The Hooke’s Law Formula, expressed as F = −kx, describes the restoring force exerted by a spring when it is stretched or compressed from its natural length. This fundamental principle appears in NCERT Class 11 Physics (Chapter 9 — Mechanical Properties of Solids) and forms the backbone of topics like Simple Harmonic Motion (SHM), elasticity, and wave mechanics. For JEE Main, JEE Advanced, and NEET aspirants, Hooke’s Law is a high-yield topic that connects spring systems, stress-strain relationships, and oscillatory motion. This article covers the formula, derivation, complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Hooke’s Law Formulas at a Glance
Quick reference for the most important spring and elasticity formulas.
- Hooke’s Law (restoring force): \( F = -kx \)
- Spring constant from force and displacement: \( k = \frac{F}{x} \)
- Elastic potential energy stored in spring: \( U = \frac{1}{2}kx^2 \)
- Time period of spring-mass system: \( T = 2\pi\sqrt{\frac{m}{k}} \)
- Springs in series: \( \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \)
- Springs in parallel: \( k_{eff} = k_1 + k_2 \)
- Young’s Modulus (stress/strain): \( Y = \frac{FL}{A\Delta L} \)
What is Hooke’s Law?
Hooke’s Law Formula is a foundational principle in classical mechanics, first stated by English physicist Robert Hooke in 1676. It governs the elastic behaviour of solid materials under small deformations. The law states that the restoring force produced in an elastic body is directly proportional to the displacement from its equilibrium position, provided the elastic limit is not exceeded.
In NCERT Class 11 Physics, Chapter 9 (Mechanical Properties of Solids), Hooke’s Law is introduced in the context of stress and strain. The law is also revisited in Chapter 14 (Oscillations) when deriving the time period of a spring-mass system undergoing Simple Harmonic Motion.
The negative sign in the formula indicates that the restoring force always acts in the direction opposite to the displacement. If you stretch a spring to the right, the spring pulls back to the left. This opposing nature is what makes the motion oscillatory. The proportionality constant k is called the spring constant or force constant. It measures the stiffness of the spring. A higher value of k means a stiffer spring that requires more force to deform by the same amount.
Hooke’s Law is valid only within the elastic limit of the material. Beyond this limit, the material undergoes plastic deformation and no longer returns to its original shape.
Hooke’s Law Formula — Expression and Variables
The standard mathematical form of Hooke’s Law is:
\[ F = -kx \]
Here, the negative sign confirms that the restoring force opposes the direction of displacement. When used to calculate the magnitude of force applied (not the restoring force), the formula is written as:
\[ F = kx \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| F | Restoring force (or applied force) | Newton (N) |
| k | Spring constant (force constant) | Newton per metre (N/m) |
| x | Displacement from equilibrium position | Metre (m) |
| U | Elastic potential energy stored in spring | Joule (J) |
| m | Mass attached to spring | Kilogram (kg) |
| T | Time period of oscillation | Second (s) |
Derivation of Hooke’s Law
Consider a spring of natural length L fixed at one end. A force F is applied at the other end, producing an extension x.
Step 1: From experimental observation, the extension x is proportional to the applied force F. So we write: \( F \propto x \).
Step 2: Introducing a proportionality constant k: \( F = kx \).
Step 3: By Newton’s Third Law, the spring exerts an equal and opposite restoring force on the object. Therefore: \( F_{restoring} = -kx \).
Step 4: The elastic potential energy stored is found by integrating the force over displacement: \( U = \int_0^x kx\, dx = \frac{1}{2}kx^2 \).
This derivation confirms that the spring stores energy as it deforms. This energy is fully recoverable within the elastic limit.
Complete Elasticity & Springs Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Hooke’s Law (restoring force) | \( F = -kx \) | F = force, k = spring constant, x = displacement | N | Class 11, Ch 9 & 14 |
| Spring Constant | \( k = \frac{F}{x} \) | F = applied force, x = extension/compression | N/m | Class 11, Ch 9 |
| Elastic Potential Energy | \( U = \frac{1}{2}kx^2 \) | k = spring constant, x = displacement | J | Class 11, Ch 6 & 14 |
| Time Period (spring-mass) | \( T = 2\pi\sqrt{\frac{m}{k}} \) | m = mass, k = spring constant | s | Class 11, Ch 14 |
| Angular Frequency (SHM) | \( \omega = \sqrt{\frac{k}{m}} \) | k = spring constant, m = mass | rad/s | Class 11, Ch 14 |
| Springs in Series | \( \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \) | k1, k2 = individual spring constants | N/m | Class 11, Ch 14 |
| Springs in Parallel | \( k_{eff} = k_1 + k_2 \) | k1, k2 = individual spring constants | N/m | Class 11, Ch 14 |
| Young’s Modulus | \( Y = \frac{F \cdot L}{A \cdot \Delta L} \) | F = force, L = original length, A = area, ΔL = change in length | N/m² (Pa) | Class 11, Ch 9 |
| Stress | \( \sigma = \frac{F}{A} \) | F = force, A = cross-sectional area | N/m² | Class 11, Ch 9 |
| Strain | \( \epsilon = \frac{\Delta L}{L} \) | ΔL = change in length, L = original length | Dimensionless | Class 11, Ch 9 |
| Frequency of Oscillation | \( f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \) | k = spring constant, m = mass | Hz | Class 11, Ch 14 |
Hooke’s Law Formula — Solved Examples
Example 1 (Class 9-10 Level — Direct Application)
Problem: A spring has a spring constant of 200 N/m. A force of 50 N is applied to stretch it. Find the extension produced in the spring.
Given: k = 200 N/m, F = 50 N, x = ?
Step 1: Write Hooke’s Law: \( F = kx \)
Step 2: Rearrange for x: \( x = \frac{F}{k} \)
Step 3: Substitute the values: \( x = \frac{50}{200} = 0.25 \) m
Answer
The extension produced in the spring is 0.25 m (25 cm).
Example 2 (Class 11-12 Level — Elastic Potential Energy)
Problem: A spring of spring constant 500 N/m is compressed by 8 cm from its natural length. Calculate (a) the restoring force and (b) the elastic potential energy stored in the spring.
Given: k = 500 N/m, x = 8 cm = 0.08 m
Step 1: Write the restoring force formula: \( F = -kx \)
Step 2: Calculate the magnitude of the restoring force: \( |F| = 500 \times 0.08 = 40 \) N
Step 3: The restoring force acts opposite to compression, so it acts in the direction of extension: \( F = 40 \) N (directed away from the compressed end).
Step 4: Write the elastic potential energy formula: \( U = \frac{1}{2}kx^2 \)
Step 5: Substitute values: \( U = \frac{1}{2} \times 500 \times (0.08)^2 = \frac{1}{2} \times 500 \times 0.0064 = 1.6 \) J
Answer
(a) Restoring force = 40 N (directed opposite to compression). (b) Elastic potential energy stored = 1.6 J.
Example 3 (JEE/NEET Level — Spring Combination and SHM)
Problem: Two springs with spring constants k1 = 300 N/m and k2 = 600 N/m are connected in series. A block of mass 2 kg is attached to the free end. Find (a) the effective spring constant, (b) the time period of oscillation, and (c) the maximum displacement if the block is given an initial velocity of 3 m/s at the equilibrium position.
Given: k1 = 300 N/m, k2 = 600 N/m, m = 2 kg, vmax = 3 m/s
Step 1: For springs in series, use: \( \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{300} + \frac{1}{600} = \frac{2+1}{600} = \frac{3}{600} = \frac{1}{200} \)
Step 2: Therefore: \( k_{eff} = 200 \) N/m
Step 3: Time period of oscillation: \( T = 2\pi\sqrt{\frac{m}{k_{eff}}} = 2\pi\sqrt{\frac{2}{200}} = 2\pi\sqrt{0.01} = 2\pi \times 0.1 = 0.628 \) s (approximately)
Step 4: At equilibrium, all energy is kinetic: \( KE = \frac{1}{2}mv_{max}^2 = \frac{1}{2} \times 2 \times 9 = 9 \) J
Step 5: At maximum displacement, all energy is potential: \( U = \frac{1}{2}k_{eff}x_{max}^2 = 9 \) J
Step 6: Solve for xmax: \( x_{max}^2 = \frac{2 \times 9}{200} = 0.09 \Rightarrow x_{max} = 0.3 \) m
Answer
(a) Effective spring constant = 200 N/m. (b) Time period = 0.628 s (≈ 0.2π s). (c) Maximum displacement = 0.3 m (30 cm).
CBSE Exam Tips 2025-26
- Always include the negative sign when writing the restoring force formula \( F = -kx \) in theory questions. CBSE examiners deduct marks for omitting it without explanation.
- State the elastic limit when answering “What is Hooke’s Law?” questions. Mentioning that the law is valid only within the elastic limit earns extra marks in 2025-26 board exams.
- Convert units carefully. Always convert cm to m before substituting into \( F = kx \) or \( U = \frac{1}{2}kx^2 \). A unit error can cost you the entire step mark.
- Draw the F-x graph. CBSE frequently asks for a graph showing force vs. displacement for a spring. This is a straight line through the origin with slope equal to k. We recommend practising this graph for 3-mark questions.
- Link to Young’s Modulus. In 5-mark questions, CBSE may ask you to connect Hooke’s Law to stress-strain curves. Practise explaining how the linear region of the stress-strain graph represents Hooke’s Law.
- Memorise spring combination formulas. Series and parallel spring questions appear almost every year in CBSE Class 11 exams and unit tests. Our experts suggest practising at least five combination problems before your exam.
Common Mistakes to Avoid
- Forgetting the negative sign: Many students write \( F = kx \) instead of \( F = -kx \) for the restoring force. The negative sign is physically meaningful — it indicates opposition to displacement. In theory answers, always include it.
- Using incorrect units for k: The spring constant k has SI units of N/m, not N/cm or N/mm. Always express x in metres before calculating k or F.
- Applying Hooke’s Law beyond the elastic limit: Students sometimes apply \( F = kx \) for very large deformations. Remember, Hooke’s Law is valid only within the elastic limit. Beyond it, the stress-strain relationship becomes non-linear.
- Confusing series and parallel formulas: Springs in series have a smaller effective k (like resistors in parallel). Springs in parallel have a larger effective k. A common error is swapping these two formulas. Use the analogy: series springs are “weaker together.”
- Ignoring the direction of restoring force: In SHM problems, students sometimes assign the wrong direction to the restoring force. Always remember — the restoring force acts towards the equilibrium position, opposing displacement.
JEE/NEET Application of Hooke’s Law Formula
In our experience, JEE aspirants encounter Hooke’s Law Formula in at least 2–3 questions per paper, spread across mechanics and oscillations. NEET aspirants see it primarily in the context of elasticity and biomechanics. Here are the key application patterns:
Pattern 1: Spring Combination Problems (JEE Main)
JEE Main frequently tests the effective spring constant when multiple springs are combined in series, parallel, or mixed configurations. The key insight is that springs in series share the same force but have different extensions, while springs in parallel share the same extension but have different forces. Mastering \( k_{eff} \) calculations is essential. Questions often involve a block on a frictionless surface connected to two springs and ask for the time period of oscillation.
Pattern 2: Energy Methods in SHM (JEE Advanced)
JEE Advanced uses Hooke’s Law to set up energy conservation equations in spring systems. A typical question gives the initial velocity at equilibrium and asks for the amplitude, or gives the amplitude and asks for the velocity at a specific displacement. The energy equation \( \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2 \) (where A is amplitude) is the workhorse formula here. Our experts suggest practising at least 10 energy-based SHM problems for JEE Advanced preparation.
Pattern 3: Stress-Strain and Young’s Modulus (NEET & JEE)
Both NEET and JEE connect Hooke’s Law to the macroscopic behaviour of materials through Young’s Modulus. Questions ask students to identify the elastic limit on a stress-strain graph, calculate elongation of a wire under load using \( Y = \frac{FL}{A\Delta L} \), or compare stiffness of two materials. Understanding that Young’s Modulus is essentially a material-level spring constant is the key conceptual link. In our experience, NEET questions in this area are straightforward substitution problems, while JEE questions involve comparing two or more materials or wires.
For additional practice with related oscillation concepts, explore our article on the Angular Speed Formula and the Average Acceleration Formula, which are closely linked to SHM analysis.
FAQs on Hooke’s Law Formula
Explore More Physics Formulas
Strengthen your understanding of related concepts with these comprehensive guides on ncertbooks.net. For oscillatory motion, study the Angular Speed Formula, which connects directly to the angular frequency in spring-mass SHM systems. To build your kinematics foundation, visit our detailed article on the Average Acceleration Formula. For optics and wave physics, our guide on the Critical Angle Formula is an excellent next step. All these formulas are part of our comprehensive Physics Formulas hub, which covers every important formula from Class 9 to Class 12 and competitive exams. For official NCERT textbook references, visit ncert.nic.in.