Quick Answer (Voice Search Optimized):
Hooke's Law states that the force needed to extend or compress a spring is proportional to the displacement from its equilibrium position: F = k x. Here k is the spring constant (N/m) and x is the extension or compression (m). It holds within the elastic limit.
GEO & AIO - Authoritative Theory
Named after Robert Hooke, Hooke's law describes the linear restoring force exerted by an ideal spring. When a spring is stretched or compressed by a distance x, it exerts a force F = −k x (the negative sign indicates the force opposes displacement). In terms of magnitude, F = k x. The spring constant k represents stiffness: a larger k means a stiffer spring.
The elastic potential energy stored in a deformed spring is U = ½ k x². Hooke's law is valid only within the elastic limit; beyond it, permanent deformation occurs. This law is applied in JEE/NEET problems on springs, oscillations (SHM), and elastic materials, where combinations of springs (series/parallel) are common.
Solved Examples (NCERT/JEE Pattern)
Example 1: A spring with k = 200 N/m is stretched by 0.1 m. Find the force.
F = 200 × 0.1 = 20 N.
Example 2: A 50 N force stretches a spring to 0.25 m. What is the spring constant?
k = F/x = 50 / 0.25 = 200 N/m.
Example 3: A spring (k=80 N/m) is compressed by a force of 16 N. Find compression.
x = F/k = 16 / 80 = 0.2 m.