11.5i: Light damping- $\gamma < 2\omega_{0}$

Since $\gamma < 2\omega_{0}$, we have to write Equations 11.5.6 as

$$k_{1}=-\frac{1}{2}\gamma+i\sqrt{\omega_{0}^{2}-\frac{1}{4}\gamma^{2}},\quad k_{2}=-\frac{1}{2}\gamma-i\sqrt{\omega_{0}^{2}-\frac{1}{4}\gamma^{2}}. \label{11.5.7}\tag{11.5.7}$$

Further, I shall write

$$\omega^{'}=\sqrt{w_{0}^{2}-\frac{1}{4}\gamma^{2}}. \label{11.5.8}\tag{11.5.8}$$

Equation 11.5.4 is therefore

$$x=Ae^{-\frac{1}{2}\gamma t+i\omega^{'}t}+Be^{-\frac{1}{2}\gamma t-i\omega^{'}t}=e^{-\frac{1}{2}\gamma t}(Ae^{+i\omega^{'}t}+Be^{-i\omega^{'}t}). \tag{11.5.9}\label{11.5.9}$$

If $x$ is to be real, $A$ and $B$ must be complex. Also, since $e^{-i\omega^{'}t}=(e^{i\omega^{'}t})$* must equal $A$*, where the asterisk denotes the complex conjugate.

Let $A=\frac{1}{2}(a-ib)$ and $B=\frac{1}{2}(a+ib)$ where $a$ and $b$ are real. Then the reader should be able to show that Equation $\ref{11.5.9}$ can be written as

$$x=e^{-\frac{1}{2}\gamma t}(a\cos\omega^{'}t+b\sin\omega^{'}t). \label{11.5.10}\tag{11.5.10}$$

And if $C=\sqrt{a^{2}+b^{2}}, \sin\alpha=\frac{a}{\sqrt{a^{2}+b^{2}}}, \cos\alpha=\frac{b}{\sqrt{a^{2}+b^{2}}},$ the equation can be written

$$x=Ce^{-\frac{1}{2}\gamma t}\sin(\omega^{'}t+\alpha). \label{11.5.11}\tag{11.5.11}$$

Equations $\ref{11.5.9}$, $\ref{11.5.10}$ or $\ref{11.5.11}$ are three equivalent ways of writing the solution. Each has two arbitrary integration constants $(A,B), (a,b)$ or $(C,\alpha)$, whose values depend on the initial conditions - i.e. on the values of $x$ and $\dot{x}$ when $t=0$. Equation $\ref{11.5.11}$ shows that the motion is a sinusoidal oscillation of period a little less than $\omega_{0}$, with an exponentially decreasing amplitude.

To find $C$ and $\alpha$ in terms of the initial conditions, differentiate Equation $\ref{11.5.11}$ with respect to the time in order to obtain an equation showing the speed as a function of the time:

$$\dot{x}=Ce^{-\frac{1}{2}\gamma t}[\omega^{'}\cos(\omega^{'}t+\alpha)-\frac{1}{2}\gamma\sin(\omega^{'}t+\alpha)]. \label{11.5.12}\tag{11.5.12}$$

By putting $t=0$ in Equations $\ref{11.5.11}$ and $\ref{11.5.12}$ we obtain

$$x_{0}=C\sin\alpha \label{11.5.13}\tag{11.5.13}$$

and

$$(\dot{x})_{0}=C(\omega^{'}\cos\alpha-\frac{1}{2}\gamma\sin\alpha). \label{11.5.14}\tag{11.5.14}$$

From these we easily obtain

$$\cot\alpha=\frac{1}{\omega^{'}}[\frac{(\dot{x})_{0}}{\dot{x}_{0}}+\frac{\gamma}{2}] \label{11.5.15}\tag{11.5.15}$$

and

$$C=x_{0}\csc\alpha. \label{11.5.16}\tag{11.5.16}$$

The quadrant of $\alpha$ can be determined from the signs of $\cot\alpha$ and $\csc\alpha$ , $C$ always being positive.

Note that the amplitude of the motion falls off with time as $e^{-\frac{1}{2}\gamma t}$, but the mechanical energy, which is proportional to the square of the amplitude, falls off as $e^{-\gamma t}$.

Figures XI.2 and XI.3 are drawn for $C=1, \alpha=0, \gamma=1$. Figure XI.2 has $\omega_{0}=25\gamma$ and hence $x=e^{-\frac{1}{2}t}\sin0.24.9949995t$, and figure XI.3 has $\omega_{0}=4\gamma$ and hence $x=e^{-\frac{1}{2}t}\sin3.968626967t$.

Exercise $\PageIndex{1}$

Draw displacement : time graphs for an oscillator with $m$ = 0.02 kg, $k$ = 0.08 N m^-1, $g$ = 1.5 s^-1, $t$ = 0 to 15 s, for the following initial conditions:

1. $x_{0}=0,\quad (\dot{x})_{0}=4$ ms^-1
2. $(\dot{x})_{0}=0,\quad x_{0}=3$ m
3. $(\dot{x})=-2$ ms^-1, $x_{0}=2$

Although the motion of a damped oscillator is not strictly "periodic", in that the motion does not repeat itself exactly, we could define a "period" $P=\frac{2\pi}{\omega^{'}}$ as the interval between two consecutive ascending zeroes. Extrema do not occur exactly halfway between consecutive zeroes, and the reader should have no difficulty in showing, by differentiation of Equation $\ref{11.5.11}$, that extrema occur at times given by $\tan(\omega^{'}t+\alpha)=\frac{2\omega^{'}}{\gamma}$. However, provided that the damping is not very large, consecutive extrema occur approximately at intervals of $\frac{P}{2}$. The ratio of consecutive maximum displacements is, then,

$$\frac{|x_{n}|}{|x_{n+1}|}=\frac{e^{-\frac{1}{2}\gamma t}}{e^{-\frac{1}{2}\gamma (t+\frac{1}{2}P)}}. \label{11.5.17}\tag{11.5.17}$$

From this, we find that the logarithmic decrement is

$$ln(\frac{|x_{n}|}{|x_{n+1}|})=\frac{P\gamma}{4}, \label{11.5.18}\tag{11.5.18}$$

from which the damping constant can be determined.