Form the auxiliary equation
9m2+km+4=0 Trying x=emt turns the model into a quadratic in m whose coefficients involve k.
State the condition for critical damping
b2−4ac=0 Critical damping is exactly the boundary case of a repeated root.
Write down the discriminant
k2−4×9×4=k2−144 Read a, b and c straight off the auxiliary equation.
Set the discriminant to zero and solve
k2=144 ⇒ k=12 The damping constant is positive, so the negative root is rejected.
Substitute the critical value back into the auxiliary equation
9m2+12m+4=0 With this value of k the quadratic should collapse to a perfect square.
Factorise the auxiliary equation
9(m+32)2=0 A perfect square is exactly what a repeated root means.
Check the repeated root
m=−32 (twice) Both roots coincide, which is the definition of critical damping.
Write down the resulting motion
x=(A+Bt)e−32t The extra factor of t is what a repeated root contributes.
Confirm the motion has no oscillation
(A+Bt)e−32t=0 at most once A linear factor has only one root, so the model crosses equilibrium at most once.
Test a smaller damping constant
k=6 ⇒ b2−4ac=−108<0 Halving k makes the discriminant negative, so the model would oscillate.
Test a larger damping constant
k=24 ⇒ b2−4ac=432>0 Doubling k makes the discriminant positive, so the model would be heavily damped.
Note what happens on either side of the critical value
k<12 light,k>12 heavy Below the critical value the model oscillates; above it, it creeps back.
Interpret the answer physically
k=12 gives the fastest return to equilibrium Critical damping is what a designer wants when overshoot must be avoided.
Recall the auxiliary equation
ax¨+bx˙+cx=0 ⇒ am2+bm+c=0 Trying x=emt turns the differential equation into a quadratic in m.
Recall how the discriminant classifies the damping
b2−4ac>0 heavy,b2−4ac=0 critical,b2−4ac<0 light The nature of the roots decides whether the model oscillates on its way to rest.
State the critical damping constant
This is the unique positive value of k that gives critical damping.