Spring oscillations: series vs parallel stiffness
Hang a mass from two expendable springs — first in series as one 'double spring', then in parallel — and time the vertical oscillation period in each. Repeat across five masses and plot √T2 against √T1: the straight line through the origin reveals how series and parallel springs trade off stiffness.
WHAT YOU'LL LEARN
- ✓Why springs in series are floppier and springs in parallel are stiffer, even though they're the same two springs
- ✓Linearising T = 2π√(m/k) by comparing √T2 against √T1 instead of T2 against T1
- ✓Reading a genuinely different restoring force (Hooke's law) from one that merely looks similar (a pendulum's timed oscillation)
ON YOUR BENCH
- □Two retort stands and bosses
- □Four expendable springs (k ≈ 25 N/m each); two pre-connected as a 'double spring'
- □Longer wooden rod (for the series/double-spring configuration) and a shorter notched rod (for parallel)
- □100 g mass hanger and slotted masses (10 g, 50 g, 100 g)
- □Stopwatch, 180° protractor, metre rule
The protocol, step by step
The same guide is printed and waiting at your bench.
- 01
Set up the double spring (series) on the longer rod
Slide the pre-connected double spring onto the longer wooden rod, held roughly 55 cm above the bench by both stands. Hang 270 g from it — this is m for your first set.
- 02
Time T1
Pull the mass down a short, consistent distance and release cleanly downward — no sideways push, or you'll get a conical wobble instead of a clean vertical oscillation. Time enough oscillations to get T1 to two decimal places.
- 03
Reconfigure to parallel and time T2
Swap to the shorter rod with two notches, and hang the same two single springs from it side by side, both taking the same mass m in parallel. Repeat the displace-and-release timing to get T2. Since two springs in parallel are four times stiffer than the same two in series, T2 should come out close to half of T1.
- 04
Vary the mass and repeat
Choose four more masses, none below 200 g, and repeat both timings for each — five sets of (m, T1, T2) in total, spread widely enough that your smallest is ≤ 220 g and your largest is ≥ 350 g.
- 05
Plot √T2 against √T1
Take √T1 and √T2 for each set and plot one against the other. T ∝ √m for a fixed k, so both quantities scale the same way with mass — the result is a straight line through the origin, with gradient P = √(k_series/k_parallel) and (ideally) zero intercept Q.
What you should see
- →T2 consistently shorter than T1 — for equal springs, T2/T1 should sit close to 0.5 across every mass you try, not just the first.
- →A straight line through the origin on the √T2–√T1 plot. T2/T1 = √(k_series/k_parallel) = √(12.5/50) = 0.5 at every mass, so the gradient of √T2 against √T1 is √(T2/T1) = √0.5 ≈ 0.707 — the paper's own accepted gradient depends on the real springs' k, which may differ slightly from the nominal 25 N/m used here.
- →A non-zero intercept usually means the series and parallel timings weren't taken on the same mass, or one of the two configurations was disturbed sideways rather than released cleanly vertically.
| m / g | 200 | 270 | 320 | 380 | 450 |
|---|---|---|---|---|---|
| T1 / s | 0.79 | 0.92 | 1.01 | 1.10 | 1.19 |
| T2 / s | 0.40 | 0.46 | 0.50 | 0.55 | 0.60 |
⚠ BEFORE YOU START
- •Secure both stands to the bench — a loaded spring assembly can pull a light, unsecured stand over.
- •Keep fingers clear when hooking or unhooking a stretched spring; it can snap back.
- •Don't overload a single expendable spring — stay within the slotted-mass range given, or it will deform permanently and stop obeying Hooke's law.
Try it, then run it for real.
Practise the whole thing on the virtual bench, then book real lab time and do it with your own hands.