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%PDF-1.5 WebAustin Community College District | Start Here. 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 xc```b``>6A /FontDescriptor 17 0 R 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 They recorded the length and the period for pendulums with ten convenient lengths. I think it's 9.802m/s2, but that's not what the problem is about. /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 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. Pendulum . /Name/F8 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 $f \propto 1/\sqrt{\ell}$. /Subtype/Type1 /BaseFont/CNOXNS+CMR10 15 0 obj Problems (4): The acceleration of gravity on the moon is $1.625\,{\rm m/s^2}$. 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. /FirstChar 33 /Subtype/Type1 /BaseFont/LFMFWL+CMTI9 /FontDescriptor 41 0 R stream /FontDescriptor 29 0 R /FontDescriptor 32 0 R 2 0 obj Attach a small object of high density to the end of the string (for example, a metal nut or a car key). 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 /Name/F10 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 endobj 277.8 500 555.6 444.4 555.6 444.4 305.6 500 555.6 277.8 305.6 527.8 277.8 833.3 555.6 843.3 507.9 569.4 815.5 877 569.4 1013.9 1136.9 877 323.4 569.4] the pendulum of the Great Clock is a physical pendulum, is not a factor that affects the period of a pendulum, Adding pennies to the pendulum of the Great Clock changes its effective length, What is the length of a seconds pendulum at a place where gravity equals the standard value of, What is the period of this same pendulum if it is moved to a location near the equator where gravity equals 9.78m/s, What is the period of this same pendulum if it is moved to a location near the north pole where gravity equals 9.83m/s. 545.5 825.4 663.6 972.9 795.8 826.4 722.6 826.4 781.6 590.3 767.4 795.8 795.8 1091 /FirstChar 33 endobj 1000 1000 1055.6 1055.6 1055.6 777.8 666.7 666.7 450 450 450 450 777.8 777.8 0 0 /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 citation tool such as, Authors: Paul Peter Urone, Roger Hinrichs. We see from Figure 16.13 that the net force on the bob is tangent to the arc and equals mgsinmgsin. xK =7QE;eFlWJA|N Oq] PB 18 0 obj Given that $g_M=0.37g$. This method isn't graphical, but I'm going to display the results on a graph just to be consistent. Note how close this is to one meter. These NCERT Solutions provide you with the answers to the question from the textbook, important questions from previous year question papers and sample papers. 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 /Name/F2 /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 Use this number as the uncertainty in the period. 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 If this doesn't solve the problem, visit our Support Center . sin Problem (7): There are two pendulums with the following specifications. Webconsider the modelling done to study the motion of a simple pendulum. Here is a set of practice problems to accompany the Lagrange Multipliers section of the Applications of Partial Derivatives chapter of the notes for Paul Dawkins Calculus III course at Lamar University. >> The pendula are only affected by the period (which is related to the pendulums length) and by the acceleration due to gravity. << 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 $f=1/T$. Solution: first find the period of this pendulum on Mars, then using relation $f=1/T$ find its frequency. Our mission is to improve educational access and learning for everyone. Webpdf/1MB), which provides additional examples. /Widths[660.7 490.6 632.1 882.1 544.1 388.9 692.4 1062.5 1062.5 1062.5 1062.5 295.1 We begin by defining the displacement to be the arc length ss. stream If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. In the late 17th century, the the length of a seconds pendulum was proposed as a potential unit definition. 8.1 Pendulum experiments Activity 1 Your intuitive ideas To begin your investigation you will need to set up a simple pendulum as shown in the diagram. 27 0 obj \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*}. % 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, $T\propto 1/\sqrt{g}$. When we discuss damping in Section 1.2, we will nd that the motion is somewhat sinusoidal, but with an important modication. That's a loss of 3524s every 30days nearly an hour (58:44). 29. Based on the equation above, can conclude that mass does not affect the frequency of the simple pendulum. 15 0 obj 4 0 obj WebPhysics 1120: Simple Harmonic Motion Solutions 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 571 285.5 314 542.4 285.5 856.5 571 513.9 571 542.4 402 405.4 399.7 571 542.4 742.3 314.8 472.2 262.3 839.5 577.2 524.7 524.7 472.2 432.9 419.8 341.1 550.9 472.2 682.1 Thus, the period is \[T=\frac{1}{f}=\frac{1}{1.25\,{\rm Hz}}=0.8\,{\rm s}\] This paper presents approximate periodic solutions to the anharmonic (i.e. Hence, the length must be nine times. g /Subtype/Type1 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 0 777.8 285.5 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 513.9 285.5 285.5 /Subtype/Type1 Websimple-pendulum.txt. Problem (1): In a simple pendulum, how much the length of it must be changed to triple its period? 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 /Type/Font Part 1 Small Angle Approximation 1 Make the small-angle approximation. Tension in the string exactly cancels the component mgcosmgcos parallel to the string. 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 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. 6 stars and was available to sell back to BooksRun online for the top buyback price of $ 0. /Name/F3 680.6 777.8 736.1 555.6 722.2 750 750 1027.8 750 750 611.1 277.8 500 277.8 500 277.8 << /Filter /FlateDecode /S 85 /Length 111 >> 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 9 0 obj <> stream 826.4 295.1 531.3] not harmonic or non-sinusoidal) response of a simple pendulum undergoing moderate- to large-amplitude oscillations. endobj Snake's velocity was constant, but not his speedD. 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. 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. endobj Pendulum 2 has a bob with a mass of 100 kg100 kg. We can solve T=2LgT=2Lg for gg, assuming only that the angle of deflection is less than 1515. WebPeriod and Frequency of a Simple Pendulum: Class Work 27. /BaseFont/WLBOPZ+CMSY10 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. << >> >> /BaseFont/AQLCPT+CMEX10 What is the period of the Great Clock's pendulum? Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . endobj 656.3 625 625 937.5 937.5 312.5 343.8 562.5 562.5 562.5 562.5 562.5 849.5 500 574.1 874 706.4 1027.8 843.3 877 767.9 877 829.4 631 815.5 843.3 843.3 1150.8 843.3 843.3 By the end of this section, you will be able to: Pendulums are in common usage. For the next question you are given the angle at the centre, 98 degrees, and the arc length, 10cm. Perform a propagation of error calculation on the two variables: length () and period (T). B. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 1515), sinsin(sinsin and differ by about 1% or less at smaller angles). 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. f = 1 T. 15.1. <> stream /BaseFont/JOREEP+CMR9 Two simple pendulums are in two different places. They recorded the length and the period for pendulums with ten convenient lengths. 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 9 0 obj % Some have crucial uses, such as in clocks; some are for fun, such as a childs swing; and some are just there, such as the sinker on a fishing line. 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 Students calculate the potential energy of the pendulum and predict how fast it will travel. <> 384.3 611.1 675.9 351.8 384.3 643.5 351.8 1000 675.9 611.1 675.9 643.5 481.5 488 680.6 777.8 736.1 555.6 722.2 750 750 1027.8 750 750 611.1 277.8 500 277.8 500 277.8 All Physics C Mechanics topics are covered in detail in these PDF files. Knowing /Name/F5 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 The reason for the discrepancy is that the pendulum of the Great Clock is a physical pendulum. <> 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. 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 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. 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. endobj /Type/Font 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. 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 << g = 9.8 m/s2. Both are suspended from small wires secured to the ceiling of a room. sin 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] >> Ze}jUcie[. Physexams.com, Simple Pendulum Problems and Formula for High Schools. Look at the equation below. 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 5. Two simple pendulums are in two different places. 13 0 obj 324.7 531.3 590.3 295.1 324.7 560.8 295.1 885.4 590.3 531.3 590.3 560.8 414.1 419.1 Exams: Midterm (July 17, 2017) and . 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 Or at high altitudes, the pendulum clock loses some time. What is the generally accepted value for gravity where the students conducted their experiment? /BaseFont/TMSMTA+CMR9 /Subtype/Type1 /FontDescriptor 14 0 R An instructor's manual is available from the authors. 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 $B$. Problem (5): To the end of a 2-m cord, a 300-g weight is hung. These Pendulum Charts will assist you in developing your intuitive skills and to accurately find solutions for everyday challenges. How about some rhetorical questions to finish things off? /Subtype/Type1 As an Amazon Associate we earn from qualifying purchases. The problem said to use the numbers given and determine g. We did that. A classroom full of students performed a simple pendulum experiment. /Name/F4 /FirstChar 33 The Island Worksheet Answers from forms of energy worksheet answers , image source: www. Bonus solutions: Start with the equation for the period of a simple pendulum. 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 /Annots [<>>> <>>> <>>> <>>> <>>> <> <> <> <> <> <> <> <> <> <> <> <> <> <> <> <>] The equation of frequency of the simple pendulum : f = frequency, g = acceleration due to gravity, l = the length of cord. Let's do them in that order. 843.3 507.9 569.4 815.5 877 569.4 1013.9 1136.9 877 323.4 569.4] If displacement from equilibrium is very small, then the pendulum of length $\ell$ approximate simple harmonic motion. The can be very accurate. Page Created: 7/11/2021. Websome mistakes made by physics teachers who retake models texts to solve the pendulum problem, and finally, we propose the right solution for the problem fashioned as on Tipler-Mosca text (2010). They attached a metal cube to a length of string and let it swing freely from a horizontal clamp. WebStudents are encouraged to use their own programming skills to solve problems. 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 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. /FontDescriptor 11 0 R [894 m] 3. 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 777.8 500 777.8 500 530.9 12 0 obj /Name/F1 %PDF-1.2 Solve it for the acceleration due to gravity. 472.2 472.2 472.2 472.2 583.3 583.3 0 0 472.2 472.2 333.3 555.6 577.8 577.8 597.2 611.1 798.5 656.8 526.5 771.4 527.8 718.7 594.9 844.5 544.5 677.8 762 689.7 1200.9 /FirstChar 33 supplemental-problems-thermal-energy-answer-key 1/1 Downloaded from engineering2. WebSimple Pendulum Calculator is a free online tool that displays the time period of a given simple. xA y?x%-Ai;R: 4 0 obj PHET energy forms and changes simulation worksheet to accompany simulation. N xnO=ll pmlkxQ(ao?7 f7|Y6:t{qOBe>`f (d;akrkCz7x/e|+v7}Ax^G>G8]S n%[SMf#lxqS> :1|%8pv(H1nb M_Z}vn_b{u= ~; sp AHs!X ,c\zn3p_>/3s]Ec]|>?KNpq n(Jh!c~D:a?FY29hAy&\/|rp-FgGk+[Io\)?gt8.Qs#pxv[PVfn=x6QM[ W3*5"OcZn\G B$ XGdO[. /Subtype/Type1 endobj Example Pendulum Problems: A. 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. If the frequency produced twice the initial frequency, then the length of the rope must be changed to. - Unit 1 Assignments & Answers Handout. Each pendulum hovers 2 cm above the floor. 33 0 obj /BaseFont/AVTVRU+CMBX12 g 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 << Notice the anharmonic behavior at large amplitude. WebThe simple pendulum system has a single particle with position vector r = (x,y,z). 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 /FirstChar 33 Pendulum 1 has a bob with a mass of 10kg10kg. 18 0 obj endstream 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 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). 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 The digital stopwatch was started at a time t 0 = 0 and then was used to measure ten swings of a <> endobj 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 /Name/F4 Thus, for angles less than about 1515, the restoring force FF is. 9 0 obj /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 Tell me where you see mass. We move it to a high altitude. (arrows pointing away from the point). ))NzX2F 9.742m/s2, 9.865m/s2, 9.678m/s2, 9.722m/s2. The linear displacement from equilibrium is, https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units, https://openstax.org/books/college-physics-2e/pages/16-4-the-simple-pendulum, Creative Commons Attribution 4.0 International License. 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 WebThe essence of solving nonlinear problems and the differences and relations of linear and nonlinear problems are also simply discussed. /LastChar 196 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 562.5 312.5 312.5 342.6 WebIn the case of the simple pendulum or ideal spring, the force does not depend on angular velocity; but on the angular frequency. First method: Start with the equation for the period of a simple pendulum. A simple pendulum shows periodic motion, and it occurs in the vertical plane and is mainly driven by the gravitational force. x DO2(EZxIiTt |"r>^p-8y:>C&%QSSV]aq,GVmgt4A7tpJ8 C |2Z4dpGuK.DqCVpHMUN j)VP(!8#n and you must attribute OpenStax. <> 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 /FontDescriptor 8 0 R /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 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. 3 0 obj /Widths[277.8 500 833.3 500 833.3 777.8 277.8 388.9 388.9 500 777.8 277.8 333.3 277.8 388.9 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 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 Electric generator works on the scientific principle. /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 '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 .p`t]>+b1Ky>%0HCW,8D/!Y6waldaZy_u1_?0-5D#0>#gb? We can discern one half the smallest division so DVVV= ()05 01 005.. .= VV V= D ()385 005.. 4. Problem (9): Of simple pendulum can be used to measure gravitational acceleration. 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. Two-fifths of a second in one 24 hour day is the same as 18.5s in one 4s period. Ever wondered why an oscillating pendulum doesnt slow down? Adding one penny causes the clock to gain two-fifths of a second in 24hours. endstream <> 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 Using this equation, we can find the period of a pendulum for amplitudes less than about 1515. 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 (b) The period and frequency have an inverse relationship. /LastChar 196 593.8 500 562.5 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 /FontDescriptor 32 0 R xcbd`g`b``8 "w ql6A$7d s"2Z RQ#"egMf`~$ O 42 0 obj 3 0 obj /BaseFont/OMHVCS+CMR8 x a&BVX~YL&c'Zm8uh~_wsWpuhc/Nh8CQgGW[k2[6n0saYmPy>(]V@:9R+-Cpp!d::yzE q >> The time taken for one complete oscillation is called the period. i.e. 525 768.9 627.2 896.7 743.3 766.7 678.3 766.7 729.4 562.2 715.6 743.3 743.3 998.9 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. How long of a simple pendulum must have there to produce a period of $2\,{\rm s}$. We recommend using a 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 endobj Websome mistakes made by physics teachers who retake models texts to solve the pendulum problem, and finally, we propose the right solution for the problem fashioned as on Tipler-Mosca text (2010). OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. 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 /Name/F1 Pendulum B is a 400-g bob that is hung from a 6-m-long string. R ))jM7uM*%? A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length L with negligible mass (Figure 15.5.1 ). << /Pages 45 0 R /Type /Catalog >>