Recipe

Solve an undamped free-vibration IVP

Use when: you have mx''+kx=0 (no damping term at all) together with initial conditions x(0)=x_0, x'(0)=v_0

Goal: the position function x(t) as a fixed-amplitude sinusoid, with C_1,C_2 pinned down by the initial conditions

Taught in Undamped free vibration, natural frequency

1. 1

   Divide k by m and take the square root: \omega_0=\sqrt{k/m}.

   Example

   For m=4 kg, k=25 N/m: \omega_0=\sqrt{25/4}=\sqrt{6.25}=2.5.

   Why: substituting x=e^{rt} into mx''+kx=0 gives mr^2+k=0, so r^2=-k/m (a negative number whenever m,k>0), meaning the roots are the purely imaginary pair r=\pm i\sqrt{k/m}, and \omega_0 is exactly that positive square root.

2. 2

   The general solution is always x(t)=C_1\cos(\omega_0t)+C_2\sin(\omega_0t), a fixed-amplitude wave that never decays, since there's no damping term to drain energy from it.

   Example

   With \omega_0=2.5: x(t)=C_1\cos(2.5t)+C_2\sin(2.5t).

   Why: a purely imaginary conjugate pair r=\pm i\omega_0 is exactly Chapter 7's complex-root case with zero real part, so the decaying envelope e^{\alpha t} collapses to e^{0}=1, leaving only the oscillating cosine/sine part.

3. 3

   Substitute t=0 into x(t); since \cos(0)=1 and \sin(0)=0, this immediately isolates C_1.

   Example

   x(0)=C_1\cos(0)+C_2\sin(0)=C_1, so C_1=3 for an initial position of 3.

   Why: this is the one substitution where the general formula collapses to a single term, reading C_1 straight off x_0.

4. 4

   Differentiate to get x'(t)=-\omega_0C_1\sin(\omega_0t)+\omega_0C_2\cos(\omega_0t), substitute t=0, and solve for C_2.

   Example

   x'(t)=-2.5C_1\sin(2.5t)+2.5C_2\cos(2.5t); with C_1=3 and an initial velocity of 5, x'(0)=2.5C_2=5 gives C_2=2, so x(t)=3\cos(2.5t)+2\sin(2.5t).

   Why: x_0 alone only pins down C_1; a second-order equation always needs a second condition, and v_0 is what fixes C_2.