Waves & Oscillationsx = A cos(ωt + φ)

Simple Harmonic Motion Simulator

Explore spring-mass oscillations with adjustable mass, spring constant, and damping. See displacement graph update in real time.

— x(t)- - envelope

Parameters

Amplitude A₀

Mass m

Spring constant k

N/m

Damping b

N·s/m

Predicted

ω₀6.32 rad/s

Period0.993 s

ωd6.32 rad/s

Time0.00 s

x0.000 m

v0.000 m/s

ω₀0.00 rad/s

Period0.000 s

Energy0.0000 J

About Simple Harmonic Motion

Simple Harmonic Motion (SHM) occurs whenever a restoring force is proportional to displacement from equilibrium. The classic example is a mass on a spring: the spring pulls or pushes the mass back toward the centre with force F = −kx.

ℹAny system where the net restoring force is proportional to displacement exhibits SHM — springs, pendulums (small angles), LC circuits, and many molecular vibrations.

Key Variables

Symbol	Name	Unit	Description
k	Spring Constant	N/m	Stiffness of the spring
m	Mass	kg	Mass of the oscillating object
b	Damping Coefficient	N·s/m	Resistance force per unit velocity
A	Amplitude	m	Initial displacement from equilibrium
ω₀	Natural Frequency	rad/s	√(k/m) — frequency without damping
ωd	Damped Frequency	rad/s	√(ω₀² − (b/2m)²)
T	Period	s	Time for one complete oscillation
γ	Damping Ratio	s⁻¹	b/m — controls decay rate

Worked Example

A 0.5 kg mass on a spring with k = 20 N/m, released from x₀ = 1.0 m at rest, with no damping:

ω_{0} = \frac{k}{m} = \frac{20}{0.5} = 40 \approx 6.32 rad/s

T = \frac{2 π}{ω _{0}} = \frac{2 π}{6.32} \approx 0.994 s

x (t) = 1.0 cos (6.32 t) m

Effect of Damping

Underdamped (b < 2√(km)): oscillation with decaying amplitude — what the simulator shows.
Critically damped (b = 2√(km)): fastest return to equilibrium without oscillating.
Overdamped (b > 2√(km)): slow exponential return, no oscillation.

💡Set b = 0 for pure SHM. Increase b to see the amplitude decay. The graph shows the envelope curve e^(−γt/2) as a dashed line.

Energy in SHM

Total mechanical energy (undamped) is conserved and equals the initial potential energy:

E = \frac{1}{2} k A^{2} = \frac{1}{2} k x^{2} + \frac{1}{2} m v^{2}

With damping, energy is dissipated as heat. The total energy decays as E(t) = ½kA² e^(−γt).

Key Formulas

Equation of Motion

m \overset{x}{¨} + b \overset{x}{˙} + k x = 0

Displacement (Underdamped)

x (t) = A_{0} e^{- γ t /2} cos (ω_{d} t + φ)

Natural and Damped Frequencies

ω_{0} = \frac{k}{m} ω_{d} = ω_{0}^{2} - (\frac{γ}{2})^{2} γ = \frac{b}{m}

Period and Frequency

T = \frac{2 π}{ω _{0}} f = \frac{1}{T} = \frac{ω _{0}}{2 π}

Energy

E = \frac{1}{2} k x^{2} + \frac{1}{2} m v^{2} = \frac{1}{2} k A^{2}

Formula	Description	Notes
ω₀ = √(k/m)	Natural angular frequency	Increases with k, decreases with m
T = 2π√(m/k)	Period	Independent of amplitude (for small oscillations)
x(t) = A cos(ω₀t)	Undamped displacement	Phase φ depends on initial conditions
v(t) = −Aω₀ sin(ω₀t)	Velocity	90° phase shift from displacement
x(t) = A e^(−γt/2) cos(ωdt)	Damped displacement	Amplitude decays exponentially
E = ½kA²	Total energy	Conserved only when b = 0

ℹThe period T = 2π√(m/k) is independent of amplitude — a heavier mass oscillates more slowly, a stiffer spring oscillates more quickly.

Frequently Asked Questions

Why does a heavier mass oscillate more slowly?

The period T = 2π√(m/k) increases with mass. A heavier mass has more inertia, so it accelerates less for the same restoring force, taking longer to complete each cycle.

Does amplitude affect the period of a spring-mass system?

No. Unlike a pendulum at large angles, a spring-mass system follows Hooke's Law exactly, so the period T = 2π√(m/k) is independent of amplitude.

What is critical damping?

Critical damping (b = 2√(km)) gives the fastest return to equilibrium without oscillating. Used in car suspensions and door closers where you want quick settling without ringing.

What happens to energy with damping?

Energy is dissipated as heat through the damping force. Total mechanical energy decays as E(t) = ½kA² · e^(−γt) where γ = b/m.

What is the difference between ω₀ and ωd?

ω₀ = √(k/m) is the natural frequency (no damping). ωd = √(ω₀² − (b/2m)²) is the damped frequency, which is slightly lower. With no damping they are equal.