A 2kg mass on a spring (k=200 N/m). What is the period?

T = 2π√(2/200) = 2π√(0.01) = 0.628 s.

How does doubling the mass affect the period?

T ∝ √m → T increases by factor √2 ≈ 1.41.

Why does a driven spring with low damping eventually break at resonance?

At resonance, energy input equals zero energy loss → amplitude grows without bound.

Pro 🔒~20 min

Masses and Springs

Explore oscillation, damping, and resonance

How it works

A mass on a spring oscillates with period T = 2π√(m/k). Unlike a pendulum, this period is independent of amplitude. Damping reduces amplitude over time; critical damping returns the system to equilibrium fastest without oscillating. Resonance occurs when a driving force matches the natural frequency — amplitude grows dramatically and is limited only by damping.

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Step-by-step

Hang a mass on the spring and release it.
Measure the period using the stopwatch.
Change mass or spring constant to verify T = 2π√(m/k).
Add damping to observe decay.
Enable a driving force and tune the frequency to find resonance.

Key formulas

$T = 2\pi\sqrt{\frac{m}{k}}$ Period of mass-spring oscillation
$f_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ Natural frequency
$x(t) = A e^{-\gamma t}\cos(\omega t + \phi)$ Damped oscillation

Frequently asked questions

A 2kg mass on a spring (k=200 N/m). What is the period?: T = 2π√(2/200) = 2π√(0.01) = 0.628 s.
How does doubling the mass affect the period?: T ∝ √m → T increases by factor √2 ≈ 1.41.
Why does a driven spring with low damping eventually break at resonance?: At resonance, energy input equals zero energy loss → amplitude grows without bound.