SHM Equations
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
Any object pulled back by a force proportional to its displacement, F = −kx, obeys d²x/dt² = −ω²x. The solution is x = A cos(ωt + φ): A is the amplitude, ω = √(k/m) is the angular frequency, and the phase φ is set by where the motion starts.
1. From force to motion
2. Reading the equations
Whenever you can write a = −(something)x, that something is ω², and the motion is SHM with T = 2π/ω.
Maximum speed vmax = Aω at x = 0; maximum acceleration amax = Aω² at x = ±A. So ω = amax/vmax.
Finding φ: plug in t = 0. x(0) = A cos φ and v(0) = −Aω sin φ. Starting at +A gives φ = 0; at x = 0 moving −x, φ = π/2; at x = 0 moving +x, φ = −π/2.
Forces like F = −kx³ aren’t SHM: the restoring force must be linear in x.
3. Example, then try it
An object moves with x(t) = 0.20 cos(5t) (SI units). Find its angular frequency, period, maximum speed and maximum acceleration.
ω = 5 rad/s, so T = 2π/5 ≈ 1.26 s.
vmax = Aω = 1.0 m/s; amax = Aω² = 5.0 m/s².
An object obeys d²x/dt² = −25x (SI units). At t = 0 it is at x = 0 moving at 0.50 m/s in the −x direction. Find ω, the period, the amplitude, and φ, and write x(t).
ω: ω² = 25, so ω = 5.0 rad/s and T = 2π/5.0 ≈ 1.26 s.
Amplitude: it passes x = 0 at top speed, so A = vmax/ω = 0.50 ÷ 5.0 = 0.10 m.
Phase: x(0) = A cos φ = 0 → φ = ±π/2. v(0) = −Aω sin φ = −0.50 → sin φ = +1, so φ = π/2.
Result: x(t) = 0.10 cos(5t + π/2), which is the same as −0.10 sin(5t).
Answer: ω = 5.0 rad/s, T ≈ 1.26 s, A = 0.10 m, φ = π/2.
4. Common mistakes
An oscillator has a maximum speed of 2.0 m/s and a maximum acceleration of 8.0 m/s². What is its angular frequency?
amax/vmax = Aω²/(Aω) = ω = 8.0 ÷ 2.0 = 4.0 rad/s.
Ready to practice?
5 questions on SHM Equations, each with a hint and an explanation for every wrong answer.
