 531.3 875 849.5 799.8 812.5 862.3 738.4 707.2 884.3 879.6 419 581 880.8 Starting at an angle of less than 1010, allow the pendulum to swing and measure the pendulums period for 10 oscillations using a stopwatch. <> 295.1 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 295.1 743.3 743.3 613.3 306.7 514.4 306.7 511.1 306.7 306.7 511.1 460 460 511.1 460 306.7 For angles less than about 1515, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator. /Contents 21 0 R /LastChar 196 Now use the slope to get the acceleration due to gravity. 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 These Pendulum Charts will assist you in developing your intuitive skills and to accurately find solutions for everyday challenges. 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. they are also just known as dowsing charts . 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. WebThe essence of solving nonlinear problems and the differences and relations of linear and nonlinear problems are also simply discussed. endobj 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 The time taken for one complete oscillation is called the period. /BaseFont/SNEJKL+CMBX12 Page Created: 7/11/2021. /FontDescriptor 17 0 R PHET energy forms and changes simulation worksheet to accompany simulation. 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 5 0 obj WebA simple pendulum is defined to have an object that has a small mass, also known as the pendulum bob, which is suspended from a light wire or string, such as shown in Figure 16.13. Web25 Roulette Dowsing Charts - Pendulum dowsing Roulette Charts PendulumDowsing101 $8. WebSimple Pendulum Calculator is a free online tool that displays the time period of a given simple. 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 777.8 500 777.8 500 530.9 <> . by 3 0 obj /FirstChar 33 /BaseFont/YBWJTP+CMMI10 xA y?x%-Ai;R: WebFor periodic motion, frequency is the number of oscillations per unit time. 2 0 obj
777.8 694.4 666.7 750 722.2 777.8 722.2 777.8 0 0 722.2 583.3 555.6 555.6 833.3 833.3 611.1 798.5 656.8 526.5 771.4 527.8 718.7 594.9 844.5 544.5 677.8 762 689.7 1200.9 896.3 896.3 740.7 351.8 611.1 351.8 611.1 351.8 351.8 611.1 675.9 546.3 675.9 546.3 >> >> Will it gain or lose time during this movement? Our mission is to improve educational access and learning for everyone. 877 0 0 815.5 677.6 646.8 646.8 970.2 970.2 323.4 354.2 569.4 569.4 569.4 569.4 569.4 593.8 500 562.5 1125 562.5 562.5 562.5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 can be important in geological exploration; for example, a map of gg over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits. What is the period of oscillations? 20 0 obj /LastChar 196 On the other hand, we know that the period of oscillation of a pendulum is proportional to the square root of its length only, $T\propto \sqrt{\ell}$. Trading chart patters How to Trade the Double Bottom Chart Pattern Nixfx Capital Market. If the length of the cord is increased by four times the initial length, then determine the period of the harmonic motion. /Type/Font 9 0 obj 571 285.5 314 542.4 285.5 856.5 571 513.9 571 542.4 402 405.4 399.7 571 542.4 742.3 Projectile motion problems and answers Problem (1): A person kicks a ball with an initial velocity of 15\, {\rm m/s} 15m/s at an angle of 37 above the horizontal (neglect the air resistance). SOLUTION: The length of the arc is 22 (6 + 6) = 10. stream /BaseFont/LQOJHA+CMR7 How long of a simple pendulum must have there to produce a period of $2\,{\rm s}$. endobj All Physics C Mechanics topics are covered in detail in these PDF files. <>/ExtGState<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 612 792] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>>
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 Websimple-pendulum.txt. /Type/Font The Island Worksheet Answers from forms of energy worksheet answers , image source: www. If you need help, our customer service team is available 24/7. 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 Thus, for angles less than about 1515, the restoring force FF is. /LastChar 196 What is the acceleration of gravity at that location? Thus, the period is \[T=\frac{1}{f}=\frac{1}{1.25\,{\rm Hz}}=0.8\,{\rm s}\] Otherwise, the mass of the object and the initial angle does not impact the period of the simple pendulum. 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. endstream For small displacements, a pendulum is a simple harmonic oscillator. The mass does not impact the frequency of the simple pendulum. /LastChar 196 750 708.3 722.2 763.9 680.6 652.8 784.7 750 361.1 513.9 777.8 625 916.7 750 777.8 << /FirstChar 33 896.3 896.3 740.7 351.8 611.1 351.8 611.1 351.8 351.8 611.1 675.9 546.3 675.9 546.3 %PDF-1.5
(7) describes simple harmonic motion, where x(t) is a simple sinusoidal function of time. /Filter[/FlateDecode] By how method we can speed up the motion of this pendulum? To Find: Potential energy at extreme point = E P =? That's a gain of 3084s every 30days also close to an hour (51:24). B. Solution: (a) the number of complete cycles $N$ in a specific time interval $t$ is defined as the frequency $f$ of an oscillatory system or \[f=\frac{N}{t}\] Therefore, the frequency of this pendulum is calculated as \[f=\frac{50}{40\,{\rm s}}=1.25\, {\rm Hz}\] 766.7 715.6 766.7 0 0 715.6 613.3 562.2 587.8 881.7 894.4 306.7 332.2 511.1 511.1 Set up a graph of period squared vs. length and fit the data to a straight line. Using this equation, we can find the period of a pendulum for amplitudes less than about 1515. /FThHh!nmoF;TSooevBFN""(+7IcQX.0:Pl@Hs (@Kqd(9)\ (jX /LastChar 196 39 0 obj endobj endobj stream /XObject <> /BaseFont/UTOXGI+CMTI10 The SI unit for frequency is the hertz (Hz) and is defined as one cycle per second: 1 Hz = 1 cycle s or 1 Hz = 1 s = 1 s 1. WebSOLUTION: Scale reads VV= 385. 777.8 777.8 1000 1000 777.8 777.8 1000 777.8] 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 sin Cut a piece of a string or dental floss so that it is about 1 m long. 27 0 obj f = 1 T. 15.1. endobj 750 758.5 714.7 827.9 738.2 643.1 786.2 831.3 439.6 554.5 849.3 680.6 970.1 803.5 /Widths[323.4 569.4 938.5 569.4 938.5 877 323.4 446.4 446.4 569.4 877 323.4 384.9 WebQuestions & Worked Solutions For AP Physics 1 2022. Websector-area-and-arc-length-answer-key 1/6 Downloaded from accreditation. 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. [894 m] 3. frequency to be doubled, the length of the pendulum should be changed to 0.25 meters. 0.5 /FirstChar 33 If the frequency produced twice the initial frequency, then the length of the rope must be changed to. Bonus solutions: Start with the equation for the period of a simple pendulum. .p`t]>+b1Ky>%0HCW,8D/!Y6waldaZy_u1_?0-5D#0>#gb? The period of a simple pendulum is described by this equation. /FirstChar 33 /Annots [<>>> <>>> <>>> <>>> <>>> <> <> <> <> <> <> <> <> <> <> <> <> <> <> <> <>] Thus, by increasing or decreasing the length of a pendulum, we can regulate the pendulum's time period. nB5- H 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 Weboscillation or swing of the pendulum. The length of the second pendulum is 0.4 times the length of the first pendulum, and the acceleration of gravity experienced by the second pendulum is 0.9 times the acceleration of gravity experienced by the first pendulum. 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 /LastChar 196 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 0 772.4 666.7 666.7 666.7 666.7 611.1 611.1 444.4 444.4 444.4 444.4 500 500 388.9 388.9 277.8 708.3 795.8 767.4 826.4 767.4 826.4 0 0 767.4 619.8 590.3 590.3 885.4 885.4 295.1 endobj WebWalking up and down a mountain. 18 0 obj /Subtype/Type1 /Widths[622.5 466.3 591.4 828.1 517 362.8 654.2 1000 1000 1000 1000 277.8 277.8 500 WebClass 11 Physics NCERT Solutions for Chapter 14 Oscillations. >> 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 Find its (a) frequency, (b) time period. 500 555.6 527.8 391.7 394.4 388.9 555.6 527.8 722.2 527.8 527.8 444.4 500 1000 500 /LastChar 196 B ased 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. x a&BVX~YL&c'Zm8uh~_wsWpuhc/Nh8CQgGW[k2[6n0saYmPy>(]V@:9R+-Cpp!d::yzE q 27 0 obj A pendulum is a massive bob attached to a string or cord and swings back and forth in a periodic motion. 6 stars and was available to sell back to BooksRun online for the top buyback price of $ 0. That way an engineer could design a counting mechanism such that the hands would cycle a convenient number of times for every rotation 900 cycles for the minute hand and 10800 cycles for the hour hand. Solution: The period and length of a pendulum are related as below \begin{align*} T&=2\pi\sqrt{\frac{\ell}{g}} \\\\3&=2\pi\sqrt{\frac{\ell}{9.8}}\\\\\frac{3}{2\pi}&=\sqrt{\frac{\ell}{9.8}} \\\\\frac{9}{4\pi^2}&=\frac{\ell}{9.8}\\\\\Rightarrow \ell&=9.8\times\left(\frac{9}{4\pi^2}\right)\\\\&=2.23\quad{\rm m}\end{align*} The frequency and periods of oscillations in a simple pendulum are related as $f=1/T$. /BaseFont/EUKAKP+CMR8 How about its frequency? 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 777.8 500 777.8 500 530.9 820.5 796.1 695.6 816.7 847.5 605.6 544.6 625.8 612.8 987.8 713.3 668.3 724.7 666.7 << They recorded the length and the period for pendulums with ten convenient lengths. Electric generator works on the scientific principle. <> We can solve T=2LgT=2Lg for gg, assuming only that the angle of deflection is less than 1515. 21 0 obj The most popular choice for the measure of central tendency is probably the mean (gbar). endobj 3 0 obj
(a) What is the amplitude, frequency, angular frequency, and period of this motion? endobj (*
!>~I33gf. 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 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, 513.9 770.7 456.8 513.9 742.3 799.4 513.9 927.8 1042 799.4 285.5 513.9] 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 g 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 %PDF-1.4 624.1 928.7 753.7 1090.7 896.3 935.2 818.5 935.2 883.3 675.9 870.4 896.3 896.3 1220.4 /Name/F1 2015 All rights reserved. 24/7 Live Expert. /Type/Font /Type/Font 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 if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-leader-2','ezslot_9',117,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-leader-2-0'); Recall that the period of a pendulum is proportional to the inverse of the gravitational acceleration, namely $T \propto 1/\sqrt{g}$. 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 As an object travels through the air, it encounters a frictional force that slows its motion called. /Name/F7 Simple Harmonic Motion describes this oscillatory motion where the displacement, velocity and acceleration are sinusoidal. Solution: The frequency of a simple pendulum is related to its length and the gravity at that place according to the following formula \[f=\frac {1}{2\pi}\sqrt{\frac{g}{\ell}}\] Solving this equation for $g$, we have \begin{align*} g&=(2\pi f)^2\ell\\&=(2\pi\times 0.601)^2(0.69)\\&=9.84\quad {\rm m/s^2}\end{align*}, Author: Ali Nemati 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 277.8 777.8 472.2 472.2 777.8 All of the methods used were appropriate to the problem and all of the calculations done were error free, so all of them. Both are suspended from small wires secured to the ceiling of a room. %PDF-1.2 Solution: Recall that the time period of a clock pendulum, which is the time between successive ticks (one complete cycle), is proportional to the inverse of the square root of acceleration of gravity, $T\propto 1/\sqrt{g}$. 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 We noticed that this kind of pendulum moves too slowly such that some time is losing. 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 In this problem has been said that the pendulum clock moves too slowly so its time period is too large. ))NzX2F To verify the hypothesis that static coefficients of friction are dependent on roughness of surfaces, and independent of the weight of the top object. 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 935.2 831.5 298.4 878 600.2 484.7 503.1 446.4 451.2 468.8 361.1 572.5 484.7 715.9 571.5 490.3 endobj Physics 1 First Semester Review Sheet, Page 2. <> 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. Restart your browser. 570 517 571.4 437.2 540.3 595.8 625.7 651.4 277.8] stream >> Set up a graph of period vs. length and fit the data to a square root curve. <> stream The masses are m1 and m2. This shortens the effective length of the pendulum. %PDF-1.5 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? 7195c96ec29f4f908a055dd536dcacf9, ab097e1fccc34cffaac2689838e277d9 Our mission is to improve educational access and A simple pendulum shows periodic motion, and it occurs in the vertical plane and is mainly driven by the gravitational force. What is the period of the Great Clock's pendulum? /FontDescriptor 11 0 R /FirstChar 33 /FirstChar 33 Example 2 Figure 2 shows a simple pendulum consisting of a string of length r and a bob of mass m that is attached to a support of mass M. The support moves without friction on the horizontal plane. endobj
% Solution: This configuration makes a pendulum. 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 7 0 obj When the pendulum is elsewhere, its vertical displacement from the = 0 point is h = L - L cos() (see diagram) /LastChar 196 What is the length of a simple pendulum oscillating on Earth with a period of 0.5 s? WebAuthor: ANA Subject: Set #4 Created Date: 11/19/2001 3:08:22 PM Its easy to measure the period using the photogate timer. /Name/F6 <> We see from Figure 16.13 that the net force on the bob is tangent to the arc and equals mgsinmgsin. /FirstChar 33 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. The pennies are not added to the pendulum bob (it's moving too fast for the pennies to stay on), but are instead placed on a small platform not far from the point of suspension. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. 6 problem-solving basics for one-dimensional kinematics, is a simple one-dimensional type of projectile motion in . N xnO=ll pmlkxQ(ao?7 f7|Y6:t{qOBe>`f (d;akrkCz7x/e|+v7}Ax^G>G8]S
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B$ XGdO[. WebMISN-0-201 7 Table1.Usefulwaverelationsandvariousone-dimensional harmonicwavefunctions.Rememberthatcosinefunctions mayalsobeusedasharmonicwavefunctions. 295.1 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 295.1 295.1 Pendulum A is a 200-g bob that is attached to a 2-m-long string. 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 /Name/F12 endobj Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . 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 Students calculate the potential energy of the pendulum and predict how fast it will travel. We will then give the method proper justication. /LastChar 196 /Subtype/Type1 endobj
Calculate the period of a simple pendulum whose length is 4.4m in London where the local gravity is 9.81m/s2. The heart of the timekeeping mechanism is a 310kg, 4.4m long steel and zinc pendulum. We know that the farther we go from the Earth's surface, the gravity is less at that altitude. (The weight mgmg has components mgcosmgcos along the string and mgsinmgsin tangent to the arc.)
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