# Spring Constant Variation > Cutting doubles stiffness, series 1/k = 1/k1 + 1/k2, parallel k = k1 + k2, and the m/3 effective mass of a massive spring — JEE Physics. - Canonical URL: https://prepelephant.com/topics/jee/physics/spring-constant-variation - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Spring Constant Variation", PrepElephant, https://prepelephant.com/topics/jee/physics/spring-constant-variation ## Direct answer A spring's stiffness belongs to its length: cutting a spring of constant k into n equal parts leaves each part with constant nk, because the same force must produce the same stress in a shorter wire, hence more extension per unit length. Combinations then follow the resistor grammar — springs end-to-end act in series with 1/k = 1/k1 + 1/k2 (the soft combination), while springs sharing a load side-by-side act in parallel with k = k1 + k2 (the stiff combination). A spring with its own mass m_s oscillating with a load m swings as if massless with an added m_s/3 at the end, giving T = 2π√((m + m_s/3)/k). Every JEE spring-variation question is one of these three transformations applied with clean bookkeeping of which spring hangs where. ## What you must remember - **Length rule:** k ∝ 1/L for a given spring material and coil geometry; cut in half and each piece has 2k, cut into n parts and each has nk; stretching a spring to double length halves its effective k. - **Series (end-to-end, one mass at the far end):** 1/k_eq = 1/k1 + 1/k2; the softer spring dominates, and the extensions add under the same force. - **Parallel (both springs attached to the mass, side by side or both vertical under one pan):** k_eq = k1 + k2; the extensions are equal and the forces add. - **Effective mass:** a spring of mass m_s contributes m_s/3 to the oscillating inertia, T = 2π√((m + m_s/3)/k); the third comes from the linear velocity profile along the spring. - **Cut-and-combine arithmetic:** half of a spring of constant k has 2k; two such halves in series restore k exactly — the self-consistency check that no other rule passes. - **Time-period consequences:** with the same load, halving the spring (k → 2k) changes T to T/√2; a second identical spring in parallel does the same, while series gives T√2. ## Why one third, worked and checked Derive the effective mass once and it never needs re-deriving. When the load at the end moves with speed v, the element of the spring a distance x from the fixed end moves with speed (x/L)v — the spring stretches uniformly. The spring's kinetic energy is then the integral of ½(dm)v_element² = ∫ from 0 to L of ½ (m_s/L dx)(v x/L)² = ½ (m_s/3) v², so as far as oscillation inertia is concerned the spring is a mass m_s/3 riding at the end. The period becomes T = 2π√((m + m_s/3)/k); for a spring heavy compared with its load (m << m_s), T approaches 2π√(m_s/3k) — measurably slower than the massless-spring textbook formula, which is why precise laboratory pendulum-spring work always applies the correction. Now the transformations on numbers. A spring of constant 100 N/m is cut into four equal parts: each has 400 N/m. Take two parts in series: 1/k = 1/400 + 1/400, k = 200 N/m; in parallel, 800 N/m. Hang a 1 kg load: the original spring oscillates with T = 2π√(1/100) ≈ 0.63 s, the series pair with 2π√(1/200) ≈ 0.44 s, the parallel pair with 2π√(1/800) ≈ 0.22 s — a factor-of-three spread from one spring's dismemberment. Notice the cutting logic's internal check: four quarters back in series must rebuild the original 100 N/m, and 1/k_eq = 4/400 does exactly that. ## Where students slip Geometry decides series versus parallel, not appearance: two springs, one above and one below a mass act in parallel; two springs on opposite walls pulling the same mass also act in parallel for horizontal oscillations; only when the same tension threads both springs in sequence is the combination series. The cutting rule trips students who halve the constant instead of doubling it — anchor with the physical reason (same force, less wire to share the extension, so the piece is stiffer). The m_s/3 result is often quoted as m_s/2 by students integrating carelessly; the 1/3 comes from the squared linear velocity profile. Advanced-level twists include springs at an angle to the motion (effective k multiplied by cos²θ of the tilt) and a spring cut in a given ratio rather than in half — a 2:1 cut of a spring k gives constants 3k and (3/2)k, and the series-recombination check should be run before answering. ## Frequently asked questions ### Why does cutting a spring in half double its spring constant? Stiffness scales inversely with length — the same force stretches half the wire half as far, so the piece as a whole extends half as much, which is the definition of double the constant. ### How do springs combine in series and parallel? Series (one spring's extension feeding the next, common tension) gives 1/k = 1/k1 + 1/k2; parallel (common extension, forces adding) gives k = k1 + k2 — identical grammar to resistors in the opposite arrangement. ### What is the effective mass of an oscillating spring? m_s/3, because points along the spring move with speeds growing linearly from zero at the fixed end, and integrating ½v_element²dm yields one-third of the spring's mass riding at the load. ### How does the time period change when the same spring is cut in half? The half-spring has constant 2k, so with the same load the period falls to T/√2, from T = 2π√(m/k). ### Two springs pull a block from opposite walls — series or parallel? Parallel: the block's displacement x extends or compresses both springs simultaneously, so the restoring force is (k1 + k2)x and the equivalent constant is the sum.