written 4.3 years ago by |

**LOGARITHMIC DECREMENT.**

In case of under damped system, we know that displacement is given by equation $x = e^{-\xi wnt} [ X \ sin (w_dt + \phi) ]$

i.e. at any time 't', the peak amplitude is given by relation $x - X \ e ^{-\xi wnt}$

Consider 2 points A and B on displacement curve as shown in above figure. Points A and B represent successive peak amplitudes on displacement curve.

Then at $t = t_A$, displacement at point A is given by relation $X_A = X. e^{-\xi wnt}$

$x_B = X . e^{-\xi w_ntB}$

then $\frac{X_A}{X_B} = \frac{Xe^{-\xi ntA}}{X_e ^{-\xi wntB}} = e^{-\xi wn} (t_A - t_B)$

then $\frac{X_A}{X_B} = e^{ \xi w_n} (t_B - t_A)$

Taking log on both sides,

In $(\frac{X_A}{X_B}) = \xi w_n (t_B - t_A)$

But $t_B - t_A = \frac{2\pi }{w_d}$

Note:

The term $\sqrt{1 - \xi ^2} w_n = w_d$ is called the **frequency of damped vibration**.

Hence in $(\frac{X_A}{X_B}) = \frac{\xi w_n}{w_d} \times 2 \pi$

$= \frac{2 \pi \xi w_n}{\sqrt{1 - \xi ^3 w_n }}$

$\delta = In \ (\frac{X_A}{X_B}) = \frac{2 \pi w_n}{\sqrt{1 - \xi ^2} w_n}$

$\delta = \frac{2 \pi \xi}{\sqrt{1 - \xi ^2}}$

Above equation is used to calculate logarithmic decrement between 2 successive peak amplitudes which is a function of only 'd'.

To find out amplitude at the end nth cycle, following relation is adopted.

Let $x_0$ = Initial peak amplitude.

$x_1$ = peak amplitude after 1st cycle.

$x_2$ = peak amplitude after 2nd cycle.

$x_n$ = peak amplitude after nth cycle.

So we can write,

$\frac{x_0}{x_n} = \frac{x_0}{x_1} \times \frac{x_1}{x_2} \times \frac{x_2}{x_3} \times . . . . \times \frac{x_n - 1}{x_n}$

Taking log on both sides:

$In \ (\frac{x_0}{x_n}) = \ In \ (\frac{x_0}{x_1} \times \frac{x_1}{x_2} \times \frac{x_2}{x_3} \times . . . . . \times \frac{x_n - 1}{x_n})$

$In \ (\frac{x_0}{x_n}) = n . \delta$

$\therefore$ $\delta = \frac{1}{n} .\ In \ (\frac{x_01}{x_n})$