The Equation of SHM

From a = −ω²x alone, a little calculus gives the complete story: where the particle is, how fast it moves, at every moment.

Speed at each position#

Write $a = \dfrac{dv}{dt} = \dfrac{dv}{dx}\dfrac{dx}{dt} = v\dfrac{dv}{dx}$. Then $v\,dv = -\omega^2x\,dx$. Integrate from the starting state (x₀, v₀) to a general (x, v):

Position at each moment#

Now $\dfrac{dx}{\sqrt{A^2 - x^2}} = \omega\,dt$. Integrating, $\sin^{-1}(x/A) = \omega t + \delta$, where δ is a constant fixed by the starting position. So:

The vocabulary of SHM#

Amplitude A

Since sin lies between −1 and +1, x lies between −A and +A: A is the greatest displacement from the centre.

Time period T

One complete oscillation brings the particle back to the same position moving the same way (Q → P → Q, or O → P → Q → O). Both sin and cos repeat when the angle grows by 2π, so $\omega T = 2\pi$:

Frequency ν and angular frequency ω

The frequency is the number of oscillations per second, $\nu = 1/T = \dfrac{\omega}{2\pi} = \dfrac1{2\pi}\sqrt{\dfrac km}$ (12.9, 12.10), measured in hertz (Hz). ω itself is the angular frequency, in rad/s.

Phase and phase constant

The angle $\phi = \omega t + \delta$ is the phase. It tells you the state of the motion: φ = 0 means passing the centre moving in the +x direction; φ = π/2 means at +A, momentarily at rest; φ = π, passing the centre moving in the −x direction; φ = 3π/2, at −A. Adding 2π brings you back to the same state.

δ, the phase constant, depends only on which instant you call t = 0. Start the clock as the particle passes the centre going forward and δ = 0, so x = A sin ωt; start it at the positive extreme and δ = π/2, so x = A cos ωt. The sine and cosine forms describe the same motion with different δ.