Playground
Animated spring-mass oscillator with sinusoidal position trace. Period adjusts with mass and k sliders.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Periodoutput Time for one complete oscillation | s | T | 0.01 – 60 | |
| Mass Mass of the oscillating object | kg | M | 0.01 – 100 | |
| Spring Constant Stiffness of the spring | N/m | MT⁻² | 1 – 1000 |
Deep dive
Derivation
From F = -kx and F = ma: ma = -kx → a = -(k/m)x. This is the SHM equation x'' + ω²x = 0 with ω = √(k/m). The period T = 2π/ω = 2π√(m/k).
Experimental verification
Spring-mass lab experiments, quartz crystal oscillator frequency measurements, MEMS resonator calibration.
Common misconceptions
- Period does NOT depend on amplitude (for ideal SHM)
- SHM is not limited to springs — pendulums, LC circuits, and molecular vibrations all exhibit it
- Damping reduces amplitude but barely affects period (for light damping)
Real-world applications
- Quartz watch timekeeping
- Seismometer design
- Vibration isolation in buildings
- Musical instrument tuning
Worked examples
Spring-mass oscillator
Given:
- m:
- 0.5
- k:
- 200
Find: T and f
Solution
T = 2π√(0.5/200) = 2π × 0.05 = 0.314 s → f = 3.18 Hz
Designing a 1 Hz oscillator
Given:
- f_target:
- 1
- m:
- 0.25
Find: k
Solution
k = m(2πf)² = 0.25 × (2π)² = 0.25 × 39.48 = 9.87 N/m
Scenarios
What if…
- scenario:
- What if mass quadruples?
- answer:
- Period doubles (√4 = 2). A 2 kg mass on the same 200 N/m spring: T = 2π√(2/200) = 0.628 s.
- scenario:
- What if you add damping?
- answer:
- Period barely changes for light damping: T_damped = T/√(1 - (b/2mω)²). Amplitude decays exponentially, but frequency is nearly the same.
- scenario:
- What if amplitude doubles?
- answer:
- Period stays exactly the same — this is the key property of SHM. Amplitude independence is what makes pendulum clocks reliable.
Limiting cases
- condition:
- m → 0
- result:
- T → 0
- explanation:
- A massless system oscillates infinitely fast (theoretical limit).
- condition:
- k → ∞
- result:
- T → 0
- explanation:
- An infinitely stiff spring snaps back instantly.
- condition:
- k → 0
- result:
- T → ∞
- explanation:
- No restoring force means the system never returns — period is infinite.
Context
Robert Hooke / Christiaan Huygens · 1673
Huygens analyzed pendulum motion in Horologium Oscillatorium; combined with Hooke's spring law, the SHM period formula emerged.
Hook
A 0.5 kg mass on a spring bounces with k = 200 N/m. How fast does it oscillate?
T = 2π√(m/k) = 2π√(0.5/200) = 2π × 0.05 = 0.314 s. That's about 3.2 oscillations per second — faster than you'd expect!
Dimensions:
- lhs:
- T → [T]
- rhs:
- 2π√(m/k) → √([M]/[MT⁻²]) = √([T²]) = [T]
- check:
- Both sides are [T] = seconds. 2π is dimensionless. ✓
Validity: Valid for ideal springs obeying Hooke's law (small oscillations). Breaks down for large amplitudes, damped systems, or nonlinear restoring forces.