Oscillatory Systems
Damped Oscillators
In real oscillating systems, friction, air resistance and other dissipative effects gradually remove mechanical energy from the system. These processes reduce the amplitude of oscillation with time, a phenomenon known as damping.
We again consider the horizontal mass-spring system, but now with a friction force due to the contact with the table. For many physical systems, the damping force can be approximated as proportional to the velocity. The friction force is
$$ \vec{F}_v = - \alpha \vec{v} $$where \( \alpha \) is a positive damping constant. The negative sign indicates that the force always opposes the motion.
The spring force remains
$$ \vec{F}_s = -kx \hat{x} $$Newton's second law gives
$$ m\ddot{x} = - \alpha \dot{x} - kx $$ $$ \ddot{x}+\frac{\alpha}{m}\dot{x}+\frac{k}{m}x=0 $$Defining the natural angular frequency
$$ \omega_0=\sqrt{\frac{k}{m}} $$we obtain
$$ \ddot{x}+\frac{\alpha}{m}\dot{x}+\omega_0^2x=0 $$Searching for a Solution
The oscillatory trial solution from the frictionless case,
$$ x(t)=A\cos(\omega t-\delta) $$no longer works directly because the damping term introduces an additional time dependence. Instead, we look for exponential solutions:
$$ x(t)=A e^{rt} $$Differentiating:
$$ \dot{x}(t)=rA e^{rt} $$ $$ \ddot{x}(t)=r^2A e^{rt} $$Substituting into the differential equation gives
$$ A\left(r^2+\frac{\alpha}{m}r+\omega_0^2\right) \, e^{rt}=0 $$Ignoring the trivial solution \(A=0\), we obtain the characteristic equation:
$$ r^2+\frac{\alpha}{m}r+\omega_0^2=0 $$Solving with the quadratic formula:
$$ r_\pm = \frac{ -\frac{\alpha}{m} \pm \sqrt{ \left( \frac{\alpha}{m} \right)^2 - 4 \omega_0^2} }{2} $$ $$ r_\pm= -\frac{\alpha}{2m} \pm \sqrt{ \left( \frac{\alpha}{2m} \right)^2 - \omega_0^2 } $$The general solution is then
$$ x(t) = A e^{r_+t} + B e^{r_-t} $$We define the damping parameter
$$ \beta= \frac{\alpha}{2m} $$so that
$$ r_\pm= -\beta \pm \sqrt{\beta^2-\omega_0^2} $$The physical behaviour depends entirely on whether the quantity inside the square root is positive, negative, or zero.
Case 1: Overdamping
Overdamping occurs when
$$ \beta^2-\omega_0^2>0 $$or equivalently
$$ \beta>\omega_0 $$In this case the roots are real:
$$ r_\pm= -\beta \pm \sqrt{\beta^2-\omega_0^2} $$Defining
$$ \omega_1= \sqrt{\beta^2 - \omega_0^2} $$gives
$$ r_\pm= -\beta \pm \omega_1 $$Therefore,
$$ x(t) = c_1 e^{(-\beta+\omega_1)t} + c_2 e^{(-\beta-\omega_1)t} $$Since \( \, \beta^2-\omega_0^2>0 \, \), we have \( \, \beta > \sqrt{\beta^2-\omega_0^2} \)
Both exponentials decay with time, meaning the motion gradually dies away without oscillating.
Physically, the system returns to equilibrium very slowly because the damping force is so strong that it prevents oscillations from occurring.
Case 2: Underdamping
Underdamping occurs when
$$ \beta^2-\omega_0^2<0 $$or
$$ \beta < \omega_0 $$The roots become complex:
$$ r_\pm= -\beta \pm i\sqrt{\omega_0^2-\beta^2} $$Defining
$$ \omega_1= \sqrt{\omega_0^2-\beta^2} $$gives
$$ r_\pm= -\beta\pm i\omega_1 $$Therefore,
$$ x(t) = c_1 e^{(-\beta+i\omega_1)t} + c_2 e^{(-\beta-i\omega_1)t} $$Using Euler's formula and rewriting in amplitude-phase form:
$$ x(t) = A e^{-\beta t} \cos(\omega_1t-\delta) $$Notice that this looks very similar to ordinary harmonic motion except for the exponential factor:
$$ e^{-\beta t} $$This exponential envelope continuously shrinks the oscillation amplitude. The system still oscillates, but each cycle becomes smaller than the previous one.
Physically, energy is steadily removed from the oscillator while it moves.
Consider the initial conditions
$$ x(0)=0 $$ $$ \dot{x}(0)=v_0 $$Applying these conditions gives
$$ \delta= \frac{\pi}{2} $$ $$ A= \frac{v_0}{\omega_1} $$Therefore
$$ x(t) = \frac{v_0}{\omega_1} \, e^{-\beta t} \sin(\omega_1t) $$The motion oscillates sinusoidally while simultaneously shrinking toward equilibrium.
Case 3: Critical Damping
Critical damping occurs when
$$ \beta^2-\omega_0^2=0 $$or
$$ \beta=\omega_0 $$The roots become identical:
$$ r_+=r_-=-\beta $$Since we only obtain one independent exponential solution, we must search for a second:
$$ x(t)=At e^{-\beta t} $$The complete solution becomes
$$ x(t) = A_1 e^{-\beta t} + A_2t e^{-\beta t} $$Two common initial conditions:
If
$$ x(0)=0 $$ $$ \dot{x}(0)=v_0 $$then
$$ x(t) = v_0t e^{-\beta t} $$If
$$ x(0)=x_0 $$ $$ \dot{x}(0)=0 $$then
$$ x(t) = x_0(1+\beta t) \, e^{-\beta t} $$Critical damping is important in engineering because it returns the system to equilibrium in the shortest possible time without overshooting.
Car suspension systems, measuring instruments, and door closers are often designed to operate close to critical damping.