{"id":604,"date":"2026-07-17T22:18:27","date_gmt":"2026-07-17T22:18:27","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=604"},"modified":"2026-08-24T13:04:01","modified_gmt":"2026-08-24T13:04:01","slug":"pendulum-period-formula","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/pendulum-period-formula\/","title":{"rendered":"The Pendulum Period Formula"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nPendulum period is the time a pendulum takes to complete one full swing \u2014 out and back again. For small swings it depends on only two things: T = 2\u03c0\u221a(L\/g), where L is the pendulum&#8217;s length and g is the local gravitational acceleration. On Earth, a 1.00 m pendulum has a period of about 2.0 seconds, whatever its mass.\n<\/p><\/div>\n<p>Legend has it that in 1583 a young Galileo Galilei sat in Pisa Cathedral, watching a lamp swing on its chain. Using his own pulse as a stopwatch, he noticed something odd: as the swings died down, each one still took the same time.<\/p>\n<p>Four centuries on, that observation runs grandfather clocks, metronomes, gravity surveys and a famous museum exhibit in Paris. The rule behind it is startlingly simple \u2014 the swing time is set by just two numbers, and neither of them is the mass.<\/p>\n<h2>What Is the Pendulum Period?<\/h2>\n<p>The pendulum period is the time taken for one complete oscillation: the bob swings from its release point, across to the far side, and all the way back again. It is measured in seconds and given the symbol T.<\/p>\n<p>Watch out for a classic slip here. One swing across is only half a cycle \u2014 the period is the full round trip, out and back. Count &#8220;over-and-back&#8221; as one, or your measured period will be half the true value.<\/p>\n<p>Closely related is the frequency, f = 1\/T, the number of complete swings per second, measured in hertz. If you want the full story on that relationship, our guide to the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/frequency-formula\/\">frequency formula<\/a> covers it in depth. Physicists model all this with a &#8220;simple pendulum&#8221;: a small, heavy bob on a light string that doesn&#8217;t stretch.<\/p>\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/pendulum-period-formula-simple-bob-string-length-l.webp\" width=\"1280\" height=\"888\" alt=\"Pendulum period - Simple pendulum diagram: a bob on a string of length L displaced by angle theta from the vertical, with its weight mg and the restoring force component mg sin theta\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:640px;display:block;margin:0 auto;\" \/><\/figure>\n<p style=\"text-align:center;font-size:14px;font-style:italic;color:#142139;\">A simple pendulum: length L runs from the pivot to the centre of the bob, and gravity supplies a restoring force of mg sin \u03b8.<\/p>\n<h2>The Pendulum Period Formula<\/h2>\n<p>The pendulum period formula is T = 2\u03c0\u221a(L\/g): the period equals two pi times the square root of the length divided by the gravitational acceleration. It holds for small swings, and it is one of the tidiest results in mechanics.<\/p>\n<div class=\"pf-formula\">T = 2\u03c0\u221a(L \/ g)<\/div>\n<p>Each symbol has a precise meaning:<\/p>\n<ul>\n<li><strong>T<\/strong> \u2014 the period, in seconds (s): the time for one complete out-and-back swing.<\/li>\n<li><strong>L<\/strong> \u2014 the length, in metres (m), measured from the pivot to the centre of mass of the bob, not just to the top of it.<\/li>\n<li><strong>g<\/strong> \u2014 the local gravitational acceleration, in metres per second squared (m\/s<sup>2<\/sup>). On Earth this is about 9.81 m\/s<sup>2<\/sup>; the internationally agreed <a href=\"https:\/\/physics.nist.gov\/cgi-bin\/cuu\/Value?gn=\" target=\"_blank\" rel=\"noopener\">standard acceleration of gravity<\/a> is defined as exactly 9.80665 m\/s<sup>2<\/sup>. Real values vary from roughly 9.78 to 9.83 m\/s<sup>2<\/sup> with latitude and altitude.<\/li>\n<\/ul>\n<p>Notice what is missing. There is no mass in the formula, and no amplitude either \u2014 a heavier bob or a wider swing changes nothing, provided the angle stays small. Since f = 1\/T, a longer pendulum also means a lower frequency.<\/p>\n<h2>How Do You Calculate the Pendulum Period?<\/h2>\n<p>To calculate the pendulum period, divide the length by the gravitational acceleration, take the square root, then multiply by 2\u03c0. In full:<\/p>\n<ol>\n<li>Measure L in metres, from the pivot down to the centre of the bob.<\/li>\n<li>Divide L by g (use 9.81 m\/s<sup>2<\/sup> on Earth). The result has units of seconds squared.<\/li>\n<li>Take the square root, giving seconds.<\/li>\n<li>Multiply by 2\u03c0, which is about 6.283.<\/li>\n<\/ol>\n<p>Try it for a 1.00 m pendulum: 1.00 \/ 9.81 = 0.1019 s<sup>2<\/sup>, the square root is 0.3193 s, and multiplying by 6.283 gives T = 2.01 s. You can check any of these steps instantly \u2014 or rearrange the formula to solve for length or gravity \u2014 with our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/pendulum-period\">Pendulum Period Calculator<\/a>.<\/p>\n<p>The table below gives a feel for the numbers on Earth:<\/p>\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Length L<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Period T (g = 9.81 m\/s<sup>2<\/sup>)<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.10 m<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.63 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.25 m<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">1.00 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">0.50 m<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">1.42 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">1.00 m<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">2.01 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">2.00 m<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">2.84 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">3.00 m<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">3.47 s<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>Spot the pattern? Going from 0.50 m to 2.00 m \u2014 four times the length \u2014 only doubles the period. That square-root behaviour catches many students out, so here it is drawn:<\/p>\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/pendulum-period-formula-against-length-square-root-curve.webp\" width=\"1280\" height=\"846\" alt=\"Graph of pendulum period against length: a square-root curve rising from the origin, with markers showing 2.01 seconds at 1 metre and 2.84 seconds at 2 metres\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:640px;display:block;margin:0 auto;\" \/><\/figure>\n<p style=\"text-align:center;font-size:14px;font-style:italic;color:#142139;\">Period grows with the square root of length: you must quadruple the length to double the period.<\/p>\n<p>In practice, a quick sanity check saves marks: any pendulum around a metre long should give a period of roughly 2 seconds on Earth. If your answer comes out as 0.2 s or 20 s, a unit slipped somewhere. Better still, test the formula yourself \u2014 drag the length and gravity sliders in the lab below and watch the period respond, then nudge the angle slider and notice how little it matters.<\/p>\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Simple Pendulum Lab<\/span><\/div><div class=\"pf-sim-slot-body\">\n<style>\n.pf-sim-frame{\nwidth:100%;\nborder:none;\nheight:560px\n}\n@media(max-width:760px){\n.pf-sim-frame{\nheight:840px\n}\n}\n<\/style>\n<iframe src=\"\/labs\/pendulum.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\">\n<\/iframe>\n<\/div><\/div>\n<h2>Why Doesn&#8217;t Mass Affect the Pendulum Period?<\/h2>\n<p>Mass cancels out: a heavier bob is pulled back towards equilibrium by a proportionally larger force, but it is exactly that much harder to accelerate, so every bob keeps the same rhythm. Double the mass and you double both the restoring force and the inertia \u2014 the two effects wipe each other out.<\/p>\n<p>You can see the cancellation in <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/newtons-second-law\/\">Newton&#8217;s second law<\/a>, a = F\/m. The restoring force on the bob is mg sin \u03b8, so its acceleration is g sin \u03b8 \u2014 the mass has vanished before the motion even starts.<\/p>\n<p>Sound familiar? It is the very same cancellation that makes a hammer and a feather fall together in a vacuum, as our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/kinematics\/free-fall-physics\/\">free fall<\/a> explains. A pendulum is really just free fall on a leash. In air, a very light bob does lose amplitude faster to drag \u2014 but its period barely shifts.<\/p>\n<h2>What Happens to the Pendulum Period on the Moon?<\/h2>\n<p>On the Moon a pendulum swings about 2.5 times more slowly: with lunar gravity of 1.62 m\/s<sup>2<\/sup>, a 1.00 m pendulum&#8217;s period stretches from 2.01 s to 4.94 s. Weaker gravity means a weaker restoring force, so each swing takes longer.<\/p>\n<p>The formula makes the dependence precise \u2014 the period varies as one over the square root of g. <a href=\"https:\/\/science.nasa.gov\/moon\/facts\/\" target=\"_blank\" rel=\"noopener\">NASA&#8217;s Moon facts page<\/a> notes that lunar surface gravity is one-sixth of Earth&#8217;s, and the slow-down factor is the square root of the gravity ratio: \u221a(9.81 \/ 1.62) = 2.46. Here is the same 1.00 m pendulum on different worlds:<\/p>\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Location<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">g (m\/s<sup>2<\/sup>)<\/th>\n<th style=\"padding:10px;border:1px solid #D9CFB8;text-align:left;\">Period of a 1.00 m pendulum<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Earth<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">9.81<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">2.01 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Mars<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">3.71<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">3.26 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">Moon<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">1.62<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">4.94 s<\/td><\/tr>\n<tr><td style=\"padding:10px;border:1px solid #D9CFB8;\">ISS (orbit \u2014 free fall)<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">0 (effective)<\/td><td style=\"padding:10px;border:1px solid #D9CFB8;\">No swing \u2014 the bob just drifts<\/td><\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>The last row is the strangest. An orbiting station is in continuous free fall, so the effective gravity inside is zero \u2014 release a pendulum bob and there is no restoring force to bring it back. The formula agrees: as g heads towards zero, T heads towards infinity.<\/p>\n<h2>When Does the Simple Formula Break Down?<\/h2>\n<p>The formula T = 2\u03c0\u221a(L\/g) is only exact in the limit of tiny swings \u2014 keep the amplitude below about 15\u00b0 and the error stays under 0.5%, which is why that is the usual rule of thumb. It comes from the small-angle approximation, sin \u03b8 \u2248 \u03b8 (in radians), used to derive the formula.<\/p>\n<p>Push the amplitude higher and the real pendulum falls behind the prediction. The true period is always longer than the simple formula says. A first correction captures most of the effect:<\/p>\n<div class=\"pf-formula\">T \u2248 T<sub>0<\/sub> \u00b7 (1 + \u03b8<sub>0<\/sub><sup>2<\/sup> \/ 16)<\/div>\n<p>Here T<sub>0<\/sub> is the small-angle value 2\u03c0\u221a(L\/g) and \u03b8<sub>0<\/sub> is the release angle in radians. The exact numbers tell the story: the period is 0.05% long at 5\u00b0, 0.43% at 15\u00b0, 1.7% at 30\u00b0, 7.3% at 60\u00b0 and about 18% at 90\u00b0.<\/p>\n<p>A common student slip follows directly. Release the pendulum with big, dramatic swings in a lab, and your measured period runs long \u2014 so the g you calculate comes out a few percent low. Small swings, accurate g.<\/p>\n<h2>What Are Real-World Examples of Pendulum Period?<\/h2>\n<p>Pendulum period sets the tick of clocks, the rhythm of playground swings, the tempo of metronomes, the stately sweep of Foucault pendulums and the classic school method for measuring g. Five examples show its range.<\/p>\n<h3>Pendulum clocks<\/h3>\n<p>Christiaan Huygens built the first pendulum clock in 1656, and the design kept the world&#8217;s best time for nearly 300 years. The famous &#8220;seconds pendulum&#8221; is 0.994 m long: its 2.00 s period means one tick every second, each way. Because length controls the period, even thermal expansion of the rod makes a clock drift \u2014 a problem we quantify in the worked problems below.<\/p>\n<h3>Playground swings<\/h3>\n<p>A swing is a pendulum with a person as the bob. Chains about 3.0 m long give a period of roughly 3.5 s, whoever is sitting on the seat \u2014 which is why an adult and a toddler swing side by side in the same rhythm. Pumping your legs in time with that natural period feeds in energy at just the right moment: resonance in action.<\/p>\n<h3>Metronomes<\/h3>\n<p>A mechanical metronome is a compound pendulum with an adjustable sliding weight on its arm. Slide the weight up, and the centre of mass moves further from the pivot: the effective length grows and the beat slows. Musicians are tuning a pendulum period every time they set a tempo.<\/p>\n<h3>The Foucault pendulum<\/h3>\n<p>Hang a pendulum long enough and its swing becomes hypnotic. The Foucault pendulum in the Panth\u00e9on in Paris hangs from about 67 m of wire, giving a period of 16.4 s \u2014 and over hours its swing plane slowly turns, direct proof that the Earth rotates beneath it. The turning is a separate effect; each individual swing still obeys T = 2\u03c0\u221a(L\/g).<\/p>\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\">\n  <img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/07\/Pantheon_Pendule_de_Foucault2.jpg\"\n       alt=\"Foucault pendulum swinging beneath the dome of the Panth\u00e9on in Paris, a 67 m pendulum with a 16.4 second period\"\n       loading=\"lazy\"\n       style=\"width:100%;height:auto;border-radius:4px;\" width=\"960\" height=\"1280\">\n  <figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">The Foucault pendulum in the Panth\u00e9on, Paris: 67 metres of wire give a stately 16.4-second period.<\/figcaption>\n<\/figure>\n<h3>Measuring g<\/h3>\n<p>Flip the formula around and a pendulum becomes a gravity meter: g = 4\u03c0<sup>2<\/sup>L\/T<sup>2<\/sup>. Nineteenth-century surveyors mapped the Earth&#8217;s gravity field this way using Kater&#8217;s reversible pendulum, and the same rearrangement is still the classic school experiment for measuring g \u2014 timed with a phone instead of a pocket watch.<\/p>\n<h2>What Are Common Misconceptions About the Pendulum Period?<\/h2>\n<p>The most common misconceptions are that mass changes the period, that bigger swings take much longer, that L is just the string length, and that the bob moves at a steady speed. Each one trips up exam answers, so let&#8217;s put them right.<\/p>\n<p><strong>Myth: a heavier bob swings slower (or faster).<\/strong> It does neither \u2014 the period is identical for any mass, because the extra weight and the extra inertia cancel exactly. The confusion usually comes from mixing up <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/weight-vs-mass\/\">weight and mass<\/a>, which are related but different quantities.<\/p>\n<p><strong>Myth: a bigger push makes each swing take longer.<\/strong> Below about 15\u00b0 the period changes by less than half a percent \u2014 the property Galileo noticed, called isochronism. Only genuinely large swings stretch the period noticeably, as the correction formula above shows.<\/p>\n<p><strong>Myth: L is the length of the string.<\/strong> L runs from the pivot to the centre of mass of the bob, so a large bob adds its own radius to the length. Measuring only to the top of the bob is one of the most frequent sources of error in the measure-g experiment.<\/p>\n<p><strong>Myth: the bob moves at a constant speed.<\/strong> In fact it is fastest at the bottom of the arc and momentarily stationary at each end, endlessly trading kinetic energy for <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/gravitational-potential-energy\/\">gravitational potential energy<\/a> and back. The period stays fixed even though the speed never does.<\/p>\n<h2>How Does the Pendulum Period Relate to Simple Harmonic Motion?<\/h2>\n<p>For small angles, a pendulum is simple harmonic motion in disguise: the restoring force is proportional to the displacement, which is the defining condition for SHM. That is precisely why the swing follows a smooth sinusoidal pattern with a constant period.<\/p>\n<div class=\"pf-formula\">\u03c9 = \u221a(g \/ L)   and   T = 2\u03c0 \/ \u03c9<\/div>\n<p>Here \u03c9 is the angular frequency. Compare the mass\u2013spring oscillator, whose period is T = 2\u03c0\u221a(m\/k): there the mass does matter, because a spring&#8217;s force depends on stretch, not on the mass it pulls. Gravity&#8217;s force scales with mass; a spring&#8217;s doesn&#8217;t \u2014 that single difference decides whether m appears in the formula.<\/p>\n<p>The full SHM framework \u2014 displacement equations, energy graphs and why the motion is sinusoidal \u2014 lives in our complete guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/simple-harmonic-motion\/\">simple harmonic motion<\/a>. This article stays with the pendulum; that one covers the general theory.<\/p>\n<h2>Worked Problems<\/h2>\n<p>Seven problems, easiest first. Work each one before reading the solution \u2014 and keep the units on every line.<\/p>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A simple pendulum is 1.00 m long. What is its period on Earth? Take g = 9.81 m\/s^2.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Use the pendulum period formula, T = 2\u03c0\u221a(L \/ g).<\/p>\n<p>Step 2: Substitute the values: T = 2\u03c0\u221a(1.00 m \/ 9.81 m\/s<sup>2<\/sup>) = 2\u03c0\u221a(0.1019 s<sup>2<\/sup>).<\/p>\n<p>Step 3: The square root is 0.3193 s, and multiplying by 2\u03c0 (= 6.283) gives T = 2.006 s.<\/p>\n<p><strong>Answer: T = 2.01 s (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A pendulum on a lab stand is 25.0 cm long. Find its period and its frequency (g = 9.81 m\/s^2).<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Convert the length to metres: L = 25.0 cm = 0.250 m.<\/p>\n<p>Step 2: Apply T = 2\u03c0\u221a(L \/ g) = 2\u03c0\u221a(0.250 \/ 9.81) = 2\u03c0\u221a(0.02548 s<sup>2<\/sup>) = 2\u03c0 \u00d7 0.1596 s = 1.003 s.<\/p>\n<p>Step 3: The frequency is the reciprocal: f = 1 \/ T = 1 \/ 1.003 s = 0.997 Hz.<\/p>\n<p><strong>Answer: T = 1.00 s and f = 0.997 Hz, or about 1.00 Hz (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">What length must a pendulum have for a period of exactly 2.00 s on Earth \u2014 the classic &#039;seconds pendulum&#039;? Take g = 9.81 m\/s^2.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Rearrange T = 2\u03c0\u221a(L \/ g) for length: L = g\u00b7T<sup>2<\/sup> \/ (4\u03c0<sup>2<\/sup>).<\/p>\n<p>Step 2: Substitute: L = (9.81 m\/s<sup>2<\/sup> \u00d7 (2.00 s)<sup>2<\/sup>) \/ 39.48 = (9.81 \u00d7 4.00) \/ 39.48 m.<\/p>\n<p>Step 3: Evaluate: L = 39.24 \/ 39.48 m = 0.994 m.<\/p>\n<p><strong>Answer: L = 0.994 m \u2014 just under one metre<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A student times 20 complete oscillations of a 0.900 m pendulum at 38.0 s. What value of g does this give?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Find the period from the timing: T = 38.0 s \/ 20 = 1.90 s. Timing many swings reduces the reaction-time error.<\/p>\n<p>Step 2: Rearrange the formula for gravity: g = 4\u03c0<sup>2<\/sup>L \/ T<sup>2<\/sup>.<\/p>\n<p>Step 3: Substitute: g = (39.48 \u00d7 0.900 m) \/ (1.90 s)<sup>2<\/sup> = 35.53 \/ 3.61 m\/s<sup>2<\/sup> = 9.84 m\/s<sup>2<\/sup>.<\/p>\n<p><strong>Answer: g = 9.84 m\/s<sup>2<\/sup> \u2014 within 0.4% of the accepted 9.81 m\/s<sup>2<\/sup><\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">The 1.00 m pendulum from Problem 1 is taken to the Moon, where g = 1.62 m\/s^2. What is its period there, and how many times slower is it than on Earth?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Apply the formula with lunar gravity: T = 2\u03c0\u221a(1.00 m \/ 1.62 m\/s<sup>2<\/sup>) = 2\u03c0\u221a(0.6173 s<sup>2<\/sup>).<\/p>\n<p>Step 2: The square root is 0.7857 s, so T = 6.283 \u00d7 0.7857 s = 4.94 s.<\/p>\n<p>Step 3: Compare with Earth: 4.94 s \/ 2.01 s = 2.46, which equals \u221a(9.81 \/ 1.62) \u2014 the slow-down is the square root of the gravity ratio.<\/p>\n<p><strong>Answer: T = 4.94 s, about 2.46 times slower than on Earth<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">On a hot day, a clock&#039;s pendulum rod expands so its length increases by 0.040%. How much time does the clock lose per day?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Since T is proportional to \u221aL, a small fractional change in length gives half that fractional change in period: \u0394T\/T = 0.5 \u00d7 \u0394L\/L.<\/p>\n<p>Step 2: Substitute: \u0394T\/T = 0.5 \u00d7 0.040% = 0.020% = 2.0 \u00d7 10<sup>\u22124<\/sup>. Each swing now takes slightly longer, so the clock runs slow.<\/p>\n<p>Step 3: Over one day of 86 400 s, the lost time is 86 400 s \u00d7 2.0 \u00d7 10<sup>\u22124<\/sup> = 17.28 s.<\/p>\n<p><strong>Answer: the clock loses about 17 s per day<\/strong><\/p>\n<\/div><\/details><\/div>\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 1.00 m pendulum is released from a large angle of 60\u00b0. Estimate its true period using the correction T = T0\u00b7(1 + \u03b80^2\/16), with \u03b80 in radians and g = 9.81 m\/s^2.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Find the small-angle period first: T<sub>0<\/sub> = 2\u03c0\u221a(1.00 \/ 9.81) = 2.006 s (Problem 1).<\/p>\n<p>Step 2: Convert the amplitude to radians: \u03b8<sub>0<\/sub> = 60\u00b0 = 1.047 rad, so \u03b8<sub>0<\/sub><sup>2<\/sup> \/ 16 = 1.096 \/ 16 = 0.0685.<\/p>\n<p>Step 3: Apply the correction: T = 2.006 s \u00d7 (1 + 0.0685) = 2.006 \u00d7 1.0685 = 2.14 s. The exact result (from the full elliptic-integral solution) is 2.15 s, so the simple formula would have been about 7% low.<\/p>\n<p><strong>Answer: T = 2.14 s by the first-order correction (exact value 2.15 s)<\/strong><\/p>\n<\/div><\/details><\/div>\n<h2>Frequently Asked Questions<\/h2>\n<details class=\"pf-faq-item\"><summary>What is the formula for the period of a pendulum?<\/summary><div class=\"pf-faq-item-answer\">\nThe period of a simple pendulum is T = 2\u03c0\u221a(L\/g), where L is the length in metres and g is the gravitational acceleration in metres per second squared. The formula is accurate for small swing angles, up to about 15\u00b0. The frequency follows directly as f = 1\/T, the number of complete swings each second.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Does the mass of the bob affect the pendulum period?<\/summary><div class=\"pf-faq-item-answer\">\nNo \u2014 the period of a simple pendulum is completely independent of the bob&#8217;s mass. A heavier bob experiences a proportionally larger restoring force but has proportionally more inertia, and the two effects cancel exactly. In air, a very light bob loses amplitude faster to drag, but even then its period is almost unchanged.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>What length gives a pendulum a period of exactly 2 seconds?<\/summary><div class=\"pf-faq-item-answer\">\nAbout 0.994 m \u2014 just under one metre \u2014 using g = 9.81 m\/s<sup>2<\/sup>. This is the classic &#8220;seconds pendulum&#8221;: with a 2.00 s period it passes the bottom of its swing once every second, which is why longcase clocks are roughly a metre tall inside. Rearranging the formula, L = gT<sup>2<\/sup>\/(4\u03c0<sup>2<\/sup>).\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Does the amplitude affect the period of a pendulum?<\/summary><div class=\"pf-faq-item-answer\">\nHardly at all for small swings: below about 15\u00b0 the period changes by less than 0.5%, a property called isochronism. At larger amplitudes the period does lengthen \u2014 by about 1.7% at 30\u00b0 and roughly 18% at 90\u00b0 \u2014 and the simple formula always underestimates it. Keep swings small if you want T = 2\u03c0\u221a(L\/g) to hold.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>How do you measure g with a pendulum?<\/summary><div class=\"pf-faq-item-answer\">\nTime at least 20 complete oscillations, divide by the count to get the period T, then use g = 4\u03c0<sup>2<\/sup>L\/T<sup>2<\/sup>. Measure L from the pivot to the centre of the bob, and keep the swings small so the simple formula applies. Timing many swings averages out your reaction-time error, typically giving g within about 1% of 9.81 m\/s<sup>2<\/sup>.\n<\/div><\/details>\n<details class=\"pf-faq-item\"><summary>Would a pendulum swing on the International Space Station?<\/summary><div class=\"pf-faq-item-answer\">\nNo \u2014 an orbiting station is in continuous free fall, so the effective gravity inside is zero and there is no restoring force to pull the bob back. Displace a pendulum on the ISS and it simply drifts, or circles the pivot if pushed. The formula agrees: as g approaches zero, the predicted period grows without limit.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>A pendulum&#8217;s period depends on just two things: length and gravity \u2014 mass never matters. Learn the T = 2\u03c0 sqrt(L\/g) formula, what changes on the Moon, and how to solve exam problems step by step.<\/p>\n","protected":false},"author":1,"featured_media":606,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-604","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mechanics"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/604","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=604"}],"version-history":[{"count":7,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/604\/revisions"}],"predecessor-version":[{"id":1560,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/604\/revisions\/1560"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/606"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=604"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=604"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=604"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}