Free oscillations · Differential Equations

Intuition

With no friction and no forcing the characteristic roots are purely imaginary, and the motion is a pure oscillation that never dies: a cosine at the natural frequency, shifted in time. Its amplitude is set by how the motion was started, and its period does not depend on the start at all.

A clock depends on its oscillator keeping the same period whether it swings wide or narrow. For a mass on a spring that is exactly true, and it is what makes an oscillator a clock.

The free oscillation $x=\tfrac32\cos(2t-1.2)$ : its amplitude $R=\tfrac32$ between the dashed lines, and successive peaks one period $\tfrac{2\pi}{\omega_{0}}=\pi$ apart.

Amplitude, phase and period

The solution of $mx''+kx=0$ with $x(0)=x_{0}$ and is , with , amplitude and period . The energy stays constant.

x=R\cos(\omega_{0}t-\delta),\qquad T=\frac{2\pi}{\omega_{0}}

What the numbers are

Released from rest at $x_{0}$ : $x=x_{0}\cos\omega_{0}t$ , with amplitude .

Without friction the energy is conserved

mx''+kx=0\ \Longrightarrow\ \frac{d}{dt}\Bigl(\tfrac12m(x')^{2}+\tfrac12kx^{2}\Bigr)=0

Differentiate the energy along a solution. The chain rule gives m times x prime times x double prime, plus k times x times x prime. Take x prime out: what is left in the bracket is the left-hand side of the equation, which is zero. So the energy never changes, and the oscillation can neither grow nor die.

Proof steps

E=\tfrac12m(x')^{2}+\tfrac12kx^{2}

The energy of the motion plus the energy stored in the spring.

E'=m\,x'x''+k\,x\,x'

Differentiate along the motion by the chain rule.

E'=x'\,(mx''+kx)

Take out the velocity.

E'=0

The bracket is the left-hand side of the equation, which is zero along a solution.

Applications