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Spring combinations look like a formula to memorise, but exam diagrams rarely look like the textbook ones. Two springs on opposite sides of a block look like series but act in parallel. A spring cut in half is not "half as strong". This post builds both rules from Hooke's law, so you can decide from the physics rather than the picture, then works four problems.
Is this in the syllabus?
The JEE (Main) 2026 syllabus lists "oscillations of a spring: restoring force and force constant" and, under work and energy, "the potential energy of a spring". The CBSE Physics curriculum for 2026-27 includes oscillations of a loaded spring and, among Class 11 practicals, finding "the force constant of a helical spring by plotting a graph between load and extension". Check your exam year's syllabus.
Why do parallel springs add?
Hooke's law says a spring pulls back with F = −kx. Two springs are in parallel when they always have the same extension x. Then each pulls back on its own, and the forces add:
F = −k1x − k2x = −(k1 + k2)x, so k = k1 + k2.
The NCERT Class 11 chapter Oscillations has a famous case in Example 13.6: a block between two identical springs fixed to walls on either side. Move the block x to the right and one spring stretches by x while the other compresses by x. Both push the block back, so NCERT gets F = −2kx and T = 2π√(m/2k). The springs are on opposite sides, but they share the same displacement, so they are in parallel.
Why do series springs add as reciprocals?
Springs are in series when they are joined end to end and carry the same force F. Each stretches by its own amount, and the extensions add:
x = x1 + x2 = F/k1 + F/k2, so 1/k = 1/k1 + 1/k2.
For two springs this is k = k1k2/(k1 + k2). The combination is always softer than the softest spring on its own.
What happens when you cut a spring?
Think of a spring of length L as many short pieces in series. A piece of length L/n has 1/n of the extension under the same force, so its constant is n times larger. In general, k × length stays constant for pieces of the same spring.
Worked example 1: cutting and rejoining
A spring of constant 600 N/m is cut into two pieces whose lengths are in the ratio 1 : 2. Find the constant of each piece, and of the two pieces joined in parallel.
The short piece is L/3 long, so k1 = 3 × 600 = 1800 N/m.
The long piece is 2L/3 long, so k2 = (3/2) × 600 = 900 N/m.
In parallel: k = 1800 + 900 = 2700 N/m.
Check: joined in series, 1/k = 1/1800 + 1/900 = 3/1800, so k = 600 N/m, the original spring. That check catches most arithmetic slips.
Worked example 2: a mixed combination holding a load
Two springs of 200 N/m each hang side by side from a ceiling, joined at the bottom by a light bar. A third spring of 400 N/m hangs from the bar, and a 0.80 kg mass hangs from it. Take g = 9.8 m/s². Find the total stretch and the period of small vertical oscillations.
The two 200 N/m springs share the same extension: parallel, 200 + 200 = 400 N/m.
That pair and the 400 N/m spring carry the same force: series, 1/k = 1/400 + 1/400, so k = 200 N/m.
Load = mg = 0.80 × 9.8 = 7.84 N. Total stretch = 7.84/200 = 0.0392 m = 3.92 cm.
Split: each part carries 7.84 N, so the pair stretches 7.84/400 = 1.96 cm and the lower spring 7.84/400 = 1.96 cm. They add to 3.92 cm.
T = 2π√(m/k) = 2π√(0.80/200) ≈ 0.40 s.
Worked example 3: a block between two different springs
A 0.75 kg block on a smooth floor sits between two springs, 120 N/m on the left and 180 N/m on the right, both attached to walls. Find the period.
As in NCERT's Example 13.6, both springs share the block's displacement, so they are in parallel: k = 120 + 180 = 300 N/m.
ω = √(300/0.75) = √400 = 20 rad/s.
T = 2π/20 = π/10 ≈ 0.31 s.
Worked example 4: which spring stores more energy?
In series, the force is the same, so the energy in each spring is U = F²/(2k). The softer spring stores more. In parallel, the extension is the same, so U = ½kx², and the stiffer spring stores more. Students often apply one rule to both cases. For energy problems with a block launched by a spring, see our post on spring-block energy and maximum speed.
Which mistakes cost marks?
Judging by the picture. Use the same-extension or same-force test.
Adding series constants. Series always gives a smaller k.
Thinking a cut spring is weaker. Each piece is stiffer.
Forgetting gravity does not change the period. A vertical spring has a shifted equilibrium, not a different T.
Missing a light bar or pulley. Ask what each spring actually carries.
Our post on simple harmonic motion problems shows how to prove SHM for less obvious systems, and the story of Robert Hooke and the law of springs gives the history.
A practice routine
Day 1: Derive both rules from F = −kx without notes.
Day 2: Three cutting problems with different ratios, each checked by rejoining in series.
Day 3: Two mixed combinations with loads; find total stretch and how it splits.
Day 4: Two block-between-walls problems, including one with three springs.
Day 5: Mix five problems from springs, pendulums and energy, and test yourself without notes. Self-testing and spacing practice over days are the two techniques rated most useful in a review of study methods by Dunlosky and colleagues (2013).
If you keep choosing the wrong combination rule, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can compare your same-force or same-extension call with a full solution.
Pick one spring diagram from your textbook tonight and write, before any algebra, whether each pair shares force or extension.
