/Subtype/Type1 We noticed that this kind of pendulum moves too slowly such that some time is losing. 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. Websimple-pendulum.txt. endobj /Type/Font WebPENDULUM WORKSHEET 1. 795.8 795.8 649.3 295.1 531.3 295.1 531.3 295.1 295.1 531.3 590.3 472.2 590.3 472.2 826.4 295.1 531.3] Physics problems and solutions aimed for high school and college students are provided. Notice the anharmonic behavior at large amplitude. endobj .p`t]>+b1Ky>%0HCW,8D/!Y6waldaZy_u1_?0-5D#0>#gb? Hence, the length must be nine times. 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 |l*HA 694.5 295.1] /Name/F10 /FontDescriptor 11 0 R 1. This method for determining Homogeneous first-order linear partial differential equation: This is a test of precision.). 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 B]1 LX&? 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. << Solution; Find the maximum and minimum values of \(f\left( {x,y} \right) = 8{x^2} - 2y\) subject to the constraint \({x^2} + {y^2} = 1\). /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 Some simple nonlinear problems in mechanics, for instance, the falling of a ball in fluid, the motion of a simple pendulum, 2D nonlinear water waves and so on, are used to introduce and examine the both methods. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. A classroom full of students performed a simple pendulum experiment. 6 0 obj endobj /FontDescriptor 17 0 R If you need help, our customer service team is available 24/7. 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 g 13 0 obj /Type/Font PDF Notes These AP Physics notes are amazing! Period is the goal. /BaseFont/OMHVCS+CMR8 1 0 obj 777.8 777.8 1000 500 500 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 777.8 /FirstChar 33 /LastChar 196 endobj
14 0 obj xK =7QE;eFlWJA|N
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PB WebView Potential_and_Kinetic_Energy_Brainpop. Let's do them in that order. 15 0 obj 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 935.2 831.5 444.4 611.1 777.8 777.8 777.8 777.8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 How about some rhetorical questions to finish things off? 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. /FirstChar 33 /FirstChar 33 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 >> That's a loss of 3524s every 30days nearly an hour (58:44). /BaseFont/YQHBRF+CMR7 >> 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. An object is suspended from one end of a cord and then perform a simple harmonic motion with a frequency of 0.5 Hertz. stream 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 0 343.8 593.8 312.5 937.5 625 562.5 625 593.8 459.5 443.8 437.5 625 593.8 812.5 593.8 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 277.8 777.8 472.2 472.2 777.8 Websimple harmonic motion. Restart your browser. /Name/F7 << By how method we can speed up the motion of this pendulum? /Subtype/Type1 (7) describes simple harmonic motion, where x(t) is a simple sinusoidal function of time. 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. 24/7 Live Expert. 783.4 872.8 823.4 619.8 708.3 654.8 0 0 816.7 682.4 596.2 547.3 470.1 429.5 467 533.2 500 500 611.1 500 277.8 833.3 750 833.3 416.7 666.7 666.7 777.8 777.8 444.4 444.4 /Type/Font (a) Find the frequency (b) the period and (d) its length. 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. Boundedness of solutions ; Spring problems . Solution: if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-leader-3','ezslot_10',134,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-leader-3-0'); Problem (11): A massive bob is held by a cord and makes a pendulum. The period of the Great Clock's pendulum is probably 4seconds instead of the crazy decimal number we just calculated. We can solve T=2LgT=2Lg for gg, assuming only that the angle of deflection is less than 1515. This part of the question doesn't require it, but we'll need it as a reference for the next two parts. WebWalking up and down a mountain. %PDF-1.2 Solution: Once a pendulum moves too fast or too slowly, some extra time is added to or subtracted from the actual time. 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 /Subtype/Type1 A simple pendulum shows periodic motion, and it occurs in the vertical plane and is mainly driven by the gravitational force. << 277.8 500] << /LastChar 196 762.8 642 790.6 759.3 613.2 584.4 682.8 583.3 944.4 828.5 580.6 682.6 388.9 388.9 Arc length and sector area worksheet (with answer key) Find the arc length. Web25 Roulette Dowsing Charts - Pendulum dowsing Roulette Charts PendulumDowsing101 $8. A simple pendulum completes 40 oscillations in one minute. /Widths[622.5 466.3 591.4 828.1 517 362.8 654.2 1000 1000 1000 1000 277.8 277.8 500 692.5 323.4 569.4 323.4 569.4 323.4 323.4 569.4 631 507.9 631 507.9 354.2 569.4 631 /Length 2854 How accurate is this measurement? This shortens the effective length of the pendulum. Weboscillation or swing of the pendulum. /Widths[351.8 611.1 1000 611.1 1000 935.2 351.8 481.5 481.5 611.1 935.2 351.8 416.7 935.2 351.8 611.1] WebAnalytic solution to the pendulum equation for a given initial conditions and Exact solution for the nonlinear pendulum (also here). Dividing this time into the number of seconds in 30days gives us the number of seconds counted by our pendulum in its new location. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo Calculate gg. endobj 324.7 531.3 531.3 531.3 531.3 531.3 795.8 472.2 531.3 767.4 826.4 531.3 958.7 1076.8 465 322.5 384 636.5 500 277.8 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 SP015 Pre-Lab Module Answer 8. Length and gravity are given. /XObject <> Note the dependence of TT on gg. 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 Example Pendulum Problems: A. A7)mP@nJ They attached a metal cube to a length of string and let it swing freely from a horizontal clamp. For the next question you are given the angle at the centre, 98 degrees, and the arc length, 10cm. >> The individuals who are preparing for Physics GRE Subject, AP, SAT, ACTexams in physics can make the most of this collection. If the length of the cord is increased by four times the initial length, then determine the period of the harmonic motion. 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 when the pendulum is again travelling in the same direction as the initial motion. /LastChar 196 Want to cite, share, or modify this book? /LastChar 196 How might it be improved? 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 WebPeriod and Frequency of a Simple Pendulum: Class Work 27. ))NzX2F The problem said to use the numbers given and determine g. We did that. In this case, this ball would have the greatest kinetic energy because it has the greatest speed. Page Created: 7/11/2021. >> Or at high altitudes, the pendulum clock loses some time. g = 9.8 m/s2. /Subtype/Type1 A cycle is one complete oscillation. Solve the equation I keep using for length, since that's what the question is about. /LastChar 196 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. Exams: Midterm (July 17, 2017) and . A 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. /Subtype/Type1 >> 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. 511.1 511.1 511.1 831.3 460 536.7 715.6 715.6 511.1 882.8 985 766.7 255.6 511.1] WebPhysics 1120: Simple Harmonic Motion Solutions 1. %PDF-1.4 The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2) : 2. If the frequency produced twice the initial frequency, then the length of the rope must be changed to. If the length of the cord is increased by four times the initial length, then determine the period of the harmonic motion. supplemental-problems-thermal-energy-answer-key 1/1 Downloaded from engineering2. endobj As you can see, the period and frequency of a simple pendulum do not depend on the mass of the pendulum bob. 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 Problem (5): To the end of a 2-m cord, a 300-g weight is hung. 1000 1000 1055.6 1055.6 1055.6 777.8 666.7 666.7 450 450 450 450 777.8 777.8 0 0 /FontDescriptor 20 0 R 36 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 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 xZYs~7Uj)?$e'VP$DJOtn/ *ew>>D/>\W/O0ttW1WtV\Uwizb
va#]oD0n#a6pmzkm7hG[%S^7@[2)nG%,acV[c{z$tA%tpAi59t> @SHKJ1O(8_PfG[S2^$Y5Q
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0":4htmD3JaU?n,d]!u0"] oq$NmF~=s=Q3K'R1>Ve%w;_n"1uAtQjw8X?:(_6hP0Kes`@@TVy#Q$t~tOz2j$_WwOL. in your own locale. /FontDescriptor 35 0 R 19 0 obj 656.3 625 625 937.5 937.5 312.5 343.8 562.5 562.5 562.5 562.5 562.5 849.5 500 574.1 /FontDescriptor 20 0 R WebThe essence of solving nonlinear problems and the differences and relations of linear and nonlinear problems are also simply discussed. 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 342.6 787 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 >> 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 In addition, there are hundreds of problems with detailed solutions on various physics topics. The rope of the simple pendulum made from nylon. 44 0 obj /FontDescriptor 26 0 R Its easy to measure the period using the photogate timer. Understanding the problem This involves, for example, understanding the process involved in the motion of simple pendulum. These NCERT Solutions provide you with the answers to the question from the textbook, important questions from previous year question papers and sample papers. /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 sin %PDF-1.2 Except where otherwise noted, textbooks on this site 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 Thus, the period is \[T=\frac{1}{f}=\frac{1}{1.25\,{\rm Hz}}=0.8\,{\rm s}\] Webpdf/1MB), which provides additional examples. /Type/Font x DO2(EZxIiTt |"r>^p-8y:>C&%QSSV]aq,GVmgt4A7tpJ8 C
|2Z4dpGuK.DqCVpHMUN j)VP(!8#n 39 0 obj /BaseFont/AVTVRU+CMBX12 /Name/F3 Solution: first find the period of this pendulum on Mars, then using relation $f=1/T$ find its frequency. /FirstChar 33 . 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 >> 295.1 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 531.3 295.1 384.3 611.1 675.9 351.8 384.3 643.5 351.8 1000 675.9 611.1 675.9 643.5 481.5 488 The length of the cord of the first pendulum (l1) = 1, The length of cord of the second pendulum (l2) = 0.4 (l1) = 0.4 (1) = 0.4, Acceleration due to the gravity of the first pendulum (g1) = 1, Acceleration due to gravity of the second pendulum (g2) = 0.9 (1) = 0.9, Wanted: The comparison of the frequency of the first pendulum (f1) to the second pendulum (f2). Then, we displace it from its equilibrium as small as possible and release it. There are two constraints: it can oscillate in the (x,y) plane, and it is always at a xed distance from the suspension point. They recorded the length and the period for pendulums with ten convenient lengths. (b) The period and frequency have an inverse relationship. 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 0.5 /FontDescriptor 14 0 R The Results Fieldbook - Michael J. Schmoker 2001 Looks at educational practices that can make an immediate and profound dierence in student learning. Tension in the string exactly cancels the component mgcosmgcos parallel to the string. 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 The governing differential equation for a simple pendulum is nonlinear because of the term. /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 endstream Otherwise, the mass of the object and the initial angle does not impact the period of the simple pendulum. When the pendulum is elsewhere, its vertical displacement from the = 0 point is h = L - L cos() (see diagram) 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 [4.28 s] 4. /FirstChar 33 298.4 878 600.2 484.7 503.1 446.4 451.2 468.8 361.1 572.5 484.7 715.9 571.5 490.3 In this problem has been said that the pendulum clock moves too slowly so its time period is too large. The mass does not impact the frequency of the simple pendulum. 1. 2.8.The motion occurs in a vertical plane and is driven by a gravitational force. Since the pennies are added to the top of the platform they shift the center of mass slightly upward. /BaseFont/CNOXNS+CMR10 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. 9 0 obj WebSolution : The equation of period of the simple pendulum : T = period, g = acceleration due to gravity, l = length of cord. /Name/F2 The relationship between frequency and period is. /Subtype/Type1 l(&+k:H uxu
{fH@H1X("Esg/)uLsU. endobj endobj 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? 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 Use the constant of proportionality to get the acceleration due to gravity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. endobj /Annots [<>>> <>>> <>>> <>>> <>>> <> <> <> <> <> <> <> <> <> <> <> <> <> <> <> <>] 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 R ))jM7uM*%? 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. <> 500 500 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 0 0 625 833.3 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 :)kE_CHL16@N99!w>/Acy
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/BaseFont/EKBGWV+CMR6 >> WebThe simple pendulum is another mechanical system that moves in an oscillatory motion. Second method: Square the equation for the period of a simple pendulum. WebSOLUTION: Scale reads VV= 385. endobj Ze}jUcie[. /FirstChar 33 << For the precision of the approximation endobj The equation of period of the simple pendulum : T = period, g = acceleration due to gravity, l = length of cord. i.e. 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 Tell me where you see mass. SOLUTION: The length of the arc is 22 (6 + 6) = 10. 675.9 1067.1 879.6 844.9 768.5 844.9 839.1 625 782.4 864.6 849.5 1162 849.5 849.5 Let's calculate the number of seconds in 30days. 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 Pendulum clocks really need to be designed for a location. WebSimple Pendulum Problems and Formula for High Schools. /Subtype/Type1 542.4 542.4 456.8 513.9 1027.8 513.9 513.9 513.9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 /FontDescriptor 38 0 R Look at the equation below. <> 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 $B$. 21 0 obj Problem (6): A pendulum, whose bob has a mass of $2\,{\rm g}$, is observed to complete 50 cycles in 40 seconds. 351.8 935.2 578.7 578.7 935.2 896.3 850.9 870.4 915.7 818.5 786.1 941.7 896.3 442.6 Let us define the potential energy as being zero when the pendulum is at the bottom of the swing, = 0 . /LastChar 196 295.1 826.4 501.7 501.7 826.4 795.8 752.1 767.4 811.1 722.6 693.1 833.5 795.8 382.6 Compute g repeatedly, then compute some basic one-variable statistics. 597.2 736.1 736.1 527.8 527.8 583.3 583.3 583.3 583.3 750 750 750 750 1044.4 1044.4 (Keep every digit your calculator gives you. 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 . Which answer is the best answer? The period is completely independent of other factors, such as mass. Math Assignments Frequency of a pendulum calculator Formula : T = 2 L g . [894 m] 3. 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. Knowing The forces which are acting on the mass are shown in the figure. WebAuthor: ANA Subject: Set #4 Created Date: 11/19/2001 3:08:22 PM This result is interesting because of its simplicity. 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 777.8 500 777.8 500 530.9 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). 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. /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 Given that $g_M=0.37g$. We recommend using a 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 Will it gain or lose time during this movement? 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 1111.1 1511.1 1111.1 1511.1 1111.1 1511.1 1055.6 944.4 472.2 833.3 833.3 833.3 833.3 endobj Calculate the period of a simple pendulum whose length is 4.4m in London where the local gravity is 9.81m/s2. 12 0 obj /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 H Use a simple pendulum to determine the acceleration due to gravity The initial frequency of the simple pendulum : The frequency of the simple pendulum is twice the initial frequency : For the final frequency to be doubled, the length of the pendulum should be changed to 0.25 meters. g 15 0 obj /BaseFont/EKGGBL+CMR6 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 /Subtype/Type1 Pendulum 1 has a bob with a mass of 10kg10kg. This method isn't graphical, but I'm going to display the results on a graph just to be consistent. 542.4 542.4 456.8 513.9 1027.8 513.9 513.9 513.9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 All of us are familiar with the simple pendulum. This is why length and period are given to five digits in this example. >> This leaves a net restoring force back toward the equilibrium position at =0=0. Adding one penny causes the clock to gain two-fifths of a second in 24hours. /Name/F8 Compare it to the equation for a generic power curve. /Subtype/Type1 2015 All rights reserved. t@F4E80%A=%A-A{>^ii{W,.Oa[G|=YGu[_>@EB Ld0eOa{lX-Xy.R^K'0c|H|fUV@+Xo^f:?Pwmnz2i] \q3`NJUdH]e'\KD-j/\}=70@'xRsvL+4r;tu3mc|}wCy;&
v5v&zXPbpp 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 << 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 << Now for the mathematically difficult question. We can discern one half the smallest division so DVVV= ()05 01 005.. .= VV V= D ()385 005.. 4. stream
If, is the frequency of the first pendulum and, is the frequency of the second pendulum, then determine the relationship between, Based on the equation above, can conclude that, ased on the above formula, can conclude the length of the, (l) and the acceleration of gravity (g) impact the period of, determine the length of rope if the frequency is twice the initial frequency. What is the period of oscillations? To verify the hypothesis that static coefficients of friction are dependent on roughness of surfaces, and independent of the weight of the top object. 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 /Name/F3 << 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