Preview
Masses and Springs: Basics
Introductory spring oscillation — no damping or driving
Hang a mass on a spring, pull it down a hand's width, and let go. Now pull it down two. Ask your gut which launch takes longer to settle into its rhythm, and the gut answers confidently: the longer trip. It is wrong, and the reason why is one of the neatest cancellations in physics. Pull the mass twice as far and the spring pulls back twice as hard at every point of the journey — the extra distance is covered at proportionally higher speed, and the clock reads the same period to the millisecond. The formula says it flatly: T = 2π√(m/k). Mass and stiffness are the only two knobs the period has. Amplitude doesn't appear. Gravity doesn't appear either: hang the same oscillator on the Moon and the equilibrium line slides toward the ceiling, but the clock keeps Earth's time — which is why NASA measures astronaut mass on the ISS with an oscillating spring chair that works in weightlessness. In this lab you will stretch the release from 0.10 m to a full metre and watch the period readout refuse to move while the energy total swings a hundredfold; sink the equilibrium line under a 5 kg load, then lift it with Moon gravity, and clock the same 0.993 seconds through all of it. By the end, a spring's period will read to you as a property of the system, not of how hard you started it.
What you'll be able to do
- Explain the hanging-spring equilibrium as the balance point where kxeq = mg (xeq = 0.245 m at the default 1.0 kg / 40 N/m / 9.8 m/s² state), and describe the net force measured from equilibrium as a pure Hooke restoring force, F_net = −k(x − xeq), with gravity only shifting where the oscillation centers — never how fast it runs
- Apply T = 2π√(m/k) to compute and clock the period of any (m, k) state — 0.993 s at the default, 2.221 s under the 5.0 kg Heavy preset, 1.987 s at k = 10 N/m — and demonstrate on the panel that the period is independent of both amplitude (0.993 s from 0.10 m to 1.00 m release) and gravity (Earth vs Moon preset)
- Track energy through an undamped cycle using the equilibrium-centered bookkeeping E = ½mv² + ½k(x − xeq)² = const = ½kA²: KE and PE trade continuously while the Total holds 3.20 J at defaults and quadruples to 12.80 J when the amplitude doubles
Formulas
Make a prediction
Default state: m = 1.0 kg, k = 40 N/m, released from rest 0.40 m below equilibrium — T (theory) reads 0.993 s and the energy Total sits at 3.20 J. Now double the release displacement to 0.80 m (same m, k, and g). What happens to the period and the energy?
No grading here — commit to a guess, then scroll down and test it yourself.
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Your prediction
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Answer: T (theory) stays pinned at 0.993 s while the energy Total jumps to 12.80 J — the period depends only on m and k; doubling the amplitude quadruples the energy (E = ½kA²) without touching the clock
The period of a spring-mass oscillator is set by exactly two quantities: T = 2π√(m/k) = 2π√(1.0/40) = 0.993459 s, and neither amplitude nor g appears in it. Doubling the release displacement does double the road, but the restoring force −k(x−xeq) is proportionally stronger at every point of that road, so the average speed doubles with it — the travel time cancels out to the same 0.993 s, which is what the d-tt row shows no matter where the Displace slider sits (the zero-cross T (measured) row agrees once two crossings complete). What does change is the energy: E = ½kA² = ½×40×0.8² = 12.80 J, four times the default 3.20 J — bigger swings, same clock. Option B is the targeted misconception, among the dominant SHM misreadings documented in PER (Somroob & Wattanakasiwich 2017 found the amplitude–frequency/period confusion near the top of their tested cohorts' SHM misconceptions; OpenStax Univ. Physics Vol 1 §15.1 counters it verbatim: 'the period T and frequency f of a simple harmonic oscillator are independent of amplitude'). Option C reads the force trade in the right direction but misses the cancellation — the stronger pull is exactly what pays for the longer road.
Quiz (0/3)
A 0.5kg mass on a spring (k=50 N/m). What is the period?
Does the period change if you double the initial stretch amplitude?
What would happen to the period on the Moon (g = 1.6 m/s²)?
You can now
- Explain the hanging-spring equilibrium as the balance point where kxeq = mg (xeq = 0.245 m at the default 1.0 kg / 40 N/m / 9.8 m/s² state), and describe the net force measured from equilibrium as a pure Hooke restoring force, F_net = −k(x − xeq), with gravity only shifting where the oscillation centers — never how fast it runs
- Apply T = 2π√(m/k) to compute and clock the period of any (m, k) state — 0.993 s at the default, 2.221 s under the 5.0 kg Heavy preset, 1.987 s at k = 10 N/m — and demonstrate on the panel that the period is independent of both amplitude (0.993 s from 0.10 m to 1.00 m release) and gravity (Earth vs Moon preset)
- Track energy through an undamped cycle using the equilibrium-centered bookkeeping E = ½mv² + ½k(x − xeq)² = const = ½kA²: KE and PE trade continuously while the Total holds 3.20 J at defaults and quadruples to 12.80 J when the amplitude doubles