The Driven Harmonic Oscillator

A symbol-by-symbol guide with a concrete worked example

The Equation

m (d²x/dt²) + γ (dx/dt) + k x = F₀ cos( ω t)

Each colored symbol is explained below

What each symbol means

Think of the equation as a tug-of-war between forces. Each symbol names one player.

Lowercase m

Mass

How heavy the object is. A heavier object is harder to move — it resists acceleration. The bigger m is, the more sluggish the system feels.

SI unit: kg (kilograms)

Greek letter gamma

Damping coefficient

How much the system is "slowed down" by friction or resistance (like air drag or a shock absorber). Larger γ = more energy lost per second.

SI unit: kg/s

Lowercase k

Spring constant (stiffness)

How stiff the restoring force is. Think of a spring: pull it further, and it pulls back harder. k dictates how much harder per meter of stretch.

SI unit: N/m (Newtons per meter)

Lowercase x

Displacement

How far the object is from its resting position right now. It changes over time — it's the thing we're ultimately trying to figure out.

SI unit: m (meters)

F₀

F-naught (F subscript zero)

Driving force amplitude

The maximum strength of the external push/pull applied to the system. The little subscript 0 (naught) just means "the peak value" — the force oscillates between +F₀ and −F₀.

SI unit: N (Newtons)

Greek letter omega

Driving frequency

How fast the external force oscillates back and forth. Measured in radians per second. When ω matches the system's natural frequency, oscillations become more prominent.

SI unit: rad/s

Reading the derivative notation

The terms d²x/dt² and dx/dt are "calculus shorthand" for rates of change.

Leibniz notation decoder

dx/dt Velocity — how fast x is changing at this instant. "The rate of change of position with respect to time."

d²x/dt² Acceleration — how fast the velocity is changing. The "²" means we've applied the rate-of-change idea twice (position → velocity → acceleration).

cos(ωt) A cosine wave oscillating at frequency ω. At t = 0 it equals 1 (maximum). It smoothly cycles between +1 and −1 forever.

What each grouped term represents physically

The equation has three terms on the left (forces acting on the object) and one on the right (the driving force). They must balance — that's what "=" means here.

m · (d²x/dt²)

Inertial force — the object's resistance to being accelerated (Newton's F = ma).

Analogy: how hard it is to get a bowling ball moving.

γ · (dx/dt)

Damping force — opposes motion, proportional to speed. Faster motion = more drag.

Analogy: running through water vs. running in air.

k · x

Restoring force — pulls the object back toward center. Bigger displacement = stronger pull-back.

Analogy: stretching a rubber band further makes it snap back harder.

F₀ · cos(ωt)

Driving force — external push/pull cycling at frequency ω. This is what "drives" the oscillation.

Analogy: pushing a child on a swing at a regular rhythm.

Worked example — a car suspension

Imagine a car wheel hitting regularly spaced bumps in the road. We'll check whether the suspension forces balance at one specific instant in time.

Our values: m = 300 kg | γ = 1500 kg/s | k = 30 000 N/m | F₀ = 800 N | ω = 10 rad/s
Snapshot at t = 0.1 s: x = 0.02 m | dx/dt = 0.15 m/s | d²x/dt² = −6.5 m/s²

Calculate the inertial term — m(d²x/dt²)

Multiply mass by acceleration. The negative sign on acceleration means the object is currently decelerating (being slowed by the spring pulling it back).

m × (d²x/dt²)
= 300 × (−6.5)
= −1950 N

💡 This is just F = ma from Newton's second law. A negative value means the net push is back toward the center.

Calculate the damping term — γ(dx/dt)

Multiply the damping coefficient by velocity. The shock absorber resists whatever direction the wheel is currently moving.

γ × (dx/dt)
= 1500 × 0.15
= +225 N

💡 Positive because the wheel is moving away from center (+0.15 m/s), so damping opposes that — but it appears as a positive contribution on the left side of the equation.

Calculate the spring restoring term — kx

Multiply stiffness by current displacement. The wheel is 2 cm out from its rest position, so the spring is pulling it back with this force.

k × x
= 30 000 × 0.02
= +600 N

💡 The further the displacement, the bigger this term. This is what makes the system oscillate — it always tries to return to zero.

Sum the three left-hand side terms

Add up all three force contributions. This is the total "net force accounting" on the left side.

m(d²x/dt²) + γ(dx/dt) + kx
= (−1950) + 225 + 600
= −1125 N

Calculate the driving force — F₀cos(ωt)

Plug in t = 0.1 s. The angle fed into cosine must be in radians — ω × t gives us radians directly.

F₀ × cos(ω × t)
= 800 × cos(10 × 0.1)
= 800 × cos(1.0 rad)
= 800 × 0.5403
= +432 N ← cos(1 rad) ≈ 0.5403

💡 The road bump is currently pushing with about 432 N — not its full 800 N strength because the cosine wave is part way through its cycle at t = 0.1 s.

Check the balance

In a perfect solution to the equation, left = right exactly. Our snapshot values were chosen to give a close (but illustrative) result. Let's see:

LHS = −1125 N
RHS = +432 N

→ The gap tells us our snapshot values aren't the exact solution — but that's fine! The equation is a recipe: given m, γ, k, F₀, ω, find x(t) so both sides match at every instant.

💡 Solving the equation means finding the function x(t) that makes both sides equal for all values of t — not just one moment. That's what the calculus machinery does.

In plain English, the equation says:

"The object's resistance to acceleration,
plus the drag slowing it down,
plus the spring pulling it back,
must always equal the external driving force."

Every term is a force (in Newtons). The equation demands they balance at every instant in time.