/Subtype/Type1 What is the length of a simple pendulum oscillating on Earth with a period of 0.5 s? 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 ). We begin by defining the displacement to be the arc length ss. Get There. 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 x a&BVX~YL&c'Zm8uh~_wsWpuhc/Nh8CQgGW[k2[6n0saYmPy>(]V@:9R+-Cpp!d::yzE q 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. Examples of Projectile Motion 1. xa ` 2s-m7k /FontDescriptor 11 0 R Except where otherwise noted, textbooks on this site How long is the pendulum? moving objects have kinetic energy. 5 0 obj The time taken for one complete oscillation is called the period. 850.9 472.2 550.9 734.6 734.6 524.7 906.2 1011.1 787 262.3 524.7] Boundedness of solutions ; Spring problems . stream /FontDescriptor 11 0 R Simplify the numerator, then divide. This method isn't graphical, but I'm going to display the results on a graph just to be consistent. Compute g repeatedly, then compute some basic one-variable statistics. Our mission is to improve educational access and learning for everyone. Exams will be effectively half of an AP exam - 17 multiple choice questions (scaled to 22. Let us define the potential energy as being zero when the pendulum is at the bottom of the swing, = 0 . 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 when the pendulum is again travelling in the same direction as the initial motion. What is its frequency on Mars, where the acceleration of gravity is about 0.37 that on Earth? 9.742m/s2, 9.865m/s2, 9.678m/s2, 9.722m/s2. 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 277.8 500] 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. 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 Calculate gg. A "seconds pendulum" has a half period of one second. We noticed that this kind of pendulum moves too slowly such that some time is losing. /Name/F2 <> The Island Worksheet Answers from forms of energy worksheet answers , image source: www. /FontDescriptor 26 0 R 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. 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 <> stream The period of a simple pendulum is described by this equation. 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 /BaseFont/NLTARL+CMTI10 endobj >> 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 /FontDescriptor 17 0 R They attached a metal cube to a length of string and let it swing freely from a horizontal clamp. >> Tension in the string exactly cancels the component mgcosmgcos parallel to the string. /Type/Font 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 Weboscillation or swing of the pendulum. 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 $f \propto 1/\sqrt{\ell}$. /LastChar 196 WebAssuming nothing gets in the way, that conclusion is reached when the projectile comes to rest on the ground. /BaseFont/LQOJHA+CMR7 >> 277.8 305.6 500 500 500 500 500 750 444.4 500 722.2 777.8 500 902.8 1013.9 777.8 384.3 611.1 675.9 351.8 384.3 643.5 351.8 1000 675.9 611.1 675.9 643.5 481.5 488 /Subtype/Type1 \(&SEc Websector-area-and-arc-length-answer-key 1/6 Downloaded from accreditation. 460 511.1 306.7 306.7 460 255.6 817.8 562.2 511.1 511.1 460 421.7 408.9 332.2 536.7 500 500 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 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. At one end of the rope suspended a mass of 10 gram and length of rope is 1 meter. 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 /LastChar 196 /BaseFont/EKBGWV+CMR6 >> 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. Let's calculate the number of seconds in 30days. 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 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. (7) describes simple harmonic motion, where x(t) is a simple sinusoidal function of time. /FirstChar 33 can be very accurate. WebAustin Community College District | Start Here. 18 0 obj OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. 481.5 675.9 643.5 870.4 643.5 643.5 546.3 611.1 1222.2 611.1 611.1 611.1 0 0 0 0 - Unit 1 Assignments & Answers Handout. B. /Filter[/FlateDecode] The mass does not impact the frequency of the simple pendulum. 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 It takes one second for it to go out (tick) and another second for it to come back (tock). The masses are m1 and m2. The rst pendulum is attached to a xed point and can freely swing about it. 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 @ @y ss~P_4qu+a" ' 9y c&Ls34f?q3[G)> `zQGOxis4t&0tC: pO+UP=ebLYl*'zte[m04743C 3d@C8"P)Dp|Y Solution; Find the maximum and minimum values of \(f\left( {x,y} \right) = 8{x^2} - 2y\) subject to the constraint \({x^2} + {y^2} = 1\). 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. 1 0 obj 874 706.4 1027.8 843.3 877 767.9 877 829.4 631 815.5 843.3 843.3 1150.8 843.3 843.3 743.3 743.3 613.3 306.7 514.4 306.7 511.1 306.7 306.7 511.1 460 460 511.1 460 306.7 xK =7QE;eFlWJA|N Oq] PB All of us are familiar with the simple pendulum. 791.7 777.8] 935.2 351.8 611.1] /Type/Font /FontDescriptor 32 0 R 2 0 obj xYK WL+z^d7 =sPd3 X`H^Ea+y}WIeoY=]}~H,x0aQ@z0UX&ks0. 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 Homogeneous first-order linear partial differential equation: >> stream /Widths[1000 500 500 1000 1000 1000 777.8 1000 1000 611.1 611.1 1000 1000 1000 777.8 endstream << Mathematically we have x2 1 + y 2 1 = l 2 1; (x2 x1) 2 + (y2 y1)2 = l22: 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 >> endobj >> /LastChar 196 What is the acceleration of gravity at that location? 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 << endobj 826.4 295.1 531.3] 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, $T\propto 1/\sqrt{g}$. In the following, a couple of problems about simple pendulum in various situations is presented. xA y?x%-Ai;R: Based on the equation above, can conclude that mass does not affect the frequency of the simple pendulum. A simple pendulum shows periodic motion, and it occurs in the vertical plane and is mainly driven by the gravitational force. /Type/Font This method for determining 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 Webpdf/1MB), which provides additional examples. /Subtype/Type1 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 What is the period of oscillations? 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 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 WebWalking up and down a mountain. Simple Pendulum: A simple pendulum device is represented as the point mass attached to a light inextensible string and suspended from a fixed support. << In addition, there are hundreds of problems with detailed solutions on various physics topics. 777.8 694.4 666.7 750 722.2 777.8 722.2 777.8 0 0 722.2 583.3 555.6 555.6 833.3 833.3 742.3 799.4 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 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). /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 Put these information into the equation of frequency of pendulum and solve for the unknown $g$ as below \begin{align*} g&=(2\pi f)^2 \ell \\&=(2\pi\times 0.841)^2(0.35)\\&=9.780\quad {\rm m/s^2}\end{align*}. if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-large-mobile-banner-2','ezslot_8',133,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-large-mobile-banner-2-0'); Problem (10): A clock works with the mechanism of a pendulum accurately. << >> <> stream /Type/Font The %PDF-1.2 21 0 obj Now for the mathematically difficult question. /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 Websimple harmonic motion. 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 0 600.2 600.2 507.9 569.4 1138.9 569.4 569.4 569.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 WebSimple Pendulum Calculator is a free online tool that displays the time period of a given simple. 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 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 The equation of frequency of the simple pendulum : f = frequency, g = acceleration due to gravity, l = the length of cord. 24/7 Live Expert. The forces which are acting on the mass are shown in the figure. 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 /FontDescriptor 23 0 R g Perform a propagation of error calculation on the two variables: length () and period (T). /BaseFont/AQLCPT+CMEX10 9 0 obj 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 Math Assignments Frequency of a pendulum calculator Formula : T = 2 L g . 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 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. Problem (6): A pendulum, whose bob has a mass of $2\,{\rm g}$, is observed to complete 50 cycles in 40 seconds. /Subtype/Type1 /MediaBox [0 0 612 792] 694.5 295.1] (b) The period and frequency have an inverse relationship. 44 0 obj 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. Adding pennies to the pendulum of the Great Clock changes its effective length. The displacement ss is directly proportional to . and you must attribute OpenStax. Physics 1: Algebra-Based If you are giving the regularly scheduled exam, say: It is Tuesday afternoon, May 3, and you will be taking the AP Physics 1: Algebra-Based Exam. 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 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 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). WebThe solution in Eq. consent of Rice University. endobj Problems (4): The acceleration of gravity on the moon is $1.625\,{\rm m/s^2}$. /FontDescriptor 38 0 R /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 << /Pages 45 0 R /Type /Catalog >> Using this equation, we can find the period of a pendulum for amplitudes less than about 1515. 24/7 Live Expert. /Subtype/Type1 /Subtype/Type1 This is the video that cover the section 7. Solve the equation I keep using for length, since that's what the question is about. If this doesn't solve the problem, visit our Support Center . The digital stopwatch was started at a time t 0 = 0 and then was used to measure ten swings of a /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 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. 812.5 875 562.5 1018.5 1143.5 875 312.5 562.5] 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 >> 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 /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 Pennies are used to regulate the clock mechanism (pre-decimal pennies with the head of EdwardVII). WebPeriod and Frequency of a Simple Pendulum: Class Work 27. /BaseFont/JMXGPL+CMR10 /Type/Font For the next question you are given the angle at the centre, 98 degrees, and the arc length, 10cm. /Subtype/Type1 5. Here is a list of problems from this chapter with the solution. Solution: /FontDescriptor 29 0 R Set up a graph of period vs. length and fit the data to a square root curve. endobj xcbd`g`b``8 "w ql6A$7d s"2Z RQ#"egMf`~$ O endstream 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 643.8 839.5 787 710.5 682.1 763 734.6 787 734.6 Students calculate the potential energy of the pendulum and predict how fast it will travel. /LastChar 196 24 0 obj /Name/F11 /Type/Font As with simple harmonic oscillators, the period TT for a pendulum is nearly independent of amplitude, especially if is less than about 1515. 7 0 obj That's a loss of 3524s every 30days nearly an hour (58:44). endobj 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 /FontDescriptor 20 0 R in your own locale. /FirstChar 33 <> stream SOLUTION: The length of the arc is 22 (6 + 6) = 10. /FontDescriptor 14 0 R <>/ExtGState<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 612 792] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>> When the pendulum is elsewhere, its vertical displacement from the = 0 point is h = L - L cos() (see diagram) /Name/F9 11 0 obj /Name/F3 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? 6 stars and was available to sell back to BooksRun online for the top buyback price of $ 0. WebAnalytic solution to the pendulum equation for a given initial conditions and Exact solution for the nonlinear pendulum (also here). 843.3 507.9 569.4 815.5 877 569.4 1013.9 1136.9 877 323.4 569.4] /Name/F2 Then, we displace it from its equilibrium as small as possible and release it. 3 0 obj 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 /Subtype/Type1 19 0 obj >> 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] /BaseFont/YQHBRF+CMR7 %PDF-1.4 PHET energy forms and changes simulation worksheet to accompany simulation. 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 << /Linearized 1 /L 141310 /H [ 964 190 ] /O 22 /E 111737 /N 6 /T 140933 >> <> 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. 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 12 0 obj 4 0 obj They recorded the length and the period for pendulums with ten convenient lengths. 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 30 0 obj 500 500 500 500 500 500 500 500 500 500 500 277.8 277.8 777.8 500 777.8 500 530.9 /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 The period is completely independent of other factors, such as mass. 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 x DO2(EZxIiTt |"r>^p-8y:>C&%QSSV]aq,GVmgt4A7tpJ8 C |2Z4dpGuK.DqCVpHMUN j)VP(!8#n endobj 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. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 642.3 856.5 799.4 713.6 685.2 770.7 742.3 799.4 314.8 524.7 524.7 524.7 524.7 524.7 524.7 524.7 524.7 524.7 524.7 524.7 314.8 314.8 Compare it to the equation for a generic power curve. endobj For small displacements, a pendulum is a simple harmonic oscillator. Knowing /Name/F8 stream 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. endobj 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. Math Assignments Frequency of a pendulum calculator Formula : T = 2 L g . These Pendulum Charts will assist you in developing your intuitive skills and to accurately find solutions for everyday challenges. WebAnalytic solution to the pendulum equation for a given initial conditions and Exact solution for the nonlinear pendulum (also here). The relationship between frequency and period is. Physics problems and solutions aimed for high school and college students are provided. WebQuestions & Worked Solutions For AP Physics 1 2022. 460.7 580.4 896 722.6 1020.4 843.3 806.2 673.6 835.7 800.2 646.2 618.6 718.8 618.8 570 517 571.4 437.2 540.3 595.8 625.7 651.4 277.8] endobj 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 (arrows pointing away from the point). Solution: Once a pendulum moves too fast or too slowly, some extra time is added to or subtracted from the actual time. Find its (a) frequency, (b) time period. % WebMISN-0-201 7 Table1.Usefulwaverelationsandvariousone-dimensional harmonicwavefunctions.Rememberthatcosinefunctions mayalsobeusedasharmonicwavefunctions. Two simple pendulums are in two different places. 812.5 875 562.5 1018.5 1143.5 875 312.5 562.5] /Type/Font 570 517 571.4 437.2 540.3 595.8 625.7 651.4 277.8] /BaseFont/VLJFRF+CMMI8 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. Starting at an angle of less than 1010, allow the pendulum to swing and measure the pendulums period for 10 oscillations using a stopwatch. << << 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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For the simple pendulum: for the period of a simple pendulum. to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about /FirstChar 33 /FirstChar 33 sin << << >> Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License .