{"id":3352,"date":"2014-12-11T02:29:26","date_gmt":"2014-12-11T02:29:26","guid":{"rendered":"https:\/\/courses.candelalearning.com\/colphysics\/?post_type=chapter&#038;p=3352"},"modified":"2016-02-19T20:34:48","modified_gmt":"2016-02-19T20:34:48","slug":"16-4-the-simple-pendulum","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-physics\/chapter\/16-4-the-simple-pendulum\/","title":{"raw":"The Simple Pendulum","rendered":"The Simple Pendulum"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n\t<li>Measure acceleration due to gravity.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"300\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20105122\/Figure_17_04_01a.jpg\" alt=\"In the figure, a horizontal bar is drawn. A perpendicular dotted line from the middle of the bar, depicting the equilibrium of pendulum, is drawn downward. A string of length L is tied to the bar at the equilibrium point. A circular bob of mass m is tied to the end of the string which is at a distance s from the equilibrium. The string is at an angle of theta with the equilibrium at the bar. A red arrow showing the time T of the oscillation of the mob is shown along the string line toward the bar. An arrow from the bob toward the equilibrium shows its restoring force asm g sine theta. A perpendicular arrow from the bob toward the ground depicts its mass as W equals to mg, and this arrow is at an angle theta with downward direction of string.\" width=\"300\" height=\"243\" \/> Figure 1.[\/caption]\r\n\r\nIn Figure 1\u00a0we see that a\u00a0simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is <em>s<\/em>, the length of the arc. Also shown are the forces on the bob, which result in a net force of \u2212<em>mg<\/em> sin<em>\u03b8<\/em>\u00a0toward the equilibrium position\u2014that is, a restoring force.\r\n\r\nPendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child\u2019s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A <em>simple pendulum<\/em> 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 1. 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.\r\n\r\nWe begin by defining the displacement to be the arc length <em>s<\/em>. We see from Figure 1\u00a0that the net force on the bob is tangent to the arc and equals \u2212<em>mg<\/em> sin<em>\u03b8<\/em>. (The weight <em>mg<\/em>\u00a0has components\u00a0<em>mg<\/em> cos<em>\u03b8<\/em>\u00a0along the string and\u00a0<em>mg<\/em> sin<em>\u03b8<\/em>\u00a0tangent to the arc.) Tension in the string exactly cancels the component\u00a0<em>mg<\/em>\u00a0cos<em>\u03b8<\/em>\u00a0parallel to the string. This leaves a <em>net<\/em> restoring force back toward the equilibrium position at <em>\u03b8\u00a0<\/em>= 0.\r\n\r\nNow, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 15\u00ba), sin<em>\u03b8\u00a0<\/em>\u2248\u00a0<em>\u03b8<\/em>\u00a0(sin<em>\u03b8<\/em>\u00a0and <em>\u03b8<\/em> differ by about 1% or less at smaller angles). Thus, for angles less than about 15\u00ba, the restoring force <em><em>F<\/em><\/em> is\r\n<p style=\"text-align: center;\"><em>F<\/em>\u00a0\u2248 \u2212<em>mg<\/em><em>\u03b8<\/em>.<\/p>\r\nThe displacement <em><em>s<\/em><\/em> is directly proportional to <em>\u03b8<\/em>. When <em>\u03b8<\/em> is expressed in radians, the arc length in a circle is related to its radius (<em><em>L<\/em><\/em> in this instance) by\u00a0<em>s<\/em> =\u00a0<em>L<em>\u03b8<\/em><\/em>, so that\r\n<p style=\"text-align: center;\">[latex]\\theta=\\frac{s}{L}\\\\[\/latex].<\/p>\r\nFor small angles, then, the expression for the restoring force is:\r\n<p style=\"text-align: center;\">[latex]F\\approx-\\frac{mg}{L}s\\\\[\/latex].<\/p>\r\nThis expression is of the form:\u00a0<em>F<\/em> =\u00a0\u2212<em>kx<\/em>,\u00a0where the force constant is given by [latex]k=\\frac{mg}{L}\\\\[\/latex]\u00a0and the displacement is given by <em>x\u00a0<\/em>=\u00a0<em>s<\/em>. For angles less than about 15\u00ba, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator.\r\n\r\nUsing this equation, we can find the period of a pendulum for amplitudes less than about 15\u00ba. For the simple pendulum:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle{T}=2\\pi\\sqrt{\\frac{m}{k}}=2\\pi\\sqrt{\\frac{m}{\\frac{mg}{L}}}\\\\[\/latex]<\/p>\r\nThus, [latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex]\u00a0for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period <em>T<\/em> for a pendulum is nearly independent of amplitude, especially if <em><em>\u03b8<\/em><\/em> is less than about 15\u00ba. Even simple pendulum clocks can be finely adjusted and accurate.\r\n\r\nNote the dependence of <em>T<\/em> on <em><em>g<\/em><\/em>. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider Example 1.\r\n<div class=\"textbox examples\">\r\n<h3>Example 1. Measuring Acceleration due to Gravity: The Period of a Pendulum<\/h3>\r\nWhat 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?\r\n<h4>Strategy<\/h4>\r\nWe are asked to find <em><em>g<\/em><\/em> given the period <em>T<\/em> and the length <em><em>L<\/em><\/em> of a pendulum. We can solve [latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex]\u00a0for <em><em>g<\/em><\/em>, assuming only that the angle of deflection is less than 15\u00ba.\r\n<h4>Solution<\/h4>\r\nSquare [latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex]\u00a0and solve for <em>g<\/em>:\r\n<p style=\"text-align: center;\">[latex]g=4\\pi^{2}\\frac{L}{T^{2}}\\\\[\/latex].<\/p>\r\nSubstitute known values into the new equation:\r\n<p style=\"text-align: center;\">[latex]g=4\\pi^{2}\\frac{0.750000\\text{ m}}{\\left(1.7357\\text{ s}\\right)^{2}}\\\\[\/latex].<\/p>\r\nCalculate to find <em>g<\/em>:\r\n<p style=\"text-align: center;\"><em>g<\/em> = 9.8281 m\/s<sup>2<\/sup>.<\/p>\r\n\r\n<h4>Discussion<\/h4>\r\nThis method for determining <em> g <\/em> can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation sin<em>\u03b8\u00a0<\/em>\u2248\u00a0<em>\u03b8<\/em>\u00a0to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5\u00ba.\r\n\r\n<\/div>\r\n<div class=\"textbox learning-objectives\">\r\n<h3>Making Career Connections<\/h3>\r\nKnowing <em> g <\/em> can be important in geological exploration; for example, a map of <em>g<\/em> over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Take Home Experiment: Determining <em>g<\/em><\/h3>\r\nUse a simple pendulum to determine the acceleration due to gravity <em> g <\/em> in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than 10\u00ba, allow the pendulum to swing and measure the pendulum\u2019s period for 10 oscillations using a stopwatch. Calculate <em>g<\/em>. How accurate is this measurement? How might it be improved?\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Check Your Understanding<\/h3>\r\nAn engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendula will differ if the bobs are both displaced by 12\u00ba.\r\n<h4>Solution<\/h4>\r\nThe movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum\u2019s length) and by the acceleration due to gravity.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h2>PhET Explorations: Pendulum Lab<\/h2>\r\nPlay 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. It\u2019s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of <em>g<\/em> on planet X. Notice the anharmonic behavior at large amplitude.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"300\"]<a style=\"text-decoration: none;\" href=\"http:\/\/phet.colorado.edu\/sims\/pendulum-lab\/pendulum-lab_en.html\" target=\"_blank\"><img style=\"border: none;\" src=\"http:\/\/phet.colorado.edu\/sims\/pendulum-lab\/pendulum-lab-600.png\" alt=\"Pendulum Lab screenshot.\" width=\"300\" height=\"197\" \/><\/a> Click to run the simulation.[\/caption]\r\n\r\n<\/div>\r\n<h2>Section Summary<\/h2>\r\n<ul>\r\n\t<li>A mass <em>m<\/em>\u00a0suspended by a wire of length <em>L<\/em>\u00a0is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15\u00ba.<\/li>\r\n\t<li>The period of a simple pendulum is\u00a0[latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex],\u00a0where <em>L<\/em>\u00a0is the length of the string and <em>g<\/em>\u00a0is the acceleration due to gravity.<\/li>\r\n<\/ul>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Conceptual Questions<\/h3>\r\n<ol>\r\n\t<li>Pendulum clocks are made to run at the correct rate by adjusting the pendulum\u2019s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Problems &amp; Exercises<\/h3>\r\n<strong>As usual, the acceleration due to gravity in these problems is taken to be <em>g\u00a0<\/em>= 9.80 m\/s<sup>2<\/sup>, unless otherwise specified.<\/strong>\r\n<ol>\r\n\t<li>What is the length of a pendulum that has a period of 0.500 s?<\/li>\r\n\t<li>Some people think a pendulum with a period of 1.00 s can be driven with \"mental energy\" or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?<\/li>\r\n\t<li>What is the period of a 1.00-m-long pendulum?<\/li>\r\n\t<li>How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?<\/li>\r\n\t<li>The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?<\/li>\r\n\t<li>Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?<\/li>\r\n\t<li>(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79 m\/s<sup>2<\/sup> is moved to a location where it the acceleration due to gravity is 9.82 m\/s<sup>2<\/sup>. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.<\/li>\r\n\t<li>A pendulum with a period of 2.00000 s in one location (<em>g<\/em> = 9.80 m\/s<sup>2<\/sup>) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?<\/li>\r\n\t<li>(a) What is the effect on the period of a pendulum if you double its length?\u00a0(b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?<\/li>\r\n\t<li>Find the ratio of the new\/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63 m\/s<sup>2<\/sup>.<\/li>\r\n\t<li>At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is 1.63 m\/s<sup>2<\/sup>, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock\u2019s hour hand to make one revolution on the Moon.<\/li>\r\n\t<li>Suppose the length of a clock\u2019s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision.<\/li>\r\n\t<li>If a pendulum-driven clock gains 5.00 s\/day, what fractional change in pendulum length must be made for it to keep perfect time?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>simple pendulum:<\/strong>\u00a0an object with a small mass suspended from a light wire or string\r\n<div class=\"textbox exercises\">\r\n<h3>Selected Solutions to\u00a0Problems &amp; Exercises<\/h3>\r\n1.\u00a06.21 cm\r\n\r\n3.\u00a02.01 s\r\n\r\n5.\u00a02.23 Hz\r\n\r\n7.\u00a0(a) 2.99541 s;\u00a0(b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by (0.01)<sup>2<\/sup>\u00a0=0.01% so it is necessary to have at least 4 digits after the decimal to see the changes.\r\n\r\n9.\u00a0(a) Period increases by a factor of 1.41 [latex]\\left(\\sqrt{2}\\right)\\\\[\/latex];\u00a0(b) Period decreases to 97.5% of old period\r\n\r\n11.\u00a0Slow by a factor of 2.45\r\n\r\n13.\u00a0length must increase by 0.0116%\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Measure acceleration due to gravity.<\/li>\n<\/ul>\n<\/div>\n<div style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/222\/2014\/12\/20105122\/Figure_17_04_01a.jpg\" alt=\"In the figure, a horizontal bar is drawn. A perpendicular dotted line from the middle of the bar, depicting the equilibrium of pendulum, is drawn downward. A string of length L is tied to the bar at the equilibrium point. A circular bob of mass m is tied to the end of the string which is at a distance s from the equilibrium. The string is at an angle of theta with the equilibrium at the bar. A red arrow showing the time T of the oscillation of the mob is shown along the string line toward the bar. An arrow from the bob toward the equilibrium shows its restoring force asm g sine theta. A perpendicular arrow from the bob toward the ground depicts its mass as W equals to mg, and this arrow is at an angle theta with downward direction of string.\" width=\"300\" height=\"243\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1.<\/p>\n<\/div>\n<p>In Figure 1\u00a0we see that a\u00a0simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is <em>s<\/em>, the length of the arc. Also shown are the forces on the bob, which result in a net force of \u2212<em>mg<\/em> sin<em>\u03b8<\/em>\u00a0toward the equilibrium position\u2014that is, a restoring force.<\/p>\n<p>Pendulums are in common usage. Some have crucial uses, such as in clocks; some are for fun, such as a child\u2019s swing; and some are just there, such as the sinker on a fishing line. For small displacements, a pendulum is a simple harmonic oscillator. A <em>simple pendulum<\/em> 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 1. 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.<\/p>\n<p>We begin by defining the displacement to be the arc length <em>s<\/em>. We see from Figure 1\u00a0that the net force on the bob is tangent to the arc and equals \u2212<em>mg<\/em> sin<em>\u03b8<\/em>. (The weight <em>mg<\/em>\u00a0has components\u00a0<em>mg<\/em> cos<em>\u03b8<\/em>\u00a0along the string and\u00a0<em>mg<\/em> sin<em>\u03b8<\/em>\u00a0tangent to the arc.) Tension in the string exactly cancels the component\u00a0<em>mg<\/em>\u00a0cos<em>\u03b8<\/em>\u00a0parallel to the string. This leaves a <em>net<\/em> restoring force back toward the equilibrium position at <em>\u03b8\u00a0<\/em>= 0.<\/p>\n<p>Now, if we can show that the restoring force is directly proportional to the displacement, then we have a simple harmonic oscillator. In trying to determine if we have a simple harmonic oscillator, we should note that for small angles (less than about 15\u00ba), sin<em>\u03b8\u00a0<\/em>\u2248\u00a0<em>\u03b8<\/em>\u00a0(sin<em>\u03b8<\/em>\u00a0and <em>\u03b8<\/em> differ by about 1% or less at smaller angles). Thus, for angles less than about 15\u00ba, the restoring force <em><em>F<\/em><\/em> is<\/p>\n<p style=\"text-align: center;\"><em>F<\/em>\u00a0\u2248 \u2212<em>mg<\/em><em>\u03b8<\/em>.<\/p>\n<p>The displacement <em><em>s<\/em><\/em> is directly proportional to <em>\u03b8<\/em>. When <em>\u03b8<\/em> is expressed in radians, the arc length in a circle is related to its radius (<em><em>L<\/em><\/em> in this instance) by\u00a0<em>s<\/em> =\u00a0<em>L<em>\u03b8<\/em><\/em>, so that<\/p>\n<p style=\"text-align: center;\">[latex]\\theta=\\frac{s}{L}\\\\[\/latex].<\/p>\n<p>For small angles, then, the expression for the restoring force is:<\/p>\n<p style=\"text-align: center;\">[latex]F\\approx-\\frac{mg}{L}s\\\\[\/latex].<\/p>\n<p>This expression is of the form:\u00a0<em>F<\/em> =\u00a0\u2212<em>kx<\/em>,\u00a0where the force constant is given by [latex]k=\\frac{mg}{L}\\\\[\/latex]\u00a0and the displacement is given by <em>x\u00a0<\/em>=\u00a0<em>s<\/em>. For angles less than about 15\u00ba, the restoring force is directly proportional to the displacement, and the simple pendulum is a simple harmonic oscillator.<\/p>\n<p>Using this equation, we can find the period of a pendulum for amplitudes less than about 15\u00ba. For the simple pendulum:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle{T}=2\\pi\\sqrt{\\frac{m}{k}}=2\\pi\\sqrt{\\frac{m}{\\frac{mg}{L}}}\\\\[\/latex]<\/p>\n<p>Thus, [latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex]\u00a0for the period of a simple pendulum. This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass. As with simple harmonic oscillators, the period <em>T<\/em> for a pendulum is nearly independent of amplitude, especially if <em><em>\u03b8<\/em><\/em> is less than about 15\u00ba. Even simple pendulum clocks can be finely adjusted and accurate.<\/p>\n<p>Note the dependence of <em>T<\/em> on <em><em>g<\/em><\/em>. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity. Consider Example 1.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 1. Measuring Acceleration due to Gravity: The Period of a Pendulum<\/h3>\n<p>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?<\/p>\n<h4>Strategy<\/h4>\n<p>We are asked to find <em><em>g<\/em><\/em> given the period <em>T<\/em> and the length <em><em>L<\/em><\/em> of a pendulum. We can solve [latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex]\u00a0for <em><em>g<\/em><\/em>, assuming only that the angle of deflection is less than 15\u00ba.<\/p>\n<h4>Solution<\/h4>\n<p>Square [latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex]\u00a0and solve for <em>g<\/em>:<\/p>\n<p style=\"text-align: center;\">[latex]g=4\\pi^{2}\\frac{L}{T^{2}}\\\\[\/latex].<\/p>\n<p>Substitute known values into the new equation:<\/p>\n<p style=\"text-align: center;\">[latex]g=4\\pi^{2}\\frac{0.750000\\text{ m}}{\\left(1.7357\\text{ s}\\right)^{2}}\\\\[\/latex].<\/p>\n<p>Calculate to find <em>g<\/em>:<\/p>\n<p style=\"text-align: center;\"><em>g<\/em> = 9.8281 m\/s<sup>2<\/sup>.<\/p>\n<h4>Discussion<\/h4>\n<p>This method for determining <em> g <\/em> can be very accurate. This is why length and period are given to five digits in this example. For the precision of the approximation sin<em>\u03b8\u00a0<\/em>\u2248\u00a0<em>\u03b8<\/em>\u00a0to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5\u00ba.<\/p>\n<\/div>\n<div class=\"textbox learning-objectives\">\n<h3>Making Career Connections<\/h3>\n<p>Knowing <em> g <\/em> can be important in geological exploration; for example, a map of <em>g<\/em> over large geographical regions aids the study of plate tectonics and helps in the search for oil fields and large mineral deposits.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Take Home Experiment: Determining <em>g<\/em><\/h3>\n<p>Use a simple pendulum to determine the acceleration due to gravity <em> g <\/em> in your own locale. Cut a piece of a string or dental floss so that it is about 1 m long. Attach a small object of high density to the end of the string (for example, a metal nut or a car key). Starting at an angle of less than 10\u00ba, allow the pendulum to swing and measure the pendulum\u2019s period for 10 oscillations using a stopwatch. Calculate <em>g<\/em>. How accurate is this measurement? How might it be improved?<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Check Your Understanding<\/h3>\n<p>An engineer builds two simple pendula. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendula will differ if the bobs are both displaced by 12\u00ba.<\/p>\n<h4>Solution<\/h4>\n<p>The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum\u2019s length) and by the acceleration due to gravity.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h2>PhET Explorations: Pendulum Lab<\/h2>\n<p>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. It\u2019s easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of <em>g<\/em> on planet X. Notice the anharmonic behavior at large amplitude.<\/p>\n<div style=\"width: 310px\" class=\"wp-caption aligncenter\"><a style=\"text-decoration: none;\" href=\"http:\/\/phet.colorado.edu\/sims\/pendulum-lab\/pendulum-lab_en.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" style=\"border: none;\" src=\"http:\/\/phet.colorado.edu\/sims\/pendulum-lab\/pendulum-lab-600.png\" alt=\"Pendulum Lab screenshot.\" width=\"300\" height=\"197\" \/><\/a><\/p>\n<p class=\"wp-caption-text\">Click to run the simulation.<\/p>\n<\/div>\n<\/div>\n<h2>Section Summary<\/h2>\n<ul>\n<li>A mass <em>m<\/em>\u00a0suspended by a wire of length <em>L<\/em>\u00a0is a simple pendulum and undergoes simple harmonic motion for amplitudes less than about 15\u00ba.<\/li>\n<li>The period of a simple pendulum is\u00a0[latex]T=2\\pi\\sqrt{\\frac{L}{g}}\\\\[\/latex],\u00a0where <em>L<\/em>\u00a0is the length of the string and <em>g<\/em>\u00a0is the acceleration due to gravity.<\/li>\n<\/ul>\n<div class=\"textbox key-takeaways\">\n<h3>Conceptual Questions<\/h3>\n<ol>\n<li>Pendulum clocks are made to run at the correct rate by adjusting the pendulum\u2019s length. Suppose you move from one city to another where the acceleration due to gravity is slightly greater, taking your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant? Explain your answer.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Problems &amp; Exercises<\/h3>\n<p><strong>As usual, the acceleration due to gravity in these problems is taken to be <em>g\u00a0<\/em>= 9.80 m\/s<sup>2<\/sup>, unless otherwise specified.<\/strong><\/p>\n<ol>\n<li>What is the length of a pendulum that has a period of 0.500 s?<\/li>\n<li>Some people think a pendulum with a period of 1.00 s can be driven with &#8220;mental energy&#8221; or psycho kinetically, because its period is the same as an average heartbeat. True or not, what is the length of such a pendulum?<\/li>\n<li>What is the period of a 1.00-m-long pendulum?<\/li>\n<li>How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?<\/li>\n<li>The pendulum on a cuckoo clock is 5.00 cm long. What is its frequency?<\/li>\n<li>Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?<\/li>\n<li>(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79 m\/s<sup>2<\/sup> is moved to a location where it the acceleration due to gravity is 9.82 m\/s<sup>2<\/sup>. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.<\/li>\n<li>A pendulum with a period of 2.00000 s in one location (<em>g<\/em> = 9.80 m\/s<sup>2<\/sup>) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?<\/li>\n<li>(a) What is the effect on the period of a pendulum if you double its length?\u00a0(b) What is the effect on the period of a pendulum if you decrease its length by 5.00%?<\/li>\n<li>Find the ratio of the new\/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63 m\/s<sup>2<\/sup>.<\/li>\n<li>At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is 1.63 m\/s<sup>2<\/sup>, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock\u2019s hour hand to make one revolution on the Moon.<\/li>\n<li>Suppose the length of a clock\u2019s pendulum is changed by 1.000%, exactly at noon one day. What time will it read 24.00 hours later, assuming it the pendulum has kept perfect time before the change? Note that there are two answers, and perform the calculation to four-digit precision.<\/li>\n<li>If a pendulum-driven clock gains 5.00 s\/day, what fractional change in pendulum length must be made for it to keep perfect time?<\/li>\n<\/ol>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>simple pendulum:<\/strong>\u00a0an object with a small mass suspended from a light wire or string<\/p>\n<div class=\"textbox exercises\">\n<h3>Selected Solutions to\u00a0Problems &amp; Exercises<\/h3>\n<p>1.\u00a06.21 cm<\/p>\n<p>3.\u00a02.01 s<\/p>\n<p>5.\u00a02.23 Hz<\/p>\n<p>7.\u00a0(a) 2.99541 s;\u00a0(b) Since the period is related to the square root of the acceleration of gravity, when the acceleration changes by 1% the period changes by (0.01)<sup>2<\/sup>\u00a0=0.01% so it is necessary to have at least 4 digits after the decimal to see the changes.<\/p>\n<p>9.\u00a0(a) Period increases by a factor of 1.41 [latex]\\left(\\sqrt{2}\\right)\\\\[\/latex];\u00a0(b) Period decreases to 97.5% of old period<\/p>\n<p>11.\u00a0Slow by a factor of 2.45<\/p>\n<p>13.\u00a0length must increase by 0.0116%<\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3352\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Physics. <strong>Authored by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\">http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Located at License<\/li><li>PhET Interactive Simulations . <strong>Provided by<\/strong>: University of Colorado Boulder . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/phet.colorado.edu\">http:\/\/phet.colorado.edu<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":5,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Physics\",\"author\":\"OpenStax College\",\"organization\":\"\",\"url\":\"http:\/\/cnx.org\/contents\/031da8d3-b525-429c-80cf-6c8ed997733a\/College_Physics\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Located at License\"},{\"type\":\"cc\",\"description\":\"PhET Interactive Simulations \",\"author\":\"\",\"organization\":\"University of Colorado Boulder \",\"url\":\"http:\/\/phet.colorado.edu\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3352","chapter","type-chapter","status-publish","hentry"],"part":7571,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/3352","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/users\/5"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/3352\/revisions"}],"predecessor-version":[{"id":11815,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/3352\/revisions\/11815"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/parts\/7571"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapters\/3352\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/media?parent=3352"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/pressbooks\/v2\/chapter-type?post=3352"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/contributor?post=3352"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-physics\/wp-json\/wp\/v2\/license?post=3352"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}