Recipe

Solve a mass-spring free-vibration IVP

Use when: you have mx''+cx'+kx=0 with initial conditions x(0)=x_0, x'(0)=v_0, covering every possible level of damping

Goal: x(t) for all three damping regimes, found by classifying the discriminant (or the damping ratio) first

Taught in Underdamping and the damping ratio · extends Solve a mass-spring free-vibration IVP

1. 1

   Substitute the trial solution x=e^{rt} into mx''+cx'+kx=0 and cancel the common factor of e^{rt} (never zero) to get mr^2+cr+k=0.

   Example

   For m=1, c=2, k=10: r^2+2r+10=0.

   Why: this collapses the differential equation to a single algebraic one in r, exactly as it does for the overdamped and critically damped cases.

2. 2

   Compute c^2-4mk; its sign decides which of the three regimes you're in.

   Example

   c^2-4mk=4-40=-36<0: a complex conjugate pair, so this is the underdamped case.

   • c^2-4mk>0 — overdamped, two distinct real roots r_{1,2}=\dfrac{-c\pm\sqrt{c^2-4mk}}{2m}, use x=C_1e^{r_1t}+C_2e^{r_2t} below.
   • c^2-4mk=0 — critically damped, one repeated root r=-c/(2m), use x=(C_1+C_2t)e^{rt} below.
   • c^2-4mk<0 — underdamped (this section), complex pair r=\alpha\pm\beta i with \alpha=-c/(2m), \beta=\sqrt{4mk-c^2}/(2m), use x=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t) below.

   ⚡ Shortcut

   Skip the discriminant entirely and compute the damping ratio \zeta=c/(2\sqrt{mk}) instead, this section's own classification theorem. \zeta<1 is underdamped, \zeta=1 is critical, and \zeta>1 is overdamped, read off one dimensionless number instead of a sign comparison. Here \zeta=2/(2\sqrt{10})=1/\sqrt{10}\approx0.316<1, confirming underdamped without ever computing c^2-4mk directly.

   Why: the discriminant's sign (equivalently, \zeta compared to 1) is what decides which of the three cases applies, and there is no fourth case: every real c\ge0 falls into exactly one.

3. 3

   Overdamped roots give x=C_1e^{r_1t}+C_2e^{r_2t}; a repeated root gives x=(C_1+C_2t)e^{rt}; a complex pair \alpha\pm\beta i gives x=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t).

   Example

   \alpha=-c/(2m)=-1, \beta=\sqrt{4mk-c^2}/(2m)=\sqrt{36}/2=3: x(t)=e^{-t}(C_1\cos3t+C_2\sin3t).

   Why: a complex pair's raw complex exponentials split into a real cosine and sine riding a decaying envelope e^{\alpha t} (this section's own technical block), because the equation's real coefficients guarantee each part is separately a real solution.

4. 4

   Differentiate x(t) for whichever of the three forms you're using, then substitute t=0 into both x and x'.

   Example

   x(0)=C_1=1. x'(t)=e^{-t}\big[(-C_1+\beta C_2)\cos\beta t+(-C_2-\beta C_1)\sin\beta t\big], so x'(0)=-C_1+3C_2=2.

   Why: this turns the free constants into a system pinned down by the actual conditions of the problem.

5. 5

   Solve for C_1,C_2, then substitute back into x(t).

   Example

   C_1=1 from x(0); substituting into -C_1+3C_2=2 gives C_2=1. x(t)=e^{-t}(\cos3t+\sin3t)

   Why: the general solution is a whole family of curves; the initial conditions pick out the one curve that actually matches the problem.