Recover oscillator phase from both initial conditions

Recover oscillator phase from both initial conditions

Newton’s law for an ideal spring gives m xddot = −kx. For m = 2 kg and k = 8 N/m, the equation is xddot + 4x = 0. Its two independent solutions are cos(2t) and sin(2t), so one initial position alone cannot determine the subsequent motion.

Match x(0) = x₀ and v(0) = v₀ in x = C cos(2t) + D sin(2t). This gives C = x₀ and D = v₀/2. The velocity is −2x₀ sin(2t) + v₀ cos(2t); a second derivative directly verifies the differential equation and both initial conditions.

For x₀ = 0.1 m and v₀ = 0.2 m/s, amplitude is sqrt(0.02) ≈ 0.141421 m and cosine phase is −π/4. At a quarter period, t = π/4 s, displacement is 0.1 m and velocity is −0.2 m/s. The same displacement can therefore occur with a different direction of motion.

Energy is 4x² + v² in joules and remains 4x₀² + v₀² = 0.08 J for this packet. In a cosine representation, phase is atan2(−v₀/2, x₀), which preserves its quadrant. If both initial conditions vanish, the solution is zero everywhere and phase is undefined, not an experimentally determined zero angle.

Changed-case check

Predict before reading the comparison. Change to x₀ = 0 and v₀ = 0.4 m/s. The amplitude becomes 0.2 m; a quarter period later the body is at +0.2 m with zero velocity and total energy 0.16 J.

Retained investigation

Derive both quadratures and evaluate the same solution through position, velocity and conserved energy. Include the zero-amplitude phase boundary. Why does one position measurement generally fail to specify oscillator phase?

Source and boundary

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