# Springs in Series and Parallel > Springs in series and parallel for JEE Physics: 1/k = 1/k1 + 1/k2 and k = k1 + k2, cutting springs, effective stiffness and oscillation periods. - Canonical URL: https://prepelephant.com/topics/jee/physics/springs-in-series-parallel - 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: "Springs in Series and Parallel", PrepElephant, https://prepelephant.com/topics/jee/physics/springs-in-series-parallel ## Direct answer Cut a spring in half and each piece is twice as stiff — spring constant is inversely proportional to length, the single most quoted spring fact in JEE. When two springs share the same force (end to end, in series), extensions add and 1/k = 1/k1 + 1/k2, so the combination is softer than the softer spring. When they share the same extension (side by side, in parallel), forces add and k = k1 + k2, stiffer than either. The series result extends to n identical springs (k/n) and parallel to nk, and the same bookkeeping governs a mass hung between two springs on opposite walls: both pull in restoring directions, so k_eff = k1 + k2, a parallel arrangement wearing a disguise. ## What you must remember - **Series (same force):** 1/k_s = 1/k1 + 1/k2; the softer spring dominates the extension, and k_s is smaller than the smaller individual k. - **Parallel (same extension):** k_p = k1 + k2; identical springs in parallel give nk, in series k/n. - **Cutting rule:** k ∝ 1/length; a spring of constant k cut into n equal pieces gives each piece constant nk — half a spring is twice as stiff. - **Opposite-wall trap:** a mass connected to springs on both sides oscillates with k_eff = k1 + k2 (parallel), because both springs restore the mass toward centre; a spring only on one side with the mass sliding freely is just k. - **Angular frequency link:** ω = √(k_eff/m) and T = 2π√(m/k_eff); every effective-spring question ends in this substitution. - **Energy sharing:** series springs store energy in ratio 1/k1 : 1/k2 (same force, U = F²/2k); parallel springs share in ratio k1 : k2 (same extension, U = ½kx²). - **Angled spring (Advanced):** a spring at angle θ to the motion direction contributes only k cos²θ to the effective stiffness along that direction. ## Numbers worth knowing Wire k1 = 200 N/m and k2 = 300 N/m. In series: 1/k = 1/200 + 1/300 gives k = 120 N/m — softer than the 200 spring, exactly as the rule promises. In parallel: k = 500 N/m. Hang a 1.2 kg mass on each combination, with g = 10 m/s²: the series period is 2π√(1.2/120) = 2π × 0.1 = 0.63 s, the parallel period 2π√(1.2/500) = 2π × 0.049 = 0.31 s — roughly a factor-two contrast that examiners reproduce with clean numbers. Now the cutting logic, because it carries more marks than the combinations. A full spring of stiffness 60 N/m carries 40 coils; cutting off 10 coils leaves 30, i.e. three-quarters of the length, so the remaining piece has stiffness 60 × 40/30 = 80 N/m. The reasoning, not the arithmetic, is the exam target: halving the length halves the stretch for the same force because each coil twists the same amount and there are fewer coils to share the extension. A fancied Advanced version glues two cut pieces side by side (parallel): piece constants 80 and the other 240 (a quarter-length piece), giving k_eff = 320 N/m, then asks for the new frequency — a three-step chain (cut, combine, oscillate) that separates the prepared from the formula-memorisers. ## What the exam tests The perennial slip is treating the opposite-wall arrangement as series because the mass sits "between" the springs; it is parallel by the same-extension criterion, and the wrong option in the paper is always the series value. Second, cutting questions are answered backwards: a shorter spring is stiffer, and students who memorised without the length-inversion write k/2 where the answer is 2k. Third, in combinations with different springs, energy questions (which spring stores more) test whether you know the invariant of each connection — force in series, extension in parallel — rather than the formula alone. Main keeps to k values and periods; Advanced adds the angled spring with k cos²θ or a spring-plus-pulley system where the effective k must first be found by displacing the mass and computing the restoring force, then inserted into √(k_eff/m). ## Frequently asked questions ### What is conserved across springs connected in series? The force is the same in every series spring (a massless spring chain transmits tension unchanged), while extensions add, giving 1/k = 1/k1 + 1/k2. ### Why does cutting a spring make it stiffer? Spring constant is inversely proportional to length: the same force stretches fewer coils through the same angle each, so total extension drops and k = force/extension rises. ### A mass hangs between two springs on opposite walls — series or parallel? Parallel, with k_eff = k1 + k2, because a displacement of the mass stretches both springs simultaneously (same extension criterion). ### How do n identical springs combine? nk in parallel (forces add at the same stretch) and k/n in series (stretches add at the same force). ### What does a spring at angle θ contribute to stiffness along a given direction? Only k cos²θ, because its force component along the motion is kx cosθ and that component's own projection adds another cosθ factor.