Simple Harmonic Motion

Acceleration that points back

A particle moves along a line, and x is its displacement from a fixed center O at time t. Its velocity is $\frac{dx}{dt}$dx/dt and its acceleration is $\frac{d^{2}x}{d{t}^2}$d²x/dt². These are often written ẋ and ẍ, with one dot for each derivative with respect to time.

In simple harmonic motion the acceleration is proportional to the displacement and points back toward O: $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$d²x/dt² = −ω²x, where $\omega$ω is a positive constant. When x is positive the acceleration is negative, and when x is negative it is positive. The pull is always toward the center, and it is stronger the farther away the particle is.

Rearranged, $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$d²x/dt² + ω²x = 0. This is a second-order equation with no $\frac{dx}{dt}$dx/dt term, and its auxiliary equation is $m^2+{\omega }^2=0$m² + ω² = 0.

Roots $\pm \omega i$±ωi

From $m^2=-{\omega }^2$m² = −ω², the roots are $m=\pm \omega i$m = ±ωi: a complex pair with real part p = 0 and imaginary part $q=\omega$q = ω. So $x=e^{0t}\left(A\cos \omega t+B\sin \omega t\right)$x = e^(0t)(A cos ωt + B sin ωt), which is $x=A\cos \omega t+B\sin \omega t$x = A cos ωt + B sin ωt.

The real part 0 means there is no shrinking or growing factor, so the swings keep the same size forever. Check by substitution: differentiating twice gives $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}A\cos \omega t-{\omega }^{2}B\sin \omega t=-{\omega }^{2}x$d²x/dt² = −ω²A cos ωt − ω²B sin ωt = −ω²x.

The starting state fixes A and B. With $\omega =2$ω = 2, a particle released from rest at x = 2 has x = 2 when t = 0, so A = 2. Its velocity is $\frac{dx}{dt}=-2A\sin 2t+2B\cos 2t$dx/dt = −2A sin 2t + 2B cos 2t, which is 2B at t = 0, so B = 0 and x = 2 cos 2t.

x = 2 cos 2t from t = 0 to $2\pi$2π. The particle swings between 2 and −2, and it is back at x = 2 at $t=\pi$t = π and again at $t=2\pi$t = 2π: every swing has the same height and takes the same time.

The harmonic form

A cosine and a sine of the same $\omega t$ωt add up to a single wave. Expanding the cosine of a difference, $R\cos \left(\omega t-\alpha \right)=R\cos \alpha \cos \omega t+R\sin \alpha \sin \omega t$R cos(ωt − α) = R cos α cos ωt + R sin α sin ωt. This equals $A\cos \omega t+B\sin \omega t$A cos ωt + B sin ωt when $R\cos \alpha =A$R cos α = A and $R\sin \alpha =B$R sin α = B.

Squaring and adding gives $R^2=A^2+B^2$R² = A² + B², so $R=\sqrt{A^2+B^2}$R = √(A² + B²). Dividing gives $\tan \alpha =\frac{B}{A}$tan α = B/A. R is the amplitude, the greatest distance from the center.

For x = 3 cos 2t + 4 sin 2t, $R=\sqrt{9+16}=5$R = √(9 + 16) = 5 and $\alpha ={\tan }^{-1}\left(\frac{4}{3}\right)\approx 0.927$α = tan⁻¹(4/3) ≈ 0.927, so x = 5 cos(2t − 0.927). The particle swings 5 either side of O, not 3 + 4 = 7: the cosine and the sine reach their peaks at different times.

Period and speed

A cosine repeats when its angle grows by $2\pi$2π. The angle $\omega t$ωt grows by $2\pi$2π in the time T where $\omega T=2\pi$ωT = 2π, so the period is $T=\frac{2\pi }{\omega }$T = 2π/ω. With $\omega =2$ω = 2, $T=\pi \approx 3.14$T = π ≈ 3.14.

R does not appear in $T=\frac{2\pi }{\omega }$T = 2π/ω, so a larger amplitude takes exactly as long. A particle pulled twice as far out is pulled back twice as hard, so it moves twice as fast and covers twice the distance in the same time.

The velocity is $\frac{dx}{dt}=-R\omega \sin \left(\omega t-\alpha \right)$dx/dt = −Rω sin(ωt − α), so the greatest speed is $R\omega$Rω, reached as the particle passes through O. For x = 5 cos(2t − 0.927) that is 5 × 2 = 10.

Since ${\sin }^2+{\cos }^2=1$sin² + cos² = 1, ${\left(\frac{dx}{dt}\right)}^2+{\omega }^2{x}^2=R^2{\omega }^2$(dx/dt)² + ω²x² = R²ω² at every moment. With v for the velocity, that is $v^2={\omega }^2\left(R^2-x^2\right)$v² = ω²(R² − x²). At x = 3 the speed is $\sqrt{4\times \left(25-9\right)}=\sqrt{64}=8$√(4 × (25 − 9)) = √64 = 8.

The motion $\frac{d^{2}x}{d{t}^2}=-4x$d²x/dt² = −4x, so $\omega =2$ω = 2, drawn as velocity v against displacement x. Released from rest at x = 1, the state goes round an ellipse that reaches $x=\pm 1$x = ±1 and $v=\pm 2$v = ±2, because the greatest speed is $R\omega =2$Rω = 2. Moved out to R = 2, the ellipse doubles in both directions and the quantity $\frac{1}{2}{v}^2+2x^2$½v² + 2x² goes from 2 to 8, while the period stays $\pi$π.

The usual mistakes

Taking ${\omega }^2$ω² for $\omega$ω. In $\frac{d^{2}x}{d{t}^2}=-9x$d²x/dt² = −9x the coefficient 9 is ${\omega }^2$ω², so $\omega =3$ω = 3 and the period is $\frac{2\pi }{3}$2π/3, not $\frac{2\pi }{9}$2π/9.

Adding the coefficients for the amplitude. 5 cos 2t + 12 sin 2t has amplitude $\sqrt{25+144}=13$√(25 + 144) = 13, not 17.

Expecting a wider swing to take longer. The period $\frac{2\pi }{\omega }$2π/ω contains no amplitude.

Working in degrees. The derivative of $\cos \omega t$cos ωt is $-\omega \sin \omega t$−ω sin ωt only when $\omega t$ωt is in radians.

An inductor and a capacitor

In the application below, the charge q on a capacitor moves in simple harmonic motion: its second derivative with respect to time is −q, with time m in milliseconds, so $\omega =1$ω = 1 and the period is $2\pi \approx 6.28$2π ≈ 6.28 milliseconds. It writes the equation as two first-order equations and finds eigenvalues $\pm i$±i. These are the roots of the auxiliary equation ${\lambda }^2+1=0$λ² + 1 = 0, written with $\lambda$λ because there m counts milliseconds.