The two constants of an undamped swing are fixed by where the mass starts and how fast it is going. A physicist reads the same motion differently: as one cosine with a height, the amplitude A, and a shift in time, the phase φ. The height squared is an exact fraction, A² = x₀² + v₀²/ω₀², and it is not the starting displacement unless the mass starts at rest.
Core ideas
Three ideas, one per symbol: what the constants are in terms of the start, what the amplitude is in terms of the constants, and what the phase records.
Put t = 0 in x = C₁cos(ω₀t) + C₂sin(ω₀t) and x(0) = C₁, so C₁ = x₀. The velocity is x′ = −ω₀C₁sin(ω₀t) + ω₀C₂cos(ω₀t), so x′(0) = ω₀·C₂ and C₂ = v₀/ω₀. The division by ω₀ is easy to forget and it is the reason the answer to the amplitude question is not x₀² + v₀².
Write C₁ = A·cos(φ) and C₂ = A·sin(φ). Squaring and adding, and using cos(φ)² + sin(φ)² = 1, gives C₁² + C₂² = A². So A² = x₀² + v₀²/ω₀², an exact fraction when the start and ω₀² are fractions. The amplitude A is a square root of it and is a surd or an integer.
Dividing the two equations gives tan(φ) = C₂/C₁, and the quadrant of φ is the quadrant of the point (C₁, C₂). The angle is rarely a rational multiple of π, so the lab prints it with ≈. It says when the cosine peaks: at t = φ/ω₀, so the motion lags a pure cos(ω₀t) by that amount of time.
The material
The identity behind the conversion, the exact formula for the height, three starts that show the different cases, and the one start where the height is the starting displacement.
The previous lesson ended with x = C₁cos(ω₀t) + C₂sin(ω₀t). Nothing is wrong with it, but it hides the two things you would want to know about the swing: how far the mass gets from rest, and when it gets there. A single cosine shows both.
A motion x = A·cos(ω₀t − φ) with A ≥ 0 has amplitude A, the largest value of x, and phase φ, the angle that shifts the cosine to the right by φ/ω₀ units of time.
Its period is still 2π/ω₀: neither A nor φ enters it.
The two forms are the same function, by the subtraction identity of trigonometry, cos(θ − φ) = cos(θ)·cos(φ) + sin(θ)·sin(φ), which this path takes as given and does not derive, as it takes the rates of sine and cosine. Applied to θ = ω₀t:
If x″ + ω₀²·x = 0 with x(0) = x₀ and x′(0) = v₀, then the swing is A·cos(ω₀t − φ) with A² = x₀² + v₀²/ω₀².
The constants are C₁ = x₀ and C₂ = v₀/ω₀, from putting t = 0 into the solution and its derivative.
Squaring C₁ = A·cos(φ) and C₂ = A·sin(φ) and adding gives C₁² + C₂² = A²·(cos(φ)² + sin(φ)²) = A². So A² = x₀² + (v₀/ω₀)² = x₀² + v₀²/ω₀².
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Take m = 1 and k = 4, so ω₀ = 2 and ω₀² = 4. The same spring, started three ways, has three amplitudes and one period.
For x(0) = 3 and x′(0) = 8: A² = 25 and A = 5. The tangent of the phase is C₂/C₁ = 4/3, so φ is the angle in the first quadrant with that tangent, about 0.927295, and rounded. The mass starts at 3 and reaches 5.
The amplitude 5 is larger than the starting displacement 3 because the mass is already moving outward when it is released.
For x(0) = 1 and x′(0) = 1: A² = 5/4 is exact. Its root A = √5/2 is exact too, and the lab prints it with its rounded value, about 1.11803, beside it. The phase has tan(φ) = 1/2.
The exact quantity to carry is A². The root and the angle are the two things in this lesson that need ≈.
The start (2, 0) is the only one of the three where the amplitude equals the starting displacement, and the reason is v₀ = 0: a mass released from rest is at its largest displacement at the moment of release, so A = |x₀| and φ = 0 when x₀ > 0. Whenever the mass has a velocity at t = 0 it travels further than where it began.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Five moves. The third is where the division by ω₀ is made.
ω₀² = k/m. Keep the fraction; take the root only if you need it for C₂. For m = 1 and k = 4 it is ω₀ = 2.
The two numbers are x₀ = x(0) and v₀ = x′(0), the displacement and the velocity, in that order. A start such as (3, 8) means displacement 3 and velocity 8.
C₁ = x₀ and C₂ = v₀/ω₀. This is the step that divides the velocity by the frequency. For (3, 8) with ω₀ = 2 the constants are 3 and 4.
A² = C₁² + C₂², exactly. For (3, 8) it is 9 + 16 = 25. Only now take a root for A, as an integer, a surd, or a rounded number marked ≈.
tan(φ) = C₂/C₁, in the quadrant of (C₁, C₂). Its value is rounded. If C₂ = 0 and C₁ > 0 the phase is 0 exactly.
The first preset: m = 1, k = 4, x(0) = 3, x′(0) = 8.
Check the amplitude from the original form: 3cos(2t) + 4sin(2t) is largest when cos(2t) = 3/5 and sin(2t) = 4/5, and then it is 9/5 + 16/5 = 5. The check uses only fractions, as it should, because A² is an exact fraction; only the angle is rounded.
You should be able to compute C₁ = x₀ and C₂ = v₀/ω₀, compute A² = C₁² + C₂² exactly, give A as an integer, a surd or a rounded number marked ≈, read the phase from tan(φ) = C₂/C₁ as a rounded number, and say why the period does not change with the start.
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