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, and the discriminant c^2-4mk is zero or positive (critically damped or overdamped)

Goal: x(t) covering the critically damped case in addition to the overdamped case

Taught in Critical damping · extends Solve a mass-spring free-vibration IVP · a later section extends this recipe with more cases

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=6, k=9: r^2+6r+9=0.

   Why: this collapses the differential equation to a single algebraic one in r, exactly as it does in the overdamped case.

2. 2

   Compute c^2-4mk before finishing the algebra.

   Example

   c^2-4mk=36-4(1)(9)=0: a repeated root, so this is the critically damped case.

   • c^2-4mk>0 — overdamped (previous section), 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 (this section), one repeated real root r=-c/(2m), use x=(C_1+C_2t)e^{rt} below.
   • Not yet: c^2-4mk<0, underdamping — covered in Underdamping and the damping ratio

   Why: a zero discriminant is exactly the boundary where the overdamped case's two roots collide into one, and c=2\sqrt{mk} (this section's own technical block) is that exact value.

3. 3

   Overdamped roots give x=C_1e^{r_1t}+C_2e^{r_2t}; a repeated root r=-c/(2m) gives x=(C_1+C_2t)e^{rt}.

   Example

   r=-c/(2m)=-6/2=-3: x(t)=(C_1+C_2t)e^{-3t}.

   Why: a repeated root only hands over one genuine exponential solution. The extra factor of t is what reduction of order (Chapter 6) contributes as the missing second solution, never a second copy of e^{rt}.

4. 4

   For the repeated-root form, differentiate to get x'(t)=(C_2+rC_1+rC_2t)e^{rt}, then substitute t=0 into both x and x'.

   Example

   x(0)=C_1=2; x'(t)=(C_2-3C_1-3C_2t)e^{-3t}, so x'(0)=C_2-3C_1=1.

   Why: this turns the two 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=2 from x(0); substituting into C_2-3C_1=1 gives C_2=7. x(t)=(2+7t)e^{-3t}

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