Equations of Motion

Consider the linear differential equation for the position, \( x(t) \), of a mass \( m \) constrained by a linear damper with coefficient \( \gamma \), and a spring of stiffness \( k_s \): \[ m \left( \frac{d^2 x}{dt^2} \right) + \gamma \left( \frac{dx}{dt} \right) + k_s x = 0 \] If the mass is given a slight displacement \( \varepsilon \) away from \( x = 0 \) and released from rest at \( t = 0 \), when is the subsequent \( x(t) \) stable, neutrally stable, or unstable when \( m > 0 \), \( k_s \geq 0 \), but \( \gamma \ne 0 \) may have either sign?

For the given initial conditions, the solution for \( x(t) \) is: \[ x(t) = A e^{\beta_- t} + B e^{\beta_+ t}, \quad \text{where } \beta_{\pm} = -\frac{\gamma}{2m} \pm \sqrt{ \frac{\gamma^2}{4m^2} - \frac{k_s}{m} } \]

There are three distinct cases:

If \( \gamma > 0 \) and \( k_s > 0 \), then \( x(t) \) is stable because both \( \beta_+ \) and \( \beta_- \) will have negative real parts and the initial displacement decays exponentially with increasing time
If \( k_s = 0 \), then \( \beta_+ \) will be zero for any value of \( \gamma \), and the solution for \( x(t) \) reduces to \( x = \varepsilon \). This represents neutral stability since a new steady solution emerges after the initial displacement
If \( \gamma < 0 \) and \( k_s > 0 \), then \( x(t) \) is unstable since both \( \beta_+ \) and \( \beta_- \) have a positive real part and the initial displacement grows exponentially with increasing time

Interestingly, real fluid flows can mimic this negative-damping destabilizing effect

¹Read more about spring-mass systems and vibration: Rao, S.S. (2006) Mechanical Vibration. 5th Edition, Pearson Education, Inc., Upper Saddle River, 603-606.