Preview

Masses and Springs: Series & Parallel

Q: Two 80 N/m springs are connected in series with 1.0 kg total hanging mass. What is the period?

K_eff = (80×80)/(80+80) = 40 N/m → T = 2π√(1/40) ≈ 0.993 s.

Q: The same two 80 N/m springs are reconnected in parallel. Is the combination stiffer or softer, and by how much?

K_eff = 80 + 80 = 160 N/m — four times the series value, so the period halves.

Q: Why do springs in series combine like resistors in parallel (and vice versa)?

Series springs share force and add stretch (compliances add); parallel springs share stretch and add force (stiffnesses add).

Combine two springs and discover what stiffness really adds up to

Hang one spring from the ceiling and it stretches under the load. Now bolt a second spring underneath it and ask your hands what should happen: two springs working together must be stronger, right? That instinct comes straight from circuits class — series components add — and it is wrong. A series pair shares one force: the full load pulls through both springs, each stretches on its own, and the stretches pile up, so the combination is softer than either spring alone. Flip the same pair side by side and the deal reverses: now they share the stretch, their forces add, and the pair is stiffer than either. Exactly the resistor rules, inverted. This lab makes the inversion impossible to miss because the combined stiffness drives a clock: hang the pair, release, and the panel reports k_eff while the oscillation period answers back — 80 N/m plus 80 N/m reads 40 N/m in series and 160 N/m in parallel, a four-fold swing from one toggle. You will watch two force readouts stay identical every frame while the stretches refuse to match, clock the period across the softest and stiffest presets, and stare down an energy total that never flinches. By the end, 'series' will mean share-the-force in your bones — and the resistor analogy will never sneak your answer again.

What you'll be able to do

Derive the effective spring constant of two-spring combinations from first principles — series springs share one force so their stretches add (1/k_eff = 1/k₁ + 1/k₂, softer than either), parallel springs share one stretch so their forces add (k_eff = k₁ + k₂, stiffer than either) — and compute k_eff for any (k₁, k₂) pair
Use T = 2π√(m/k_eff) to turn combined stiffness into a clockable period, and verify on the panel that two equal 80 N/m springs act as 40 N/m in series (T = 0.993 s at 1.0 kg) but 160 N/m in parallel
Track energy through an undamped oscillation: kinetic and potential trade continuously while the total stays pinned at the release energy E = ½·k_eff·A² (1.80 J at the default state)

Formulas

\frac{1}{k_{series}} = \frac{1}{k_1} + \frac{1}{k_2}

Series springs — softer than either spring

k_{parallel} = k_1 + k_2

Parallel springs — stiffer than either spring

T = 2\pi\sqrt{\frac{m}{k_{eff}}}

Period of the combined oscillator

Make a prediction

Default setup: two identical 80 N/m springs hang in series carrying 1.0 kg total (0.5 kg at the junction, 0.5 kg at the bottom), released from rest 0.3 m below equilibrium. Before you peek at the LIVE DATA panel — what does the k_eff readout show, and what period does T (theory) report?

No grading here — commit to a guess, then scroll down and test it yourself.

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At the default state, read the panel: k_eff = 40.00 N/m — exactly half of either 80 N/m spring — and T (theory) = 0.993 s. Then keep your eyes on the two force rows for a full cycle: d-f1 and d-f2 are identical every frame while the two springs visibly split the stretch. Same force, added stretch — that is the entire series rule.
Flip the Mode toggle to Parallel with the sliders untouched (80/80 N/m, 0.5/0.5 kg): k_eff jumps 40.00 → 160.00 N/m — four times stiffer, not double. Watch what else changed: the pink m₂ block leaves the oscillator (in parallel the engine swings m₁ = 0.5 kg alone), and T drops 0.993 → 0.351 s. Two numbers moved at once; untangle how much each contributed.
Hit the Series Soft preset (20/20 N/m): k_eff = 10.00 N/m and the clock stretches to T = 1.987 s. Now hit Parallel Stiff (200/200 N/m): k_eff = 400.00 N/m, T = 0.314 s. Forty times the stiffness buys a ≈6.3× faster clock — verify 6.3 ≈ √40 straight from T ∝ 1/√k_eff.
Watch the ENERGY panel through any full cycle: the blue PE₁ and green KE bars trade height back and forth while the amber Total bar never twitches — at defaults it sits on 1.80 J (= ½·k_eff·A² = ½×40×0.3², the fixed 0.3 m release). Freeze mid-swing with the Step button: the Total still has not moved.

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Check

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Your prediction

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Answer: k_eff reads 40.00 N/m and T shows 0.993 s — in series both springs carry the full force, each stretches under it, and the stretches add: the pair is softer than either spring alone

The series rule is share-the-force: the same tension runs through both springs, so each one stretches by F/k on its own and the total stretch is the sum — 1/k_eff = 1/80 + 1/80 = 2/80, giving k_eff = 40 N/m, softer than either spring alone. The panel confirms it two ways at once: d-keff reads 40.00 N/m, and the force rows d-f1 and d-f2 are identical every single frame (that equality is the series condition made visible). The period follows from the combination, not the parts: T = 2π√(m_eff/k_eff) = 2π√(1.0/40) = 0.993459 s, so d-tt shows 0.993 s. Option B is the targeted misconception: it imports the series-resistor rule R = R₁ + R₂ into a domain where the roles are inverted — and its numbers are not even random, they are this sim's parallel-mode state (flip Mode and d-keff really does read 160.00 N/m; T then reads 0.351 s because the parallel engine also swings m1 = 0.5 kg alone). Students have carried exactly this intuition into documented research interviews (Clement, Preconceptions in Mechanics, Unit 4). Option C misses that each spring still stretches fully under the shared force: a series pair is longer and softer, never identical — 80 N/m with 1.0 kg would give T = 2π√(1/80) = 0.702 s, which the panel never shows in this configuration.

Quiz (0/3)

Two 80 N/m springs are connected in series with 1.0 kg total hanging mass. What is the period?

The same two 80 N/m springs are reconnected in parallel. Is the combination stiffer or softer, and by how much?

Why do springs in series combine like resistors in parallel (and vice versa)?

You can now

Derive the effective spring constant of two-spring combinations from first principles — series springs share one force so their stretches add (1/k_eff = 1/k₁ + 1/k₂, softer than either), parallel springs share one stretch so their forces add (k_eff = k₁ + k₂, stiffer than either) — and compute k_eff for any (k₁, k₂) pair
Use T = 2π√(m/k_eff) to turn combined stiffness into a clockable period, and verify on the panel that two equal 80 N/m springs act as 40 N/m in series (T = 0.993 s at 1.0 kg) but 160 N/m in parallel
Track energy through an undamped oscillation: kinetic and potential trade continuously while the total stays pinned at the release energy E = ½·k_eff·A² (1.80 J at the default state)