Displacement, Velocity & Acceleration in SHM

Start from the solution the equation of motion supplies, x(t) = A cos(ωt + φ), and differentiate with the chain rule. Each derivative acts on the cosine and brings one factor of ω out of its argument: v = dx/dt = −ωA sin(ωt + φ), then a = dv/dt = −ω²A cos(ωt + φ). Comparing the last result with x itself gives a = −ω²x, the equation of motion recovered, now serving as a check rather than a starting point. The powers of ω double as a units check: A is in m, ωA in m s⁻¹ and ω²A in m s⁻². They also say how the peaks scale. An oscillation with A = 0.025 m and ω = 8.0 rad s⁻¹ tops out at 0.20 m s⁻¹ and 1.6 m s⁻². Double ω at the same amplitude and the peak speed doubles to 0.40 m s⁻¹, but the peak acceleration quadruples to 6.4 m s⁻², which is why small, fast vibrations are the ones that shake parts loose.

A minus sign and a sine hide the timing, so rewrite both results as shifted cosines. Because −sin θ = cos(θ + π/2) and −cos θ = cos(θ + π), v = ωA cos(ωt + φ + π/2) and a = ω²A cos(ωt + φ + π). To lead is to reach each stage of the cycle sooner: velocity hits its positive maximum a quarter cycle before displacement reaches x = +A, and acceleration, half a cycle away, is the mirror image of x. Convert with Δt = (Δφ/2π)T. For T = 2.0 s, the v = +ωA peak comes 0.50 s before the x = +A peak. On a graph, beware the peak that comes after. The next positive v peak is 1.50 s after the x peak, while the moment of greatest speed 0.50 s after it has v = −ωA. The magnitudes match there, but a phase lead is defined by the signed curves, and a lead of T/4 is not a lag of T/4.

Picture a point moving round a circle of radius A at angular speed ω, its angle from the +x axis equal to the phase ωt + φ. Its projection on the x axis is x(t). Its velocity is a vector of length ωA drawn 90° ahead of the radius, and its acceleration is a vector of length ω²A pointing at the centre, 180° ahead; their x projections are v(t) and a(t). Now the signs follow from the quadrant of the phase. With the phase between 0 and π/2, x > 0, v < 0 and a < 0: the body is on the positive side heading inward, and because v and a share a sign it is speeding up. Between π/2 and π, x < 0, v < 0 and a > 0: still moving in −x, but now a opposes v and the body slows. The pattern is general. A body in SHM speeds up exactly when it moves toward equilibrium and slows down exactly when it moves away.

Square the two time functions and add them, and the time drops out: (x/A)² + (v/ωA)² = cos²(ωt + φ) + sin²(ωt + φ) = 1, so v = ±ω√(A² − x²), a relation the next topic rederives from energy. Speed is greatest, ωA, at x = 0 and zero at x = ±A, while |a| = ω²|x| does the reverse, so the two peaks are always a quarter period apart and never coincide. The same functions show the body lingering at the ends. At x = A/2 the speed is still √3/2 of its maximum, about 87%. Starting from equilibrium, the body reaches x = A/2 when ωt = arcsin(½) = π/6, which takes T/12; the remaining stretch from A/2 to A takes T/4 − T/12 = T/6, twice as long for the same distance. A strobe photograph of an oscillating glider shows exactly this: images crowded near the turning points and spread thin through the middle.

Suppose all you know is ω = 2.5 rad s⁻¹ and, at t = 0, x₀ = −0.040 m and v₀ = +0.075 m s⁻¹. Setting t = 0 in the time functions gives x₀ = A cos φ and v₀ = −ωA sin φ, two equations in two unknowns. Square and add for the amplitude: A = √(x₀² + (v₀/ω)²) = √(0.040² + 0.030²) m = 0.050 m. For the phase use both ratios: cos φ = x₀/A = −0.80 and sin φ = −v₀/(ωA) = −0.60, so φ lies in the third quadrant and φ = −2.50 rad. A calculator asked for arctan(−v₀/(ωx₀)) = arctan(0.75) returns +0.64 rad, which describes a body at +0.040 m moving at −0.075 m s⁻¹: every sign reversed, half a cycle out. With A, ω and φ in hand, x(t), v(t) and a(t) are known at every time, forward and backward.

A snapshot is not the only route in. Two peak readings fix the scale with no clock at all: ω = aₘₐₓ/vₘₐₓ and A = vₘₐₓ²/aₘₐₓ, so a sensor reporting 0.30 m s⁻¹ and 1.2 m s⁻² implies ω = 4.0 rad s⁻¹ and A = 0.075 m. A displacement graph gives T from two successive upward zero crossings and φ from the first maximum: if x first reaches +A at time tₚ, then ωtₚ + φ = 0, so φ = −ωtₚ. A reconstruction is a model, and a model is worth having only if it could fail. Once A, ω and φ are fixed, every further reading is a prediction: the ratio a/x must equal −ω² at every position, and the speed at any x must be ω√(A² − x²). A ratio that drifts with position means the restoring force is not linear, and peaks that shrink cycle by cycle mean damping; either way the sinusoid was only an approximation.