🕷️ Crawler Inspector

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Raw Queries and Responses

1. Shard Calculation

Query:

Response:

Calculated Shard: 62 (from laksa002)

2. Crawled Status Check

Query:

curl -X POST \
  'http://laksa062.int.ahrefs:8124/' \
  -H 'Content-Type: text/plain' \
  -H 'X-ClickHouse-Database: crawler3' \
  -H 'Authorization: Basic YXBpOg==' \
  -d 'SELECT getAhrefsURLFromUnparsed(src_unparsed) AS found_url, ifNull(toUnixTimestamp(download_stamp), 0) AS crawl_time, ifNull(toUnixTimestamp(props_url_first_seen), 0) AS first_indexed_time, download_http_code AS http_code, src_unparsed AS src_unparsed, src_root_hash AS src_root_hash, history_drop_reason AS history_drop_reason, meta_title AS meta_title, meta_descriptions AS meta_descriptions, attrs_boilerpipe_text AS attrs_boilerpipe_text, attrs_markdown AS attrs_markdown, attrs_readable_markdown AS attrs_readable_markdown, meta_canonical AS meta_canonical, ml_categories_json AS ml_categories_json, ml_types_json AS ml_types_json, ml_intent_types_json AS ml_intent_types_json, meta_language AS meta_language, attrs_author AS attrs_author, ifNull(toUnixTimestamp(attrs_publish_time), 0) AS attrs_publish_time, ifNull(toUnixTimestamp(attrs_original_publish_time), 0) AS attrs_original_publish_time, ifNull(attrs_is_republished, 0) AS attrs_is_republished, ifNull(attrs_nr_words, 0) AS attrs_nr_words, ifNull(attrs_boilerpipe_nr_words, 0) AS attrs_boilerpipe_nr_words, ifNull(body_ext_links_number, 0) AS body_ext_links_number, ifNull(body_int_links_number, 0) AS body_int_links_number, ifNull(meta_nofollow, 0) AS meta_nofollow, ifNull(meta_noarchive, 0) AS meta_noarchive, ifNull(props_was_rendered, 0) AS props_was_rendered, ifNull(src_redirect, \'\') AS src_redirect, ifNull(download_time_msec, 0) AS download_time_msec, ifNull(download_ttfb_msec, 0) AS download_ttfb_msec, ifNull(download_size, 0) AS download_size FROM crawler3.page_info_local FINAL PREWHERE (src_root_hash, src_unparsed) IN ((getAhrefsRootHashFromUnparsed(getAhrefsUnparsedNoserviceFromURL(\'https://www.vedantu.com/physics/spring-constant\')), getAhrefsUnparsedNoserviceFromURL(\'https://www.vedantu.com/physics/spring-constant\'))) FORMAT JSONEachRow'

Response:

{"found_url":"https:\/\/www.vedantu.com\/physics\/spring-constant","crawl_time":1776162687,"first_indexed_time":1609304639,"http_code":200,"src_unparsed":"com,vedantu!www,\/physics\/spring-constant s443","src_root_hash":"7124968212008936662","history_drop_reason":null,"meta_title":"Spring Constant: Definition, Formula & Solved Examples","meta_descriptions":["Master spring constant concepts-learn formulas, see step-by-step problems, and boost your physics grades with Vedantu."],"attrs_boilerpipe_text":"Simple Harmonious motion i.e. SHM is a veritably intriguing type of stir. It's constantly applied in the oscillatory motion of the objects. Springs generally have SHM. Springs have their own native “ spring constants'' which define how stiff they are. Hooke's law is a notorious law that explains the SHM and gives a formula for the force applied using spring constant.\nFor example, ask if a group of car designers will knock on your door and help develop a suspension system. \"Of course,\" you say. They say the car weighs 1,000 kilograms and requires the use of four shock absorbers, each 0.5 meters long. How strong should the feathers be? Assuming these shock absorbers use springs, each must support a weight of at least 250 kilograms: \n F = mg = (250 kg) (9.8 m \/ s\n2\n) = 2,450 N \nWhere F is the force, m is the mass of the object, g is the gravitational acceleration, 9.8 meters \/ sec\n2\n. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke`s law says:\nF = –kx\nNow, we need to rework the equation so that we are calculating for the missing metric, which is the spring constant, or k.\nThe springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter.\nSpring Constant Units\nSpring constant definition is related to simple harmonic motions and Hooke's law. So, before we try to define the spring constant and understand the workings of the spring constant, we need to look at Hooke's law. According to the theory of elasticity, when a load is applied to spring it will naturally extend proportionally, as long as the load applied is less than the elastic limit. Now we know that when force is applied to an object, it tends to deform in some way. \nConsider a spring, when we apply force on one side of the spring, it will get compressed, as they are elastic. At this time the spring exerts its force in the direction opposite to the applied force, to expand to its original size. Therefore, to define the spring constant, we first define Hooke's law. Hooke's law is defined as the force required by the spring to revert to its size is directly proportional to the distance of the compression of the spring.\n(Image to be added soon)\nThe image shows the movement of the spring when force is applied to one side.\nSpring Constant Definition\nTo understand the spring constant definition, we will look at Hooke’s law formula. Hooke’s law formula is also known as the spring constant formula. The formula is given below.\nFormula\nF = -kx\nSpring Constant Units\nNm\n-1\nWhere F represents the restoring force of the spring, x is the displacement of the spring, and k is known as the spring constant. The spring constant units are given as Newton per meter. \nNow that we know that k is the spring constant, we will look at the spring constant definition. We define spring constant as the stiffness of the spring. In other words, when the displacement of the spring is one unit, we can define the spring constant as the force applied to cause that said displacement. Therefore, it is clear to say that, the stiffer the spring is, the higher will be its spring constant.\nSpring Constant Dimensional Formula\nAccording to Hooke’s law, we know that,\nF= -kx\nTherefore, \nk= -F\/x\nNow we know that the\nunit of force\nis given as Newton (N), or as kg m\/s\n2\n.\nTherefore, we can write the dimensional unit as [MLT\n-2\n].\nWe also know the dimensional unit of x is given as L\nApplying spring constant formula, we get, \n[\nM\nL\nT\n−\n2\nL\n]\nk= - [MT\n-2\n]\nThe spring constant unit is in terms of Newton per meter (N\/m).\nSolved Problems\nQuestion 1)\nA spring is stretched by 40cm when a load of 5kg is added to it. Find the spring constant.\nAnswer 1)\nGiven,\nMass m = 5kg,\nDisplacement x = 40cm = 0.4m\nTo find the spring constant, we first need to find the force that is acting on the spring. \nWe know that F = m * x\nTherefore, F = 5 * 0.4\nF = 2N\nThe load applies a force of 2N on the spring. Hence, the spring will apply an equal and opposite force of – 2N.\nNow, by substituting the values in the spring constant formula we get,\nk = -F\/x\nk =\n−\n2\n0.4\nk = 5 N\/m\nTherefore, the spring constant of the spring is 5N\/m\nQuestion 2)\nConsider a spring with a spring constant of 14000N\/m. A force of 3500N is applied to the spring. What will be the displacement of the spring?\nAnswer 2)\nGiven,\nForce F = 3500N,\nSpring constant k = 14000N\/m,\nWe can calculate the displacement of the spring by using the spring constant formula.\nx=\n−\nF\nk\nThe load applies a force of 3500N on the spring. Hence the spring will apply an equal and opposite force of – 3500N.\nThus,\nx =\n−\n(\n−\n3500\n14000\n)\nx = 0.25 m\nx = 25 cm\nTherefore, the spring is displaced by a distance of 25cm.\nConclusion\nThis is how the spring constant is determined and calculated. The formula is defined using the terms to determine the various physical quantities. Understand how this constant is utilized in solving different problems from the given examples.","attrs_markdown":"![ask Ved](https:\/\/commons.vedantu.com\/cdn\/images\/askVedSearch.svg)\n\nSign In\n\nCourses\n\nExplore all Vedantu courses by class or target exam, starting at ₹9000\n\n![](https:\/\/growth-cdn.vedantu.com\/vmkt\/PROD\/svg%2Bxml\/5bb2c400-af38-473d-94b5-a0f0633c1257-1661769224426-4102418410874681.svg%2Bxml#fcbc)\n\nFind courses by class\n\n![](https:\/\/growth-cdn.vedantu.com\/vmkt\/PROD\/svg%2Bxml\/5bb2c400-af38-473d-94b5-a0f0633c1257-1661769224426-4102418410874681.svg%2Bxml#fcbt)\n\nFind courses by target\n\n[![](https:\/\/www.vedantu.com\/seo\/content-images\/84891ea1-58f0-4d45-858b-14b1e6ba2652.svg) Long Term Courses Guaranteed improvement in marks or get your fees back](https:\/\/www.vedantu.com\/course\/ai-live?show_bn=t1&couponCode=VNAV20&itm_source=new_nav_menu&itm_medium=new_nav_menu&itm_campaign=new_nav_menu)\n\n[![](https:\/\/growth-cdn.vedantu.com\/vmkt\/PROD\/svg%2Bxml\/5bb2c400-af38-473d-94b5-a0f0633c1257-1661769224426-4102418410874681.svg%2Bxml#oto) One-to-one LIVE classes Learn one-to-one with a teacher for a personalised experience](https:\/\/courses.vedantu.com\/one-to-one-live\/?itm_source=new_nav_menu&itm_medium=new_nav_menu&itm_campaign=new_nav_menu&utm_source=new_nav_menu&utm_medium=new_nav_menu&utm_campaign=new_nav_menu)\n\n[![](https:\/\/vmkt.vedantu.com\/vmkt\/PROD\/svg%2Bxml\/b8988f51-df03-4450-88a7-ca66fbf931a0-1674547460687-4102418410874681.svg%2Bxml) Topic specific courses Master any topic at just ₹1](https:\/\/www.vedantu.com\/microcourses-list)\n\n![](https:\/\/growth-cdn.vedantu.com\/vmkt\/PROD\/svg%2Bxml\/5bb2c400-af38-473d-94b5-a0f0633c1257-1661769224426-4102418410874681.svg%2Bxml#vsk)\n\nCourses for Kids\n\nCourses for Kids\n\nConfidence-building & personalised learning courses for Class LKG-8 students\n\n[![](https:\/\/www.vedantu.com\/cdn\/images\/new-header\/icons\/header-vsk-icons.svg#english-superstar) English Superstar Age 4 - 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Examples\nSimple Harmonious motion i.e. SHM is a veritably intriguing type of stir. It's constantly applied in the oscillatory motion of the objects. Springs generally have SHM. Springs have their own native “ spring constants'' which define how stiff they are. Hooke's law is a notorious law that explains the SHM and gives a formula for the force applied using spring constant.\n\nFor example, ask if a group of car designers will knock on your door and help develop a suspension system. \"Of course,\" you say. They say the car weighs 1,000 kilograms and requires the use of four shock absorbers, each 0.5 meters long. How strong should the feathers be? Assuming these shock absorbers use springs, each must support a weight of at least 250 kilograms:\n\nF = mg = (250 kg) (9.8 m \/ s2) = 2,450 N\n\nWhere F is the force, m is the mass of the object, g is the gravitational acceleration, 9.8 meters \/ sec2. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke\\`s law says:\n\nF = –kx\n\nNow, we need to rework the equation so that we are calculating for the missing metric, which is the spring constant, or k.\n\nThe springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter.\n\n### Spring Constant Units\nSpring constant definition is related to simple harmonic motions and Hooke's law. So, before we try to define the spring constant and understand the workings of the spring constant, we need to look at Hooke's law. According to the theory of elasticity, when a load is applied to spring it will naturally extend proportionally, as long as the load applied is less than the elastic limit. Now we know that when force is applied to an object, it tends to deform in some way.\n\nConsider a spring, when we apply force on one side of the spring, it will get compressed, as they are elastic. At this time the spring exerts its force in the direction opposite to the applied force, to expand to its original size. Therefore, to define the spring constant, we first define Hooke's law. Hooke's law is defined as the force required by the spring to revert to its size is directly proportional to the distance of the compression of the spring.\n\n(Image to be added soon)\n\nThe image shows the movement of the spring when force is applied to one side.\n\n### Spring Constant Definition\nTo understand the spring constant definition, we will look at Hooke’s law formula. Hooke’s law formula is also known as the spring constant formula. The formula is given below.\n\n|   |   |\n|---|---|\n| Formula | F = -kx |\n| Spring Constant Units | Nm\\-1 |\n\nWhere F represents the restoring force of the spring, x is the displacement of the spring, and k is known as the spring constant. The spring constant units are given as Newton per meter.\n\nNow that we know that k is the spring constant, we will look at the spring constant definition. We define spring constant as the stiffness of the spring. In other words, when the displacement of the spring is one unit, we can define the spring constant as the force applied to cause that said displacement. Therefore, it is clear to say that, the stiffer the spring is, the higher will be its spring constant.\n\n### Spring Constant Dimensional Formula\nAccording to Hooke’s law, we know that,\n\nF= -kx\n\nTherefore,\n\nk= -F\/x\n\nNow we know that the [unit of force](https:\/\/www.vedantu.com\/physics\/unit-of-force) is given as Newton (N), or as kg m\/s2.\n\nTherefore, we can write the dimensional unit as \\[MLT\\-2\\].\n\nWe also know the dimensional unit of x is given as L\n\nApplying spring constant formula, we get,\n\n\\[ M L T − 2 L \\]\n\nk= - \\[MT\\-2\\]\n\nThe spring constant unit is in terms of Newton per meter (N\/m).\n\n### Solved Problems\n**Question 1)** A spring is stretched by 40cm when a load of 5kg is added to it. Find the spring constant.\n\n**Answer 1)** Given,\n\nMass m = 5kg,\n\nDisplacement x = 40cm = 0.4m\n\nTo find the spring constant, we first need to find the force that is acting on the spring.\n\nWe know that F = m \\* x\n\nTherefore, F = 5 \\* 0.4\n\nF = 2N\n\nThe load applies a force of 2N on the spring. Hence, the spring will apply an equal and opposite force of – 2N.\n\nNow, by substituting the values in the spring constant formula we get,\n\nk = -F\/x\n\nk = − 2 0\\.4\n\nk = 5 N\/m\n\nTherefore, the spring constant of the spring is 5N\/m\n\n**Question 2)** Consider a spring with a spring constant of 14000N\/m. A force of 3500N is applied to the spring. What will be the displacement of the spring?\n\n**Answer 2)** Given,\n\nForce F = 3500N,\n\nSpring constant k = 14000N\/m,\n\nWe can calculate the displacement of the spring by using the spring constant formula.\n\nx= − F k\n\nThe load applies a force of 3500N on the spring. Hence the spring will apply an equal and opposite force of – 3500N.\n\nThus,\n\nx = − ( − 3500 14000 )\n\nx = 0.25 m\n\nx = 25 cm\n\nTherefore, the spring is displaced by a distance of 25cm.\n\n### Conclusion\nThis is how the spring constant is determined and calculated. The formula is defined using the terms to determine the various physical quantities. Understand how this constant is utilized in solving different problems from the given examples.\n\n## FAQs on Spring Constant Explained: Definition, Formula & Problems\n1\\. What is the spring constant (k)?\n\nThe spring constant, represented by the symbol **k**, is a measure of a spring's stiffness. It quantifies the amount of force required to stretch or compress a spring by a certain distance. A higher spring constant indicates a stiffer spring, meaning more force is needed for displacement, while a lower value signifies a more flexible spring.\n\n2\\. What is the formula for the spring constant based on Hooke's Law?\n\nThe spring constant formula is derived from **Hooke's Law**. The law is stated as **F = -kx**, where F is the restoring force exerted by the spring, x is the displacement from its equilibrium position, and k is the spring constant. By rearranging this formula, the spring constant can be calculated as:  \n**k = -F \/ x**\n\n3\\. What is the SI unit and dimensional formula for the spring constant?\n\nThe SI unit and dimensional formula for the spring constant are as follows:\n\n- **SI Unit:** The SI unit for the spring constant is **newtons per meter (N\/m)**.\n- **Dimensional Formula:** The dimensional formula is derived from the ratio of force (\\[MLT\\-2\\]) to distance (\\[L\\]), resulting in **\\[MT\\-2\\]**.\n\n4\\. What are some real-world examples where the spring constant is important?\n\nThe spring constant is a critical property in many everyday applications. Some key examples include:\n\n- **Vehicle Suspension Systems:** Springs in cars and motorcycles must have a specific spring constant to absorb shocks from the road effectively, providing a smooth ride.\n- **Weighing Scales:** Mechanical weighing scales use the compression or extension of a calibrated spring to measure weight. The accuracy depends on a precise and stable spring constant.\n- **Ballpoint Pens:** The simple click mechanism in a retractable pen uses a small spring with a specific constant to extend and retract the ink cartridge.\n- **Mattresses:** The coils inside a spring mattress are designed with varying spring constants to provide support and comfort.\n\n5\\. How is the equivalent spring constant calculated for springs in series and parallel combinations?\n\nWhen multiple springs are combined, their overall stiffness changes. The equivalent spring constant (keq) is calculated differently depending on the arrangement:\n\n- **Springs in Series:** When connected end-to-end, the springs become more flexible. The reciprocal of the equivalent spring constant is the sum of the reciprocals of individual constants: **1\/keq = 1\/k1 + 1\/k2 + ...**\n- **Springs in Parallel:** When connected side-by-side, the combination becomes stiffer. The equivalent spring constant is the sum of the individual constants: **keq = k1 + k2 + ...**\n\n6\\. Why is there a negative sign in the Hooke's Law formula, F = -kx?\n\nThe negative sign in Hooke's Law (**F = -kx**) is fundamentally important as it signifies the direction of the spring's force. This force, known as the **restoring force**, always acts in the opposite direction to the displacement (x). If you pull a spring to the right (positive displacement), it pulls back to the left (negative force). If you compress it to the left (negative displacement), it pushes back to the right (positive force). The sign ensures the force always tries to return the spring to its equilibrium position.\n\n7\\. What would happen if a spring had a negative or zero spring constant?\n\nA spring cannot physically have a negative or zero spring constant, but considering it hypothetically reveals why.\n\n- **Zero Spring Constant (k=0):** If k were zero, it would mean that no force is required to stretch or compress the object. It would not be a spring but an object with no resistance to deformation, like a loose chain.\n- **Negative Spring Constant (k\\<0):** If k were negative, the restoring force would act in the same direction as the displacement. A slight push would cause the spring to push further in that direction, and a slight pull would make it pull itself further apart, leading to unstable and infinite expansion or compression.\n\n8\\. How does the spring constant affect the time period of a mass oscillating on a spring?\n\nThe spring constant directly influences the speed of oscillation for a mass attached to a spring in Simple Harmonic Motion (SHM). The relationship is given by the formula for the time period (T):  \n**T = 2π√(m\/k)**  \nHere, T is the time for one full oscillation, m is the mass, and k is the spring constant. This shows that the time period is inversely proportional to the square root of the spring constant. A **stiffer spring (higher k)** will oscillate faster, resulting in a **shorter time period**.\n\n9\\. What does the graph of applied force versus extension for a spring look like, and what does its slope represent?\n\nFor a spring that obeys Hooke's Law, a graph of the applied force (F) on the y-axis versus the resulting extension or displacement (x) on the x-axis is a **straight line passing through the origin**. The slope of this line is calculated as the change in force divided by the change in displacement (ΔF\/Δx). According to Hooke's law (F=kx), this slope directly represents the value of the **spring constant, k**.\n\n10\\. Does the spring constant change if you cut a spring in half?\n\nYes, the spring constant changes significantly. When you cut a spring in half, each of the two new, shorter springs becomes **stiffer** than the original. The spring constant of each half will be **double the spring constant of the original spring**. 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Admission![arrow-right](https:\/\/www.vedantu.com\/cdn\/images\/seo-templates\/arrow-right.png)](https:\/\/www.vedantu.com\/jee-main\/cut-off-marks-for-nit)\n\n[NEET 2026 Exam Dates: Complete Details on Syllabus, Eligibility, Admit Card, Result, Counselling Process![arrow-right](https:\/\/www.vedantu.com\/cdn\/images\/seo-templates\/arrow-right.png)](https:\/\/www.vedantu.com\/neet)\n\n[BSc Nursing Application Form 2026: Exam Dates, Eligibility Criteria![arrow-right](https:\/\/www.vedantu.com\/cdn\/images\/seo-templates\/arrow-right.png)](https:\/\/www.vedantu.com\/neet\/bsc-nursing-form)\n\n[NEET 2026 Exam Date (OUT)![arrow-right](https:\/\/www.vedantu.com\/cdn\/images\/seo-templates\/arrow-right.png)](https:\/\/www.vedantu.com\/neet\/neet-exam-dates)\n\n- 1\n- 2\n\n[![]()![](https:\/\/marketing-assets.vedantu.com\/vmkt\/image\/png\/97d562be-e456-4e8a-b58b-d1bdecb3e256-1770019394248-4102638261224925.png)](https:\/\/www.vedantu.com\/bookademo?&utm_source=mau_crm&utm_medium=seoSideBanner&utm_campaign=pagelanding&utm_term=11neet_bod&couponCode=)\n\n\\<h2\\>Simple Harmonic Motion and Spring Constant: Understanding Hooke's Law\\<\/h2\\> \\<p\\>\\<strong\\>Summary:\\<\/strong\\> Simple Harmonic Motion (SHM) is fundamental in oscillatory motion, particularly in springs. Springs possess unique spring constants that define their stiffness. Hooke's law explains SHM through the formula F = -kx, where F is restoring force, k is spring constant, and x is displacement. The spring constant represents stiffness - stiffer springs have higher constants. According to elasticity theory, springs extend proportionally when loaded within elastic limits. When compressed, springs exert opposite force to return to original size. Spring constant units are Newton per meter (N\/m), with dimensional formula \\[MT⁻²\\]. Practical applications include car suspension systems where designers calculate required spring constants based on weight distribution and compression distances.\\<\/p\\> \\<h3\\>Questions\/Concepts Covered:\\<\/h3\\> \\<ul\\> \\<li\\>How to calculate spring constant using Hooke's law formula and displacement measurements?\\<\/li\\> \\<li\\>What is the relationship between spring stiffness and spring constant in SHM applications?\\<\/li\\> \\<li\\>How to determine spring displacement when force and spring constant are given?\\<\/li\\> \\<\/ul\\> \\<h3\\>Keywords:\\<\/h3\\> \\<ul\\> \\<li\\>Simple harmonic motion spring constant\\<\/li\\> \\<li\\>Hooke's law formula calculation\\<\/li\\> \\<li\\>Spring constant dimensional formula\\<\/li\\> \\<li\\>Newton per meter spring units\\<\/li\\> \\<li\\>Spring displacement force relationship\\<\/li\\> \\<\/ul\\>\n\nStudy material\n\nNCERT\n\n- 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All rights reserved\n\n[Privacy policy](https:\/\/www.vedantu.com\/privacy-policy)[Terms and conditions](https:\/\/www.vedantu.com\/tnc)","attrs_readable_markdown":"Simple Harmonious motion i.e. SHM is a veritably intriguing type of stir. It's constantly applied in the oscillatory motion of the objects. Springs generally have SHM. Springs have their own native “ spring constants'' which define how stiff they are. Hooke's law is a notorious law that explains the SHM and gives a formula for the force applied using spring constant.\n\nFor example, ask if a group of car designers will knock on your door and help develop a suspension system. \"Of course,\" you say. They say the car weighs 1,000 kilograms and requires the use of four shock absorbers, each 0.5 meters long. How strong should the feathers be? Assuming these shock absorbers use springs, each must support a weight of at least 250 kilograms:\n\nF = mg = (250 kg) (9.8 m \/ s2) = 2,450 N\n\nWhere F is the force, m is the mass of the object, g is the gravitational acceleration, 9.8 meters \/ sec2. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke\\`s law says:\n\nF = –kx\n\nNow, we need to rework the equation so that we are calculating for the missing metric, which is the spring constant, or k.\n\nThe springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter.\n\n### Spring Constant Units\nSpring constant definition is related to simple harmonic motions and Hooke's law. So, before we try to define the spring constant and understand the workings of the spring constant, we need to look at Hooke's law. According to the theory of elasticity, when a load is applied to spring it will naturally extend proportionally, as long as the load applied is less than the elastic limit. Now we know that when force is applied to an object, it tends to deform in some way.\n\nConsider a spring, when we apply force on one side of the spring, it will get compressed, as they are elastic. At this time the spring exerts its force in the direction opposite to the applied force, to expand to its original size. Therefore, to define the spring constant, we first define Hooke's law. Hooke's law is defined as the force required by the spring to revert to its size is directly proportional to the distance of the compression of the spring.\n\n(Image to be added soon)\n\nThe image shows the movement of the spring when force is applied to one side.\n\n### Spring Constant Definition\nTo understand the spring constant definition, we will look at Hooke’s law formula. Hooke’s law formula is also known as the spring constant formula. The formula is given below.\n\n|   |   |\n|---|---|\n| Formula | F = -kx |\n| Spring Constant Units | Nm\\-1 |\n\nWhere F represents the restoring force of the spring, x is the displacement of the spring, and k is known as the spring constant. The spring constant units are given as Newton per meter.\n\nNow that we know that k is the spring constant, we will look at the spring constant definition. We define spring constant as the stiffness of the spring. In other words, when the displacement of the spring is one unit, we can define the spring constant as the force applied to cause that said displacement. Therefore, it is clear to say that, the stiffer the spring is, the higher will be its spring constant.\n\n### Spring Constant Dimensional Formula\nAccording to Hooke’s law, we know that,\n\nF= -kx\n\nTherefore,\n\nk= -F\/x\n\nNow we know that the [unit of force](https:\/\/www.vedantu.com\/physics\/unit-of-force) is given as Newton (N), or as kg m\/s2.\n\nTherefore, we can write the dimensional unit as \\[MLT\\-2\\].\n\nWe also know the dimensional unit of x is given as L\n\nApplying spring constant formula, we get,\n\n\\[ M L T − 2 L \\]\n\nk= - \\[MT\\-2\\]\n\nThe spring constant unit is in terms of Newton per meter (N\/m).\n\n### Solved Problems\n**Question 1)** A spring is stretched by 40cm when a load of 5kg is added to it. Find the spring constant.\n\n**Answer 1)** Given,\n\nMass m = 5kg,\n\nDisplacement x = 40cm = 0.4m\n\nTo find the spring constant, we first need to find the force that is acting on the spring.\n\nWe know that F = m \\* x\n\nTherefore, F = 5 \\* 0.4\n\nF = 2N\n\nThe load applies a force of 2N on the spring. Hence, the spring will apply an equal and opposite force of – 2N.\n\nNow, by substituting the values in the spring constant formula we get,\n\nk = -F\/x\n\nk = − 2 0\\.4\n\nk = 5 N\/m\n\nTherefore, the spring constant of the spring is 5N\/m\n\n**Question 2)** Consider a spring with a spring constant of 14000N\/m. A force of 3500N is applied to the spring. What will be the displacement of the spring?\n\n**Answer 2)** Given,\n\nForce F = 3500N,\n\nSpring constant k = 14000N\/m,\n\nWe can calculate the displacement of the spring by using the spring constant formula.\n\nx= − F k\n\nThe load applies a force of 3500N on the spring. Hence the spring will apply an equal and opposite force of – 3500N.\n\nThus,\n\nx = − ( − 3500 14000 )\n\nx = 0.25 m\n\nx = 25 cm\n\nTherefore, the spring is displaced by a distance of 25cm.\n\n### Conclusion\nThis is how the spring constant is determined and calculated. The formula is defined using the terms to determine the various physical quantities. Understand how this constant is utilized in solving different problems from the given examples.","meta_canonical":null,"ml_categories_json":"{\"\/Science\":968,\"\/Science\/Physics\":954}","ml_types_json":"{\"\/Article\":994,\"\/Article\/Tutorial_or_Guide\":820}","ml_intent_types_json":"{\"Informational\":999}","meta_language":"en","attrs_author":null,"attrs_publish_time":1594166400,"attrs_original_publish_time":1594166400,"attrs_is_republished":0,"attrs_nr_words":"3578","attrs_boilerpipe_nr_words":"960","body_ext_links_number":7,"body_int_links_number":235,"meta_nofollow":0,"meta_noarchive":0,"props_was_rendered":1,"src_redirect":"","download_time_msec":1363,"download_ttfb_msec":1297,"download_size":40654}

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Simple Harmonious motion i.e. SHM is a veritably intriguing type of stir. It's constantly applied in the oscillatory motion of the objects. Springs generally have SHM. Springs have their own native “ spring constants'' which define how stiff they are. Hooke's law is a notorious law that explains the SHM and gives a formula for the force applied using spring constant. For example, ask if a group of car designers will knock on your door and help develop a suspension system. "Of course," you say. They say the car weighs 1,000 kilograms and requires the use of four shock absorbers, each 0.5 meters long. How strong should the feathers be? Assuming these shock absorbers use springs, each must support a weight of at least 250 kilograms: F = mg = (250 kg) (9.8 m / s 2 ) = 2,450 N Where F is the force, m is the mass of the object, g is the gravitational acceleration, 9.8 meters / sec 2 . The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke`s law says: F = –kx Now, we need to rework the equation so that we are calculating for the missing metric, which is the spring constant, or k. The springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter. Spring Constant Units Spring constant definition is related to simple harmonic motions and Hooke's law. So, before we try to define the spring constant and understand the workings of the spring constant, we need to look at Hooke's law. According to the theory of elasticity, when a load is applied to spring it will naturally extend proportionally, as long as the load applied is less than the elastic limit. Now we know that when force is applied to an object, it tends to deform in some way. Consider a spring, when we apply force on one side of the spring, it will get compressed, as they are elastic. At this time the spring exerts its force in the direction opposite to the applied force, to expand to its original size. Therefore, to define the spring constant, we first define Hooke's law. Hooke's law is defined as the force required by the spring to revert to its size is directly proportional to the distance of the compression of the spring. (Image to be added soon) The image shows the movement of the spring when force is applied to one side. Spring Constant Definition To understand the spring constant definition, we will look at Hooke’s law formula. Hooke’s law formula is also known as the spring constant formula. The formula is given below. Formula F = -kx Spring Constant Units Nm -1 Where F represents the restoring force of the spring, x is the displacement of the spring, and k is known as the spring constant. The spring constant units are given as Newton per meter. Now that we know that k is the spring constant, we will look at the spring constant definition. We define spring constant as the stiffness of the spring. In other words, when the displacement of the spring is one unit, we can define the spring constant as the force applied to cause that said displacement. Therefore, it is clear to say that, the stiffer the spring is, the higher will be its spring constant. Spring Constant Dimensional Formula According to Hooke’s law, we know that, F= -kx Therefore, k= -F/x Now we know that the unit of force is given as Newton (N), or as kg m/s 2 . Therefore, we can write the dimensional unit as [MLT -2 ]. We also know the dimensional unit of x is given as L Applying spring constant formula, we get, [ M L T − 2 L ] k= - [MT -2 ] The spring constant unit is in terms of Newton per meter (N/m). Solved Problems Question 1) A spring is stretched by 40cm when a load of 5kg is added to it. Find the spring constant. Answer 1) Given, Mass m = 5kg, Displacement x = 40cm = 0.4m To find the spring constant, we first need to find the force that is acting on the spring. We know that F = m * x Therefore, F = 5 * 0.4 F = 2N The load applies a force of 2N on the spring. Hence, the spring will apply an equal and opposite force of – 2N. Now, by substituting the values in the spring constant formula we get, k = -F/x k = − 2 0.4 k = 5 N/m Therefore, the spring constant of the spring is 5N/m Question 2) Consider a spring with a spring constant of 14000N/m. A force of 3500N is applied to the spring. What will be the displacement of the spring? Answer 2) Given, Force F = 3500N, Spring constant k = 14000N/m, We can calculate the displacement of the spring by using the spring constant formula. x= − F k The load applies a force of 3500N on the spring. Hence the spring will apply an equal and opposite force of – 3500N. Thus, x = − ( − 3500 14000 ) x = 0.25 m x = 25 cm Therefore, the spring is displaced by a distance of 25cm. Conclusion This is how the spring constant is determined and calculated. The formula is defined using the terms to determine the various physical quantities. Understand how this constant is utilized in solving different problems from the given examples.

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It's constantly applied in the oscillatory motion of the objects. Springs generally have SHM. Springs have their own native “ spring constants'' which define how stiff they are. Hooke's law is a notorious law that explains the SHM and gives a formula for the force applied using spring constant. For example, ask if a group of car designers will knock on your door and help develop a suspension system. "Of course," you say. They say the car weighs 1,000 kilograms and requires the use of four shock absorbers, each 0.5 meters long. How strong should the feathers be? Assuming these shock absorbers use springs, each must support a weight of at least 250 kilograms: F = mg = (250 kg) (9.8 m / s2) = 2,450 N Where F is the force, m is the mass of the object, g is the gravitational acceleration, 9.8 meters / sec2. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke\`s law says: F = –kx Now, we need to rework the equation so that we are calculating for the missing metric, which is the spring constant, or k. The springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter. ### Spring Constant Units Spring constant definition is related to simple harmonic motions and Hooke's law. So, before we try to define the spring constant and understand the workings of the spring constant, we need to look at Hooke's law. According to the theory of elasticity, when a load is applied to spring it will naturally extend proportionally, as long as the load applied is less than the elastic limit. Now we know that when force is applied to an object, it tends to deform in some way. Consider a spring, when we apply force on one side of the spring, it will get compressed, as they are elastic. At this time the spring exerts its force in the direction opposite to the applied force, to expand to its original size. Therefore, to define the spring constant, we first define Hooke's law. Hooke's law is defined as the force required by the spring to revert to its size is directly proportional to the distance of the compression of the spring. (Image to be added soon) The image shows the movement of the spring when force is applied to one side. ### Spring Constant Definition To understand the spring constant definition, we will look at Hooke’s law formula. Hooke’s law formula is also known as the spring constant formula. The formula is given below. | | | |---|---| | Formula | F = -kx | | Spring Constant Units | Nm\-1 | Where F represents the restoring force of the spring, x is the displacement of the spring, and k is known as the spring constant. The spring constant units are given as Newton per meter. Now that we know that k is the spring constant, we will look at the spring constant definition. We define spring constant as the stiffness of the spring. In other words, when the displacement of the spring is one unit, we can define the spring constant as the force applied to cause that said displacement. Therefore, it is clear to say that, the stiffer the spring is, the higher will be its spring constant. ### Spring Constant Dimensional Formula According to Hooke’s law, we know that, F= -kx Therefore, k= -F/x Now we know that the [unit of force](https://www.vedantu.com/physics/unit-of-force) is given as Newton (N), or as kg m/s2. Therefore, we can write the dimensional unit as \[MLT\-2\]. We also know the dimensional unit of x is given as L Applying spring constant formula, we get, \[ M L T − 2 L \] k= - \[MT\-2\] The spring constant unit is in terms of Newton per meter (N/m). ### Solved Problems **Question 1)** A spring is stretched by 40cm when a load of 5kg is added to it. Find the spring constant. **Answer 1)** Given, Mass m = 5kg, Displacement x = 40cm = 0.4m To find the spring constant, we first need to find the force that is acting on the spring. We know that F = m \* x Therefore, F = 5 \* 0.4 F = 2N The load applies a force of 2N on the spring. Hence, the spring will apply an equal and opposite force of – 2N. Now, by substituting the values in the spring constant formula we get, k = -F/x k = − 2 0\.4 k = 5 N/m Therefore, the spring constant of the spring is 5N/m **Question 2)** Consider a spring with a spring constant of 14000N/m. A force of 3500N is applied to the spring. What will be the displacement of the spring? **Answer 2)** Given, Force F = 3500N, Spring constant k = 14000N/m, We can calculate the displacement of the spring by using the spring constant formula. x= − F k The load applies a force of 3500N on the spring. Hence the spring will apply an equal and opposite force of – 3500N. Thus, x = − ( − 3500 14000 ) x = 0.25 m x = 25 cm Therefore, the spring is displaced by a distance of 25cm. ### Conclusion This is how the spring constant is determined and calculated. The formula is defined using the terms to determine the various physical quantities. Understand how this constant is utilized in solving different problems from the given examples. ## FAQs on Spring Constant Explained: Definition, Formula & Problems 1\. What is the spring constant (k)? The spring constant, represented by the symbol **k**, is a measure of a spring's stiffness. It quantifies the amount of force required to stretch or compress a spring by a certain distance. A higher spring constant indicates a stiffer spring, meaning more force is needed for displacement, while a lower value signifies a more flexible spring. 2\. What is the formula for the spring constant based on Hooke's Law? The spring constant formula is derived from **Hooke's Law**. The law is stated as **F = -kx**, where F is the restoring force exerted by the spring, x is the displacement from its equilibrium position, and k is the spring constant. By rearranging this formula, the spring constant can be calculated as: **k = -F / x** 3\. What is the SI unit and dimensional formula for the spring constant? The SI unit and dimensional formula for the spring constant are as follows: - **SI Unit:** The SI unit for the spring constant is **newtons per meter (N/m)**. - **Dimensional Formula:** The dimensional formula is derived from the ratio of force (\[MLT\-2\]) to distance (\[L\]), resulting in **\[MT\-2\]**. 4\. What are some real-world examples where the spring constant is important? The spring constant is a critical property in many everyday applications. Some key examples include: - **Vehicle Suspension Systems:** Springs in cars and motorcycles must have a specific spring constant to absorb shocks from the road effectively, providing a smooth ride. - **Weighing Scales:** Mechanical weighing scales use the compression or extension of a calibrated spring to measure weight. The accuracy depends on a precise and stable spring constant. - **Ballpoint Pens:** The simple click mechanism in a retractable pen uses a small spring with a specific constant to extend and retract the ink cartridge. - **Mattresses:** The coils inside a spring mattress are designed with varying spring constants to provide support and comfort. 5\. How is the equivalent spring constant calculated for springs in series and parallel combinations? When multiple springs are combined, their overall stiffness changes. The equivalent spring constant (keq) is calculated differently depending on the arrangement: - **Springs in Series:** When connected end-to-end, the springs become more flexible. The reciprocal of the equivalent spring constant is the sum of the reciprocals of individual constants: **1/keq = 1/k1 + 1/k2 + ...** - **Springs in Parallel:** When connected side-by-side, the combination becomes stiffer. The equivalent spring constant is the sum of the individual constants: **keq = k1 + k2 + ...** 6\. Why is there a negative sign in the Hooke's Law formula, F = -kx? The negative sign in Hooke's Law (**F = -kx**) is fundamentally important as it signifies the direction of the spring's force. This force, known as the **restoring force**, always acts in the opposite direction to the displacement (x). If you pull a spring to the right (positive displacement), it pulls back to the left (negative force). If you compress it to the left (negative displacement), it pushes back to the right (positive force). The sign ensures the force always tries to return the spring to its equilibrium position. 7\. What would happen if a spring had a negative or zero spring constant? A spring cannot physically have a negative or zero spring constant, but considering it hypothetically reveals why. - **Zero Spring Constant (k=0):** If k were zero, it would mean that no force is required to stretch or compress the object. It would not be a spring but an object with no resistance to deformation, like a loose chain. - **Negative Spring Constant (k\<0):** If k were negative, the restoring force would act in the same direction as the displacement. A slight push would cause the spring to push further in that direction, and a slight pull would make it pull itself further apart, leading to unstable and infinite expansion or compression. 8\. How does the spring constant affect the time period of a mass oscillating on a spring? The spring constant directly influences the speed of oscillation for a mass attached to a spring in Simple Harmonic Motion (SHM). The relationship is given by the formula for the time period (T): **T = 2π√(m/k)** Here, T is the time for one full oscillation, m is the mass, and k is the spring constant. This shows that the time period is inversely proportional to the square root of the spring constant. A **stiffer spring (higher k)** will oscillate faster, resulting in a **shorter time period**. 9\. What does the graph of applied force versus extension for a spring look like, and what does its slope represent? For a spring that obeys Hooke's Law, a graph of the applied force (F) on the y-axis versus the resulting extension or displacement (x) on the x-axis is a **straight line passing through the origin**. The slope of this line is calculated as the change in force divided by the change in displacement (ΔF/Δx). According to Hooke's law (F=kx), this slope directly represents the value of the **spring constant, k**. 10\. Does the spring constant change if you cut a spring in half? Yes, the spring constant changes significantly. When you cut a spring in half, each of the two new, shorter springs becomes **stiffer** than the original. The spring constant of each half will be **double the spring constant of the original spring**. This is because the stiffness of a spring is inversely proportional to its length; a shorter length of the same material offers more resistance to the same displacement. 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Admission![arrow-right](https://www.vedantu.com/cdn/images/seo-templates/arrow-right.png)](https://www.vedantu.com/jee-main/cut-off-marks-for-nit) [NEET 2026 Exam Dates: Complete Details on Syllabus, Eligibility, Admit Card, Result, Counselling Process![arrow-right](https://www.vedantu.com/cdn/images/seo-templates/arrow-right.png)](https://www.vedantu.com/neet) [BSc Nursing Application Form 2026: Exam Dates, Eligibility Criteria![arrow-right](https://www.vedantu.com/cdn/images/seo-templates/arrow-right.png)](https://www.vedantu.com/neet/bsc-nursing-form) [NEET 2026 Exam Date (OUT)![arrow-right](https://www.vedantu.com/cdn/images/seo-templates/arrow-right.png)](https://www.vedantu.com/neet/neet-exam-dates) - 1 - 2 [![]()![](https://marketing-assets.vedantu.com/vmkt/image/png/97d562be-e456-4e8a-b58b-d1bdecb3e256-1770019394248-4102638261224925.png)](https://www.vedantu.com/bookademo?&utm_source=mau_crm&utm_medium=seoSideBanner&utm_campaign=pagelanding&utm_term=11neet_bod&couponCode=) \<h2\>Simple Harmonic Motion and Spring Constant: Understanding Hooke's Law\</h2\> \<p\>\<strong\>Summary:\</strong\> Simple Harmonic Motion (SHM) is fundamental in oscillatory motion, particularly in springs. Springs possess unique spring constants that define their stiffness. Hooke's law explains SHM through the formula F = -kx, where F is restoring force, k is spring constant, and x is displacement. The spring constant represents stiffness - stiffer springs have higher constants. According to elasticity theory, springs extend proportionally when loaded within elastic limits. When compressed, springs exert opposite force to return to original size. Spring constant units are Newton per meter (N/m), with dimensional formula \[MT⁻²\]. Practical applications include car suspension systems where designers calculate required spring constants based on weight distribution and compression distances.\</p\> \<h3\>Questions/Concepts Covered:\</h3\> \<ul\> \<li\>How to calculate spring constant using Hooke's law formula and displacement measurements?\</li\> \<li\>What is the relationship between spring stiffness and spring constant in SHM applications?\</li\> \<li\>How to determine spring displacement when force and spring constant are given?\</li\> \</ul\> \<h3\>Keywords:\</h3\> \<ul\> \<li\>Simple harmonic motion spring constant\</li\> \<li\>Hooke's law formula calculation\</li\> \<li\>Spring constant dimensional formula\</li\> \<li\>Newton per meter spring units\</li\> \<li\>Spring displacement force relationship\</li\> \</ul\> Study material NCERT - [NCERT](https://www.vedantu.com/ncert?utm_source=SEO&utm_medium=Footer&utm_campaign=SEO&utm_content=Study_Materials) - [NCERT Solutions](https://www.vedantu.com/ncert-solutions?utm_source=SEO&utm_medium=Footer&utm_campaign=SEO&utm_content=Study_Materials) - 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Simple Harmonious motion i.e. SHM is a veritably intriguing type of stir. It's constantly applied in the oscillatory motion of the objects. Springs generally have SHM. Springs have their own native “ spring constants'' which define how stiff they are. Hooke's law is a notorious law that explains the SHM and gives a formula for the force applied using spring constant. For example, ask if a group of car designers will knock on your door and help develop a suspension system. "Of course," you say. They say the car weighs 1,000 kilograms and requires the use of four shock absorbers, each 0.5 meters long. How strong should the feathers be? Assuming these shock absorbers use springs, each must support a weight of at least 250 kilograms: F = mg = (250 kg) (9.8 m / s2) = 2,450 N Where F is the force, m is the mass of the object, g is the gravitational acceleration, 9.8 meters / sec2. The spring in the shock absorber will, at a minimum, have to give you 2,450 newtons of force at the maximum compression of 0.5 meters. What does this mean the spring constant should be? In order to figure out how to calculate the spring constant, we must remember what Hooke\`s law says: F = –kx Now, we need to rework the equation so that we are calculating for the missing metric, which is the spring constant, or k. The springs used in the shock absorbers must have spring constants of at least 4,900 newtons per meter. ### Spring Constant Units Spring constant definition is related to simple harmonic motions and Hooke's law. So, before we try to define the spring constant and understand the workings of the spring constant, we need to look at Hooke's law. According to the theory of elasticity, when a load is applied to spring it will naturally extend proportionally, as long as the load applied is less than the elastic limit. Now we know that when force is applied to an object, it tends to deform in some way. Consider a spring, when we apply force on one side of the spring, it will get compressed, as they are elastic. At this time the spring exerts its force in the direction opposite to the applied force, to expand to its original size. Therefore, to define the spring constant, we first define Hooke's law. Hooke's law is defined as the force required by the spring to revert to its size is directly proportional to the distance of the compression of the spring. (Image to be added soon) The image shows the movement of the spring when force is applied to one side. ### Spring Constant Definition To understand the spring constant definition, we will look at Hooke’s law formula. Hooke’s law formula is also known as the spring constant formula. The formula is given below. | | | |---|---| | Formula | F = -kx | | Spring Constant Units | Nm\-1 | Where F represents the restoring force of the spring, x is the displacement of the spring, and k is known as the spring constant. The spring constant units are given as Newton per meter. Now that we know that k is the spring constant, we will look at the spring constant definition. We define spring constant as the stiffness of the spring. In other words, when the displacement of the spring is one unit, we can define the spring constant as the force applied to cause that said displacement. Therefore, it is clear to say that, the stiffer the spring is, the higher will be its spring constant. ### Spring Constant Dimensional Formula According to Hooke’s law, we know that, F= -kx Therefore, k= -F/x Now we know that the [unit of force](https://www.vedantu.com/physics/unit-of-force) is given as Newton (N), or as kg m/s2. Therefore, we can write the dimensional unit as \[MLT\-2\]. We also know the dimensional unit of x is given as L Applying spring constant formula, we get, \[ M L T − 2 L \] k= - \[MT\-2\] The spring constant unit is in terms of Newton per meter (N/m). ### Solved Problems **Question 1)** A spring is stretched by 40cm when a load of 5kg is added to it. Find the spring constant. **Answer 1)** Given, Mass m = 5kg, Displacement x = 40cm = 0.4m To find the spring constant, we first need to find the force that is acting on the spring. We know that F = m \* x Therefore, F = 5 \* 0.4 F = 2N The load applies a force of 2N on the spring. Hence, the spring will apply an equal and opposite force of – 2N. Now, by substituting the values in the spring constant formula we get, k = -F/x k = − 2 0\.4 k = 5 N/m Therefore, the spring constant of the spring is 5N/m **Question 2)** Consider a spring with a spring constant of 14000N/m. A force of 3500N is applied to the spring. What will be the displacement of the spring? **Answer 2)** Given, Force F = 3500N, Spring constant k = 14000N/m, We can calculate the displacement of the spring by using the spring constant formula. x= − F k The load applies a force of 3500N on the spring. Hence the spring will apply an equal and opposite force of – 3500N. Thus, x = − ( − 3500 14000 ) x = 0.25 m x = 25 cm Therefore, the spring is displaced by a distance of 25cm. ### Conclusion This is how the spring constant is determined and calculated. The formula is defined using the terms to determine the various physical quantities. Understand how this constant is utilized in solving different problems from the given examples.

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