Why Does Critical Damping Reach Zero Faster Than Overdamping?

[MathewsMD]

Based on the image provided, why exactly does critical damping allow the system's amplitude to reach 0 more quickly than the case of overdamping? In overdamping, isn't the term in the NEGATIVE exponential greater in absolute value? Wouldn't this cause the system to approach 0 more rapidly? That is, $e^{-( \frac {rt}{2m} + w_d)}$ (overdamping) approaches 0 more rapidly than $e^{- \frac {rt}{2m}}$ (critical damping) since the terms $r, t, m, w_d$ are all positive. Could someone please clarify what's incorrect with my interpretation?

 

[Orodruin]

A critically damped system has two solutions which both decay with the same exponential factor while the over-damped system has two solutions which decay with one exponential factor which decays faster than and one that decays slower than that of the critically damped system. The coefficients of these solutions have to be adapted to the initial conditions and generally your solution will be a linear combination of the two solutions for each case. This means you will practically always have a component of the slower decaying solution to the over-damped system, which will therefore decay slower than the critically damped one.

 

[MathewsMD]

A critically damped system has two solutions which both decay with the same exponential factor while the over-damped system has two solutions which decay with one exponential factor which decays faster than and one that decays slower than that of the critically damped system. The coefficients of these solutions have to be adapted to the initial conditions and generally your solution will be a linear combination of the two solutions for each case. This means you will practically always have a component of the slower decaying solution to the over-damped system, which will therefore decay slower than the critically damped one.

Thank you! I was overlooking both solutions and only included one.