The SHM equation and its general solution

Start from Newton's second law with the one force this section cares about, a restoring force proportional to displacement: m d²x/dt² = −kx. Divide by m and the physics of the particular apparatus disappears: d²x/dt² = −(k/m)x. Check the coefficient dimensionally. A stiffness in N m⁻¹ is kg s⁻², so k/m has units s⁻², a reciprocal time squared, and its positive square root is an angular frequency. Name it ω, and the equation becomes d²x/dt² = −ω²x. Three words describe it, and each one earns its place later. Linear: x and its derivatives appear only to the first power. Homogeneous: every term contains x, so x = 0 is a solution and equilibrium stays equilibrium. Constant coefficient: ω does not depend on t or on x. For a 0.20 kg glider on a spring of stiffness 45 N m⁻¹, ω = √(45/0.20) = 15 rad s⁻¹, so the coefficient in the equation is 225 s⁻² whatever the glider is later asked to do.

A solution is a claim you can test in two lines. Take x(t) = A cos(ωt + φ). Differentiating once gives dx/dt = −Aω sin(ωt + φ), and again gives d²x/dt² = −Aω² cos(ωt + φ), which is −ω² times the original x. The equation is satisfied identically — for every value of A and every value of φ, with no condition imposed on either. That is the first important result, and it is a negative one: the differential equation determines ω and nothing else. Notice also that x(t) = A sin(ωt + φ′) passes the same test, because it is the same family with φ′ = φ + π/2; the choice between cosine and sine is bookkeeping about where you start the clock, not physics. What the substitution cannot tell you is whether there are other solutions hiding elsewhere, and that question is settled by counting.

A second-order equation needs two integrations, so its general solution carries exactly two arbitrary constants. A cos(ωt + φ) already has two, so the family is complete: there is nothing else to find. The two numbers that select one member are the state at one instant, conventionally t = 0. Setting t = 0 gives x₀ = A cos φ and v₀ = −Aω sin φ. Square and add, using cos² + sin² = 1, and A drops out of the trigonometry: A = √(x₀² + (v₀/ω)²). Divide the second by the first and A cancels instead: tan φ = −v₀/(ω x₀). Take ω = 12 rad s⁻¹, x₀ = 0.040 m and v₀ = −0.60 m s⁻¹. Then v₀/ω = −0.050 m, so A = √(0.0016 + 0.0025) = 0.0640 m and tan φ = 1.25, giving φ = 0.896 rad. Check it: 0.0640 cos(0.896) = 0.0400 m and −0.0640 × 12 × sin(0.896) = −0.600 m s⁻¹, both restored. The arctangent alone is ambiguous by π, so settle the quadrant with the signs of cos φ = x₀/A and sin φ = −v₀/(ωA).

Expand the cosine of a sum: A cos(ωt + φ) = (A cos φ) cos ωt − (A sin φ) sin ωt. So x(t) = C cos ωt + D sin ωt is the same general solution wearing different constants, with C = A cos φ and D = −A sin φ, and going back the other way A = √(C² + D²). Fitted to initial conditions this form is even quicker: at t = 0 it gives x₀ = C directly, and dx/dt = −Cω sin ωt + Dω cos ωt gives v₀ = Dω, so C = x₀ and D = v₀/ω with no trigonometry at all. Its advantage is that it is linear in C and D, which is what makes least-squares fitting to data and the adding of solutions straightforward. A third dress is worth meeting now because the wave and driven-oscillator topics live in it: write x(t) = Re[ e^(iωt)] with the complex amplitude  = A e^(iφ). One complex number then carries both constants, and adding oscillations becomes adding arrows in the complex plane.

Sometimes the clock is the wrong variable: you know where the body is, not when. Write θ = ωt + φ, so x = A cos θ and v = −Aω sin θ. Then x/A = cos θ and v/(Aω) = −sin θ, and squaring and adding kills θ: (x/A)² + (v/(Aω))² = 1, or v² = ω²(A² − x²). Hence v = ±ω√(A² − x²). Dimensions check: rad s⁻¹ multiplied by a length gives m s⁻¹. Limits check: at x = 0 the speed is its maximum ωA, and at x = ±A it is zero, which is what a turning point means. The ± is not sloppiness — it is the statement that the body passes each position twice per cycle, once outbound and once returning, and only the sign distinguishes them. Test it on the earlier numbers: with A = 0.0640 m and ω = 12 rad s⁻¹, the speed at x = 0.040 m is 12√(0.0041 − 0.0016) = 12 × 0.050 = 0.60 m s⁻¹, the starting speed given, as it must be.

If x₁ and x₂ both satisfy d²x/dt² = −ω²x, then so does a x₁ + b x₂ for any constants a and b, because differentiation is linear and no term in the equation squares x or feeds it to a sine. That licence is what makes the two-constant family closed, and it is why adding two oscillations of the same ω returns an oscillation of the same ω with a new amplitude and phase — amplitudes combining as phasors, not as numbers: 0.030 m and 0.040 m a quarter cycle apart give 0.050 m, never 0.070 m. The licence is withdrawn the moment the equation stops being linear. A pendulum's restoring torque goes as sin θ, not θ; keep the next term and the period grows with amplitude, T ≈ T₀(1 + θ₀²/16), which is 0.76% at θ₀ = 20° and no longer negligible at 60°. A spring with F = −kx − βx³ behaves the same way: the frequency depends on the amplitude, so the sum of two solutions is not a solution at all.