The SHM Differential Equation

Choose the origin where the net force on the block is zero and call the displacement x. Hooke's law gives the spring force as −kx: its size grows in proportion to x, and the minus sign makes it point back toward the origin on either side. Newton's second law turns that into m d²x/dt² = −kx, and dividing by m gives d²x/dt² = −(k/m)x. For a 0.300 kg glider on a 14.7 N m⁻¹ spring, k/m = 49.0 s⁻²; the unit is right, because N m⁻¹ ÷ kg = s⁻². Read the equation as an instruction: at x = +0.020 m the acceleration is −0.98 m s⁻², at x = −0.020 m it is +0.98 m s⁻², and at x = 0 it vanishes. The unknown here is not a number but the entire history x(t), and the equation constrains it at every instant. Three labels describe it. It is second order, since the highest derivative is the second. It is linear, since x and its derivatives appear only to the first power. It is homogeneous, since every term contains x. Each label has a consequence below.

The most direct way to solve an equation like this one is to propose a function and test it. We need a function whose second derivative is a negative multiple of itself. A polynomial cannot do it: t² differentiates twice to the constant 2, never to −49.0t². A real exponential cannot either: x = e^(λt) has second derivative λ²x, and λ² = −49.0 s⁻² has no real solution. That is the minus sign speaking; with a plus sign the exponentials would work and the block would run away. Try x = cos(pt) with p unknown. Then dx/dt = −p sin(pt) and d²x/dt² = −p² cos(pt), so the equation demands −p² cos(pt) = −49.0 cos(pt) at every t. The cosine cancels and leaves p² = 49.0 s⁻², so p = 7.00 rad s⁻¹. The same test passes sin(7.00t) and fails any other positive p. Define ω = √(k/m). Notice what cancelled: had the trial been 5 cos(pt) or 0.01 cos(pt), the factor in front would have dropped out too. The equation fixes how fast the argument of the cosine advances, and says nothing about how big the motion is.

Because the equation is linear and homogeneous, solutions can be scaled and added. If x₁ and x₂ each satisfy d²x/dt² = −ω²x, then for C₁x₁ + C₂x₂ the second derivative is C₁(−ω²x₁) + C₂(−ω²x₂) = −ω²(C₁x₁ + C₂x₂), so the combination satisfies it as well. Taking x₁ = cos ωt and x₂ = sin ωt gives the family x(t) = C₁ cos ωt + C₂ sin ωt. That this family is complete reflects the order of the equation: recovering x from its second derivative takes two integrations, and each leaves one constant undetermined. You have met this before. For constant acceleration, d²x/dt² = a integrates to x = x₀ + v₀t + ½at², and the two constants are the initial position and velocity. The oscillator is the same story with a different right-hand side. Newton's second law fixes accelerations, so a motion is determined only when a position and a velocity at one instant are supplied, and then it is determined uniquely: two oscillators with the same ω, x₀ and v₀ move identically for all time.

Put t = 0 into x = C₁ cos ωt + C₂ sin ωt. The sine vanishes and the cosine is 1, so C₁ = x₀. Differentiate to get v = −ωC₁ sin ωt + ωC₂ cos ωt and put t = 0 again: v₀ = ωC₂, so C₂ = v₀/ω. Two pure cases are worth recognising on sight. A block pulled aside and released from rest has v₀ = 0, so x = x₀ cos ωt. A block struck while it sits at equilibrium has x₀ = 0, so x = (v₀/ω) sin ωt, and the combination v₀/ω is a length: the distance that launch speed carries it. A general start mixes the two. Give the glider with ω = 7.00 rad s⁻¹ the start x₀ = +0.024 m, v₀ = −0.49 m s⁻¹. Then C₁ = 0.024 m and C₂ = −0.49/7.00 = −0.070 m, so x(t) = 0.024 cos(7.00t) − 0.070 sin(7.00t) in metres. Check it against the equation at t = 0: differentiating twice gives a(0) = −49.0 × 0.024 = −1.18 m s⁻², which is exactly −(k/m)x₀.

A sum of a cosine and a sine of the same ω is a single shifted cosine. Expand A cos(ωt + φ) = A cos φ cos ωt − A sin φ sin ωt and match terms: C₁ = A cos φ and C₂ = −A sin φ. Squaring and adding gives A = √(C₁² + C₂²). For the glider, A = √(0.024² + 0.070²) = 0.074 m, about three times the starting displacement, because most of the motion came from the launch speed. Then cos φ = 0.024/0.074 = 0.324 and sin φ = +0.070/0.074 = 0.946, both positive, so φ = 1.24 rad and x(t) = 0.074 cos(7.00t + 1.24). Use both signs. The shortcut tan φ = −v₀/(ωx₀) cannot tell φ from φ + π, which is a block on the other side of equilibrium moving the other way. The phase constant is a statement about timing. The maximum belongs to phase zero, which here occurred at t = −φ/ω = −0.177 s, before the clock started; the next one comes when the phase reaches 2π, at t = (2π − 1.24)/7.00 = 0.720 s. The gap between them is T = 0.898 s.

The mathematics never used the word spring. Any coordinate q obeying d²q/dt² = −(positive constant) × q moves sinusoidally with ω² equal to that constant. For small swings a pendulum of length 0.500 m obeys d²θ/dt² = −(g/L)θ = −19.6θ, so ω = 4.43 rad s⁻¹ without any further work; later topics derive such equations, and this one tells you what to do once you have them. A constant extra force adds a term without x, so the equation stops being homogeneous, but the cure is simple. The constant only moves the equilibrium point, and measuring x from the new equilibrium restores d²x/dt² = −ω²x with the same ω. What does spoil the form is a force whose graph against x is not a straight line, or one that depends on velocity. A drag force adds a term in dx/dt. A spring that stiffens as it stretches adds a term in x³, and then doubling a solution no longer gives a solution, so the period starts to depend on amplitude. That suggests the experimental test of the model: time the oscillation at two quite different amplitudes. Equal periods support F = −kx over that range; a period that drifts with amplitude falsifies it.