Recipe

Solve a mass-spring free-vibration IVP

Use when: you have the free-vibration equation mx''+cx'+kx=0 (no external forcing) together with initial conditions x(0)=x_0, x'(0)=v_0

Goal: the position function x(t) for the overdamped case, built from the characteristic equation's roots

Taught in Overdamping · 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=8: r^2+6r+8=0.

   Why: this collapses the differential equation to a single algebraic one in r, exactly Chapter 7's own derivation, with m,c,k standing in for a,b,c.

2. 2

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

   Example

   c^2-4mk=36-32=4>0: two distinct real roots, so this is the overdamped case this section covers.

   • Not yet: c^2-4mk=0, critical damping — covered in Critical damping
   • Not yet: c^2-4mk<0, underdamping — covered in Underdamping and the damping ratio

   Why: the discriminant's sign is exactly what decides whether the roots come out real and distinct, repeated, or complex. Because m,k>0 here, a positive discriminant always gives two roots that are both negative (this section's own technical block), never one positive and one negative.

3. 3

   Use the quadratic formula, r_{1,2}=\dfrac{-c\pm\sqrt{c^2-4mk}}{2m}, and write the general solution x(t)=C_1e^{r_1t}+C_2e^{r_2t}.

   Example

   r_{1,2}=\dfrac{-6\pm\sqrt4}{2}=\dfrac{-6\pm2}{2}, so r_1=-2, r_2=-4: x(t)=C_1e^{-2t}+C_2e^{-4t}.

   Why: both exponential terms decay to zero with no oscillation, since both roots are negative. This is precisely what "overdamped" means.

4. 4

   Differentiate to get x'(t)=r_1C_1e^{r_1t}+r_2C_2e^{r_2t}, then substitute t=0 into both x and x'.

   Example

   x(0)=C_1+C_2=2; x'(t)=-2C_1e^{-2t}-4C_2e^{-4t}, so x'(0)=-2C_1-4C_2=-1.

   Why: this turns the two free constants into a system pinned down by the actual problem, not just the equation.

5. 5

   Solve the 2\times2 linear system for C_1,C_2, then substitute back into x(t).

   Example

   From C_1+C_2=2, C_1=2-C_2. Substituting: -2(2-C_2)-4C_2=-1\Rightarrow-4-2C_2=-1\Rightarrow C_2=-1.5, so C_1=3.5. x(t)=3.5e^{-2t}-1.5e^{-4t}

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