Velocity, acceleration and phase in SHM

Start from x(t) = A cos(ωt + φ), with A, ω and φ all constants — only t varies. The chain rule gives v(t) = dx/dt = −ωA sin(ωt + φ). The ω out front comes from differentiating the argument, not from the amplitude, which is why a higher-frequency oscillator is faster at the same amplitude. Differentiating once more gives a(t) = dv/dt = −ω²A cos(ωt + φ), and the bracket is x(t) again, so a = −ω²x. That is the equation of motion you started from, recovered — a free check that the algebra survived two derivatives. Check dimensions as well: [ωA] = T⁻¹ × L = L T⁻¹ and [ω²A] = L T⁻², so the ladder converts metres to m s⁻¹ to m s⁻² exactly as it must. With ω = 9.0 rad s⁻¹ and A = 0.12 m this reads v(t) = −1.08 sin(9.0t) m s⁻¹ and a(t) = −9.72 cos(9.0t) m s⁻².

Writing the derivatives as sines hides the pattern; writing them as cosines shows it. Since −sin θ = cos(θ + π/2), the velocity is v = ωA cos(ωt + φ + π/2) and the acceleration is a = ω²A cos(ωt + φ + π). Every differentiation adds π/2 to the phase. So velocity leads displacement by a quarter cycle, acceleration leads velocity by a quarter cycle, and acceleration is a full half cycle — antiphase — from displacement. On a graph that reads as three rules: wherever x is at a peak, v crosses zero; wherever x crosses zero, v is at a peak; and a is the mirror image of x through the time axis. A quarter cycle is T/4 = π/(2ω) in seconds, so the shift shrinks as the frequency rises — 5.0 ms at 50 Hz, only 0.57 ms at 440 Hz.

Each derivative multiplies the peak by ω, so the three peaks form a ladder: A in m, ωA in m s⁻¹, ω²A in m s⁻². The peak speed ωA is reached at x = 0, where the sine is ±1; the peak acceleration magnitude ω²A is reached at x = ±A, where the cosine is ±1. The two never coincide, and they sit a quarter period apart. Because ω enters squared, acceleration is the quantity a high frequency punishes. A loudspeaker cone of amplitude 0.24 mm at 250 Hz has ω = 2πf = 1.57 × 10³ rad s⁻¹, so its peak speed is a modest ωA = 0.377 m s⁻¹, while its peak acceleration is ω²A = 592 m s⁻², roughly 60 times the acceleration of free fall. That is why voice coils are built light: the inertial load scales as ω², not as the amplitude you can see.

Sometimes the question gives a position, not a time. Divide through: x/A = cos(ωt + φ) and v/(ωA) = −sin(ωt + φ). Square both and add, and the Pythagorean identity kills the time: (x/A)² + (v/ωA)² = 1. That is an ellipse in the phase plane of x against v, traced once per period, and rearranged it reads v = ±ω√(A² − x²). Test the ends: at x = 0 it returns v = ±ωA, and at x = ±A it returns v = 0, both as they should be. With ω = 9.0 rad s⁻¹ and A = 0.12 m, the speed at x = 0.050 m is 9.0 × √(0.0144 − 0.0025) = 9.0 × 0.1091 = 0.982 m s⁻¹. The ± is the price of throwing away the time: the relation knows the speed but not the direction, so it cannot tell the outward half of the cycle from the inward half. Only the time-domain solution, or an extra statement that the body is moving towards the centre, puts the sign back.

Let x₁ = A₁ cos(ωt + φ₁) and x₂ = A₂ cos(ωt + φ₂). Because both carry the same ω, the difference of their arguments, Δφ = φ₂ − φ₁, never changes: it is a property of the pair, not of the instant. Convert it to time by dividing by ω, since phase accumulates at ω radians per second, giving Δt = Δφ/ω = (Δφ/2π)T. Three cases have names — Δφ = 0 is in phase, Δφ = π/2 is quadrature, Δφ = π is antiphase — and note that x and v of one oscillator are in quadrature while x and a are antiphase. The picture is phasors: two arrows of lengths A₁ and A₂ rotating together at ω with the angle Δφ locked between them, each displacement being an arrow's projection on the horizontal axis. Two accelerometers on a bridge deck both read 4.0 Hz, so T = 250 ms, and one peaks 25 ms after the other: that lag is 0.10 of a cycle, so Δφ = 0.10 × 2π = 0.63 rad, or 36°. If the two frequencies differ even slightly, Δφ drifts with time and no single phase difference exists to quote.

Given x and v at one instant you can place the oscillator in its cycle, and a follows for free from a = −ω²x. Suppose x > 0 and v > 0: the body lies between the centre and +A and is still heading outwards, and since a = −ω²x is negative it opposes v, so the speed is falling. Now take x > 0 and v < 0: the body is returning, a is still negative, now along v, so the speed is rising. The rule underneath is worth memorising — speed rises whenever the body moves towards the centre and falls whenever it moves away, in every quarter of every cycle. Two more checks close any answer. Dimensions: ωA must come out in m s⁻¹ and ω²A in m s⁻². And a limiting case: hold A fixed and let ω → 0, whereupon both v and a go to zero, a frozen oscillator parked at a displacement, which is exactly what a vanishing frequency should give.