(Note: this question is for a spring mass system moving through air)
Intuitively it makes sense to me that they would be equal to each other, but I have not found a clear answer referencing the forces. All I have found after searching the internet have been different versions of this statement:
Which makes sense to me mathematically when I look at exponents of the general functions compared to the exponents of an overdamped case.
EDIT:
$c = $ damping constant $\left( \frac{kg}{s^{2}} \right)$, $v = $ velocity $\left( \frac{m}{s} \right)$.
In this case damping force is air resistance
Force of spring $= -k x$
$x = $ extension of spring (meter),
$m = $ mass, $k = $ spring constant
The general function that is referred to is:
Best Answer
No.
Regardless of whether a damped harmonic oscillator is underdamped, critically damped, or overdamped, the damping force is proportional to velocity and the spring force is still to spring length. Imagine someone pulls the attached mass beyond the resting length, holds it there for a moment, and then releases the mass. The initial velocity is zero, so the initial damping force is zero. The initial spring force on the other hand is non-zero. The two forces are not equal.
Consider a spring with an attached mass that is oriented vertically in a uniform gravity field. Denoting down as positive, the force on the mass is $$F = m\ddot l = mg - kl - c\dot l\tag{1}$$ where $m$ is the mass, $k$ is the spring force, $c$ is the damping force, and $l$ is the length of the spring. Defining $x=l-l_0$ where $l_0 = mg/k$ simplifies equation (1) to $$m\ddot x + c\dot x + kx = 0\tag{2}$$ This is the canonical form for a damped harmonic oscillator.
The resulting motion is a sine wave in the case $c=0$. This is an undamped harmonic oscillator. Increasing $c$ a little still results in oscillatory motion, but the motion damps out over time. This is an underdamped harmonic oscillator. Increasing $c$ by a lot results in the sum of two decaying exponentials, one slow and the other fast. The coefficients for the slow and fast exponentials depend on initial conditions.
There must be some intermediate value of $c$ where the motion changes from oscillatory to decaying exponential. This is $c_c = 2\sqrt{km}$, the critical damping coefficient. At this value, the fast and slow decaying exponentials become equal. Smaller values of $c$ result in exponentially damped oscillatory motion, larger values in exponential decay. That's all critical damping means. It's the transition point from exponentially damped oscillatory motion to non-oscillatory exponential decay.