{"id":649,"date":"2026-07-25T06:14:32","date_gmt":"2026-07-25T06:14:32","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=649"},"modified":"2026-08-24T13:03:58","modified_gmt":"2026-08-24T13:03:58","slug":"spring-constant","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/spring-constant\/","title":{"rendered":"Spring Constant (k = F\/x): Formula &amp; Examples"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\nThe spring constant (k) is a measure of a spring&#8217;s stiffness: the force in newtons needed to stretch or compress it by one metre. It is calculated as k&nbsp;=&nbsp;F\/x \u2014 applied force divided by extension \u2014 and measured in newtons per metre (N\/m). Stiffer springs have larger k values.\n<\/p><\/div>\n \n<p>Click the top of a retractable pen and your thumb squashes a small steel spring by about seven millimetres. The push you feel \u2014 roughly two newtons \u2014 is set by a single number built into that spring.<\/p>\n \n<p>That number is the <strong>spring constant<\/strong>, and it quietly runs daily life: how firm your mattress feels, how a car soaks up potholes, how a weighing scale turns stretch into kilograms. This guide shows you how to measure k for any spring, what realistic values look like, and how to read the number the way an engineer does.<\/p>\n \n<h2>What Is the Spring Constant?<\/h2>\n \n<p>The spring constant, written k, is the force needed to stretch or compress a spring by one unit of length \u2014 in SI units, the newtons required per metre of extension. It is a single number that captures how stiff one particular spring is.<\/p>\n \n<p>Think of k as a price list. A soft spring charges you almost nothing per centimetre of stretch; a stiff one demands a serious push for the same distance. Two springs can look identical on the bench and still have wildly different constants.<\/p>\n \n<p>One boundary before we start: k comes from the proportional relationship between force and extension, which our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/hookes-law\/\">Hooke&#8217;s law<\/a> covers in depth \u2014 the law itself, its discovery, and where it breaks down. This article stays on the constant: measuring it, typical values, and what it tells you.<\/p>\n \n<h2>The Spring Constant Formula: k = F\/x<\/h2>\n \n<p>To calculate the spring constant, divide the applied force by the extension it produces.<\/p>\n \n<div class=\"pf-formula\">k = F \/ x<\/div>\n \n<ul>\n  <li><strong>k<\/strong> \u2014 spring constant, in newtons per metre (N\/m)<\/li>\n  <li><strong>F<\/strong> \u2014 applied force, in newtons (N)<\/li>\n  <li><strong>x<\/strong> \u2014 extension or compression, in metres (m): the <em>change<\/em> from the spring&#8217;s natural length, never the spring&#8217;s total length<\/li>\n<\/ul>\n \n<p>The formula rearranges both ways: <strong>F = kx<\/strong> gives the force for a chosen stretch, and <strong>x = F\/k<\/strong> predicts how far a spring gives under a known load. You can also skip the algebra and get any of the three instantly with our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/spring-constant\">Spring Constant Calculator<\/a>, which solves for k, F, or x from the other two.<\/p>\n \n<p>A note on units. Newtons per metre follows directly from the <a href=\"https:\/\/www.nist.gov\/pml\/owm\/metric-si\/si-units\" target=\"_blank\" rel=\"noopener\">SI system of units<\/a>, since k is a force divided by a length. Engineers usually call the same quantity a <em>spring rate<\/em> and quote it in N\/mm or lb\/in \u2014 handy conversions are 1 N\/mm = 1,000 N\/m and 1 lb\/in &asymp; 175 N\/m.<\/p>\n \n<h2>How Do You Measure a Spring Constant?<\/h2>\n \n<p>Measure a spring constant by hanging known weights from the spring, recording the extension each load produces, plotting force against extension, and taking the slope of the best-fit straight line \u2014 that slope is k. A second, independent route uses timing: set a mass bouncing on the spring and calculate k from the oscillation period.<\/p>\n \n<h3>Method 1: The Load\u2013Extension Graph (Static Method)<\/h3>\n \n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/spring-constant-experiment-hangs-clamp-stand-slotted.webp\" width=\"1280\" height=\"974\" alt=\"Diagram of a spring constant experiment: a spring hangs from a clamp stand, slotted masses stretch it, and a metre rule measures the extension x between the unloaded and loaded positions\" 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;\">The static method: each added mass raises the stretching force by F&nbsp;=&nbsp;mg, and the metre rule records the matching extension x.<\/p>\n \n<p>This is the classic school practical, and done carefully it gives k to within a few percent. You need a clamp stand, the spring, a metre rule, and a set of slotted masses.<\/p>\n \n<ol>\n  <li>Hang the spring and note the ruler reading of its lower end with <strong>no load<\/strong>.<\/li>\n  <li>Add a known mass and convert it to force using F&nbsp;=&nbsp;mg \u2014 the distinction matters, as our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/weight-vs-mass\/\">weight vs mass<\/a> explains, because the spring feels newtons, not grams.<\/li>\n  <li>Record the new reading and subtract the unloaded reading to get the extension x.<\/li>\n  <li>Repeat for five or six loads, then remove the masses one by one and record the readings again on the way down.<\/li>\n  <li>Plot force F on the vertical axis against extension x on the horizontal axis.<\/li>\n  <li>Draw the best-fit straight line through the origin. Its slope is the spring constant.<\/li>\n<\/ol>\n \n<p>Why bother unloading? It is a built-in honesty check. If the unloading readings sit above the loading ones, the spring has been permanently stretched and the later data points are worthless.<\/p>\n \n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/spring-constant-force-against-extension-four-gold.webp\" width=\"1280\" height=\"910\" alt=\"Spring constant - Force against extension graph for a spring: four gold data points lie on a straight best-fit line through the origin, with a slope triangle showing k equals delta F over delta x equals 250 newtons per metre\" 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;\">The spring constant is the slope of the force\u2013extension graph. Here k&nbsp;=&nbsp;(5.0&nbsp;N)\/(0.020&nbsp;m)&nbsp;=&nbsp;250&nbsp;N\/m.<\/p>\n \n<p>Take the slope from two well-separated points <em>on the line<\/em>, never from a single raw data point. The best-fit line averages out reading errors; one point carries them all.<\/p>\n \n<h3>Method 2: The Oscillation (Dynamic) Method<\/h3>\n \n<p>Hang a known mass on the spring, pull it down a centimetre or two, release it, and time the bouncing. The period T links directly to k:<\/p>\n \n<div class=\"pf-formula\">k = 4&pi;<sup>2<\/sup>m \/ T<sup>2<\/sup><\/div>\n \n<p>Time 20 complete oscillations with a stopwatch and divide by 20 to get T \u2014 a common student slip is timing a single bounce, which makes the human reaction-time error enormous. The mass on a spring is a textbook case of <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/simple-harmonic-motion\/\">simple harmonic motion<\/a>, which is exactly why this shortcut works.<\/p>\n \n<p>In practice the dynamic method often beats the ruler. A stopwatch over 20 cycles is a far more forgiving instrument than a millimetre scale read by eye, and the two methods agreeing within a few percent is the strongest evidence you can offer that your value of k is right.<\/p>\n \n<p>Want to rehearse the whole experiment before touching real apparatus? Load the virtual spring below, pin your readings, and watch the slope build.<\/p>\n \n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Spring Constant Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/spring-constant.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n \n<h2>Typical Spring Constant Values for Everyday Springs<\/h2>\n \n<p>Everyday spring constants span nearly five powers of ten, from about 1\u20132 N\/m for a slinky to well over 60,000 N\/m for a car or mountain-bike suspension spring. The table below gives typical, order-of-magnitude values \u2014 real parts vary, and manufacturers quote an exact spring rate for each one.<\/p>\n \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=\"background:#0A1628;color:#FAF6EE;border:1px solid #D9CFB8;padding:10px;text-align:left;\">Spring<\/th>\n  <th style=\"background:#0A1628;color:#FAF6EE;border:1px solid #D9CFB8;padding:10px;text-align:left;\">Typical spring constant k<\/th>\n  <th style=\"background:#0A1628;color:#FAF6EE;border:1px solid #D9CFB8;padding:10px;text-align:left;\">What that feels like<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">Slinky<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 1\u20132 N\/m<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 0.01\u20130.02 N per cm \u2014 it stretches under its own weight<\/td>\n<\/tr>\n<tr>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">School newton meter (0\u201310 N tube)<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 100 N\/m<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 1 N per cm, by design: 10 N spread over a 10 cm scale<\/td>\n<\/tr>\n<tr>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">Retractable pen spring<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 200\u2013400 N\/m<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 2\u20134 N per cm \u2014 the firm click under your thumb<\/td>\n<\/tr>\n<tr>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">Mattress pocket coil (single)<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 1,000\u20135,000 N\/m<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 10\u201350 N per cm; firmness grades vary widely<\/td>\n<\/tr>\n<tr>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">Car suspension coil (per corner)<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 20,000\u201360,000 N\/m<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 200\u2013600 N per cm \u2014 like resting a 20\u201360 kg load on it<\/td>\n<\/tr>\n<tr>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">Mountain-bike coil shock spring<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 60,000\u201395,000 N\/m<\/td>\n  <td style=\"border:1px solid #D9CFB8;padding:10px;\">&asymp; 600\u2013950 N per cm; sold as &ldquo;350\u2013550 lb\/in&rdquo;<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n \n<p>Here is a sanity check worth memorising: divide any k in N\/m by 100 and you get newtons per centimetre. A 30,000 N\/m car spring therefore takes about 300 N \u2014 a 30 kg load \u2014 to squash by a single centimetre, which is exactly why you can barely compress one by hand.<\/p>\n \n<p>Notice the surprise in the last two rows. A bike&#8217;s shock spring is <em>stiffer<\/em> than a car&#8217;s, even though the bike is far lighter, because the rear linkage levers the wheel&#8217;s force onto the spring at roughly two-to-three times its size. The spring never knows how heavy the vehicle is \u2014 it only feels the force actually delivered to it.<\/p>\n \n<h2>What Does the Spring Constant Actually Tell You?<\/h2>\n \n<p>The spring constant tells you how much force a spring returns for every metre you deform it \u2014 a compact statement of its stiffness, and the slope of its force\u2013extension graph. A steep graph is a stiff spring; a shallow graph is a soft one.<\/p>\n \n<p>Crucially, k belongs to the <em>whole spring<\/em>, not just its material. Wire thickness matters enormously (stiffness grows with the fourth power of wire diameter), while a wider coil or extra turns make a spring softer \u2014 the full geometry is worked through in this Physics LibreTexts treatment of <a href=\"https:\/\/phys.libretexts.org\/Courses\/Prince_Georges_Community_College\/General_Physics_I%3A_Classical_Mechanics\/42%3A__Simple_Harmonic_Motion\/42.06%3A_More_on_the_Spring_Constant\" target=\"_blank\" rel=\"noopener\">the spring constant and spring geometry<\/a>. Two steel springs can differ in k by a factor of a thousand.<\/p>\n \n<h3>Springs in Series and Parallel<\/h3>\n \n<p>Combine springs and the constants combine in opposite ways. Side by side (parallel), they share the load and the stiffnesses simply add: k<sub>total<\/sub>&nbsp;=&nbsp;k<sub>1<\/sub>&nbsp;+&nbsp;k<sub>2<\/sub>.<\/p>\n \n<p>End to end (series), the combination is <em>softer<\/em>: 1\/k<sub>total<\/sub>&nbsp;=&nbsp;1\/k<sub>1<\/sub>&nbsp;+&nbsp;1\/k<sub>2<\/sub>. Why? Each spring in the chain carries the full pull \u2014 the same principle that governs <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/tension-force\/\">tension in a rope<\/a> \u2014 so every spring stretches fully and the extensions stack up.<\/p>\n \n<h3>Why Cutting a Spring in Half Doubles k<\/h3>\n \n<p>Chop a spring in half and each half becomes twice as stiff. The full spring is effectively two half-springs in series, so under a given force each half supplies only half the total stretch.<\/p>\n \n<p>Apply the same force to one half alone and you get half the extension \u2014 and F divided by half of x is double the original k. Counter-intuitive, easy to test, and a favourite exam twist.<\/p>\n \n<h2>Real-World Examples of the Spring Constant<\/h2>\n \n<p>Once you can read k, spring numbers start appearing everywhere \u2014 here are four places engineers choose them deliberately.<\/p>\n \n<h3>Car Suspension Tuning<\/h3>\n \n<p>Each corner of a family car sits on a coil of roughly 20,000\u201360,000 N\/m. Softer constants soak up bumps but let the body wallow and roll; stiffer ones sharpen handling at the cost of comfort.<\/p>\n \n<p>Run the numbers yourself: a 350 kg corner load (about 3,400 N) on a 40,000 N\/m spring sags around 8.5 cm at rest. That &ldquo;static sag&rdquo; is a real quantity suspension tuners measure with a tape.<\/p>\n \n<h3>Spring Scales and Newton Meters<\/h3>\n \n<p>A spring scale is just k made visible. Because force and extension stay proportional, the maker can print an evenly spaced scale and let the pointer convert stretch straight into newtons or kilograms. If k drifted with load, the markings would bunch up and the instrument would lie.<\/p>\n \n<h3>Mattress Firmness<\/h3>\n \n<p>A pocket-sprung mattress is hundreds of small springs in parallel, so its overall feel is the per-coil constant multiplied up by coil count. &ldquo;Zoned&rdquo; mattresses go further, fitting stiffer coils under the hips and softer ones under the shoulders \u2014 different k values doing different jobs in one product.<\/p>\n \n<h3>Keyboard Switches and Pen Clicks<\/h3>\n \n<p>The springs under keyboard keys and pen buttons run at a few hundred newtons per metre, tuned so a fingertip force of about half a newton gives a satisfying couple of millimetres of travel. Change k by a fraction and typists genuinely feel the difference \u2014 switch makers sell the same design in several spring weights for exactly this reason.<\/p>\n \n<h2>Common Misconceptions About the Spring Constant<\/h2>\n \n<h3>&ldquo;x is the spring&#8217;s total length&rdquo;<\/h3>\n \n<p>No \u2014 x is the <strong>extension<\/strong>: the change from the spring&#8217;s natural, unloaded length. If a 20 cm spring stretches to 26 cm, x is 0.06 m, not 0.26 m. This single slip is the most common cause of wildly wrong k values in student work, throwing the answer off by a factor of four or more.<\/p>\n \n<h3>&ldquo;A heavier load changes the spring constant&rdquo;<\/h3>\n \n<p>It doesn&#8217;t. Doubling the load doubles the extension, and the ratio F\/x \u2014 which is k \u2014 stays put. The constant only genuinely shifts if you overload the spring so far that it deforms permanently and stops behaving linearly.<\/p>\n \n<h3>&ldquo;A bigger spring is always a stiffer spring&rdquo;<\/h3>\n \n<p>Size and stiffness are different things. A slinky is huge yet scores barely 1 N\/m, while the tiny spring in a pen sits near 300 N\/m. Wire thickness, coil diameter, and turn count decide k \u2014 overall size alone tells you almost nothing.<\/p>\n \n<h3>&ldquo;The spring constant is a property of the material&rdquo;<\/h3>\n \n<p>Steel doesn&#8217;t have a spring constant; a particular steel spring does. As covered above, geometry dominates \u2014 the same wire wound differently gives a completely different k.<\/p>\n \n<h2>How the Spring Constant Relates to Other Physics Concepts<\/h2>\n \n<p>The constant k is the gateway number for everything a spring can do. Stretch a spring by x and it stores energy equal to half of k times x squared \u2014 the subject of our guide to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/mechanics\/elastic-potential-energy\/\">elastic potential energy<\/a> \u2014 which is why doubling the stretch quadruples the stored energy.<\/p>\n \n<p>Set a mass bouncing and k reappears in the oscillation: a stiff spring snaps its load back quickly, giving the short period that the dynamic method in this article exploits. That timing behaviour is the core of simple harmonic motion, one of the most important models in all of physics.<\/p>\n \n<p>And behind it all sits the proportionality between force and extension itself \u2014 Hooke&#8217;s law \u2014 including the elastic limit beyond which the neat straight line fails. For materials rather than springs, the equivalent stiffness idea is Young&#8217;s modulus, which strips geometry away and describes the substance alone.<\/p>\n \n<h2>Worked Problems<\/h2>\n \n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">A force of 12 N stretches a spring by 4.8 cm. Calculate the spring constant.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Use k = F \/ x, converting the extension to metres: x = 4.8 cm = 0.048 m.<\/p>\n<p>Step 2: Substitute: k = 12 N \/ 0.048 m.<\/p>\n<p>Step 3: Solve: k = 250 N\/m.<\/p>\n<p><strong>Answer: k = 250 N\/m (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n \n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A car suspension spring has k = 45,000 N\/m. What force compresses it by 2.5 cm, and what mass would that force support? 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 to F = k x, with x = 2.5 cm = 0.025 m.<\/p>\n<p>Step 2: Substitute: F = 45,000 N\/m &times; 0.025 m = 1,125 N.<\/p>\n<p>Step 3: The supported mass is m = F \/ g = 1,125 N \/ 9.81 m\/s<sup>2<\/sup> = 114.7 kg.<\/p>\n<p><strong>Answer: F = 1,130 N and m = 115 kg (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n \n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A 250 g mass hangs at rest from a spring, stretching it by 3.5 cm. Find the spring constant. 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: The stretching force is the weight: F = m g = 0.250 kg &times; 9.81 m\/s<sup>2<\/sup> = 2.4525 N.<\/p>\n<p>Step 2: Convert the extension: x = 3.5 cm = 0.035 m.<\/p>\n<p>Step 3: Apply k = F \/ x = 2.4525 N \/ 0.035 m = 70.07 N\/m.<\/p>\n<p><strong>Answer: k = 70.1 N\/m (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n \n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A student loads a spring and records: 1.0 N gives 0.8 cm, 2.0 N gives 1.7 cm, 3.0 N gives 2.4 cm, and 4.0 N gives 3.3 cm. Use a best-fit approach to find the spring constant.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Plot F against x; the points scatter slightly about a straight line through the origin, so take the slope between two well-separated points that sit on the best-fit line: (0.008 m, 1.0 N) and (0.033 m, 4.0 N).<\/p>\n<p>Step 2: Compute the slope: k = &Delta;F \/ &Delta;x = (4.0 N &minus; 1.0 N) \/ (0.033 m &minus; 0.008 m) = 3.0 N \/ 0.025 m.<\/p>\n<p>Step 3: Solve: k = 120 N\/m. Using any single data point instead would carry that point&#8217;s full reading error.<\/p>\n<p><strong>Answer: k = 120 N\/m (best-fit slope, 2 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n \n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">Two identical springs, each with k = 320 N\/m, are mounted side by side (in parallel) and together support a 4.0 kg mass. Find the combined spring constant and the extension. 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: In parallel the constants add: k<sub>total<\/sub> = 320 N\/m + 320 N\/m = 640 N\/m.<\/p>\n<p>Step 2: The load is F = m g = 4.0 kg &times; 9.81 m\/s<sup>2<\/sup> = 39.24 N.<\/p>\n<p>Step 3: Extension: x = F \/ k = 39.24 N \/ 640 N\/m = 0.0613 m = 6.13 cm.<\/p>\n<p><strong>Answer: k<sub>total<\/sub> = 640 N\/m; x = 6.1 cm (2 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n \n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">The same two 320 N\/m springs are now joined end to end (in series) and support the same 4.0 kg mass. Find the combined spring constant and the extension.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: In series the reciprocals add: 1\/k<sub>total<\/sub> = 1\/320 + 1\/320 = 2\/320, so k<sub>total<\/sub> = 160 N\/m.<\/p>\n<p>Step 2: The load is unchanged: F = 39.24 N.<\/p>\n<p>Step 3: Extension: x = F \/ k = 39.24 N \/ 160 N\/m = 0.245 m = 24.5 cm \u2014 four times the parallel case, since the combination is four times softer.<\/p>\n<p><strong>Answer: k<sub>total<\/sub> = 160 N\/m; x = 24.5 cm (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n \n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A 0.50 kg mass hanging on a spring completes 20 oscillations in 12.6 s. Use the dynamic method to find the spring constant.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Find the period: T = 12.6 s \/ 20 = 0.63 s.<\/p>\n<p>Step 2: Use k = 4&pi;<sup>2<\/sup>m \/ T<sup>2<\/sup> with T<sup>2<\/sup> = (0.63 s)<sup>2<\/sup> = 0.3969 s<sup>2<\/sup>.<\/p>\n<p>Step 3: Substitute: k = (4 &times; 9.8696 &times; 0.50 kg) \/ 0.3969 s<sup>2<\/sup> = 19.739 \/ 0.3969 = 49.7 N\/m.<\/p>\n<p><strong>Answer: k = 49.7 N\/m, i.e. about 50 N\/m (3 s.f.)<\/strong><\/p>\n<\/div><\/details><\/div>\n \n<h2>Frequently Asked Questions<\/h2>\n \n<details class=\"pf-faq-item\"><summary>What is the spring constant in simple terms?<\/summary><div class=\"pf-faq-item-answer\">\nThe spring constant is a stiffness score: it tells you how many newtons of force a spring pushes or pulls back with for every metre you stretch or squash it. A soft slinky scores around 1\u20132 N\/m, while a car suspension spring scores tens of thousands. The bigger the number, the harder the spring fights any change in its length.\n<\/div><\/details>\n \n<details class=\"pf-faq-item\"><summary>What are the units of the spring constant?<\/summary><div class=\"pf-faq-item-answer\">\nThe SI unit of the spring constant is the newton per metre (N\/m), because k is a force divided by a length. Engineers often quote the same quantity as a spring rate in newtons per millimetre (N\/mm) or pounds per inch (lb\/in). To convert, 1 N\/mm equals 1,000 N\/m and 1 lb\/in is roughly 175 N\/m.\n<\/div><\/details>\n \n<details class=\"pf-faq-item\"><summary>How do you find the spring constant from a graph?<\/summary><div class=\"pf-faq-item-answer\">\nPlot force on the vertical axis against extension on the horizontal axis, draw the best-fit straight line through the origin, and take its gradient: the slope equals the spring constant k. Pick two well-separated points on the line, not raw data points, and divide the change in force by the change in extension. A slope of 6 N over 0.02 m gives k = 300 N\/m.\n<\/div><\/details>\n \n<details class=\"pf-faq-item\"><summary>Does the spring constant change if you hang a heavier mass?<\/summary><div class=\"pf-faq-item-answer\">\nNo, the spring constant stays the same; only the extension changes. Doubling the hanging mass doubles the stretching force and therefore doubles the extension, but the ratio F\/x, which is k, is unchanged. The constant only appears to shift if you overload the spring so far that it deforms permanently and stops behaving linearly.\n<\/div><\/details>\n \n<details class=\"pf-faq-item\"><summary>What happens to the spring constant if you cut a spring in half?<\/summary><div class=\"pf-faq-item-answer\">\nCutting a spring in half doubles its spring constant, so each half is twice as stiff as the original. The full spring behaves like two half-springs joined end to end, and each half contributes only half of the total stretch under a given force. Keep the force the same with half the stretch and F\/x, the value of k, doubles.\n<\/div><\/details>\n \n<details class=\"pf-faq-item\"><summary>What is a typical spring constant for a car suspension spring?<\/summary><div class=\"pf-faq-item-answer\">\nA typical passenger-car suspension coil has a spring constant of roughly 20,000\u201360,000 N\/m, often quoted by manufacturers as 20\u201360 N\/mm. In plain terms, compressing one corner of the car by a single centimetre takes roughly 200\u2013600 newtons, like resting a 20\u201360 kg load on it. Sports and performance cars sit at the stiffer end of the range.\n<\/div><\/details>\n \n<details class=\"pf-faq-item\"><summary>Is a higher spring constant stiffer or softer?<\/summary><div class=\"pf-faq-item-answer\">\nA higher spring constant means a stiffer spring: more force is needed to produce the same extension. On a force\u2013extension graph, a stiff spring shows up as a steep line and a soft spring as a shallow one. That is why a pen spring (a few hundred N\/m) feels firm between your fingers while a slinky (about 1 N\/m) offers almost no resistance.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>The spring constant k measures a spring&#8217;s stiffness in newtons per metre. Learn how to measure k from a force-extension graph, see typical values from slinky to car spring, and try the interactive lab and calculator.<\/p>\n","protected":false},"author":1,"featured_media":650,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"class_list":["post-649","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\/649","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=649"}],"version-history":[{"count":7,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/649\/revisions"}],"predecessor-version":[{"id":1544,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/649\/revisions\/1544"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/650"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=649"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=649"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=649"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}