Question

What is the effect of damping in an oscillatory system? Differentiate between heavy
and critical damping. Show that the displacement of a weakly damped oscillator is
given by
x(t) = a exp(−bt) cos(w t − φ)
where symbols have their usual meanings

Answer 1

Answer on Question #73115, Physics / Mechanics | Relativity

Question. What is the effect of damping in an oscillatory system? Differentiate between heavy and critical damping. Show that the displacement of a weakly damped oscillator is given by $x(t) = A \cdot \exp(-\beta t) \cdot \cos(\omega t - \varphi)$ where symbols have their usual meanings.

Solution.

1. What is the effect of damping in an oscillatory system?

As a result of the damping, the oscillatory system loses energy and the amplitude is decreased.

2. Differentiate between heavy and critical damping.

Damping can be light, in which case the system oscillates about the midpoint (a), heavy, in which the system takes a long time to reach equilibrium (b) or critical, where the system reaches equilibrium in a short time compared with $T$ with no overshoot, where $T$ is the natural period of vibration of the system (c).

3. Show that the displacement of a weakly damped oscillator is given by $x(t) = A \cdot \exp(-\beta t) \cdot \cos(\omega t - \varphi)$ where symbols have their usual meanings.

The unforced damped harmonic oscillator has equation

$$\frac {d ^ {2} x}{d t ^ {2}} + 2 \beta \frac {d x}{d t} + \omega_ {0} ^ {2} x = 0.$$

It has characteristic equation

$$\lambda^ {2} + 2 \beta \lambda + \omega_ {0} ^ {2} \lambda = 0$$

with characteristics roots

$$\lambda_{1,2} = \frac{-2\beta \pm \sqrt{4\beta^2 - 4\omega_0^2}}{2} = -\beta \pm \sqrt{\beta^2 - \omega_0^2}.$$

There are three cases depending on the sign of the expression under the square root:

i) $\beta^2 < \omega_0^2$ (this will be *light* damping, $\beta$ is small relative to $\omega_0$).

ii) $\beta^2 > \omega_0^2$ (this will be *heavy* damping, $\beta$ is large relative to $\omega_0$).

iii) $\beta^2 = \omega_0^2$ (this will be *critical* damping, $\beta$ is just between *heavy* and *light* damping).

Case (i) *Light* damping (non-real complex roots)

If $\beta^2 < \omega_0^2$ then the term under the square root is negative and the characteristic roots are not real. In order for $\beta^2 < \omega_0^2$ the constant $\beta$ must be relatively small. Let $\omega = \sqrt{\omega_0^2 - \beta^2}$. Then we have characteristic roots

$$-\beta \pm i\omega$$

leading to complex exponential solutions:

$$e^{(-\beta + i\omega)t}, e^{(-\beta - i\omega)t}.$$

The basic real solution are

$$e^{-\beta t} \cos \omega t \quad \text{and} \quad e^{-\beta t} \sin \omega t.$$

The general real solution is found by taking linear combinations of the two basic solutions, that is

$$x(t) = C_1 e^{-\beta t} \cos \omega t + C_2 e^{-\beta t} \sin \omega t$$

or

$$x(t) = e^{-\beta t} (C_1 \cos \omega t + C_2 \sin \omega t) = A e^{-\beta t} \cos (\omega t - \phi).$$

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