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# Under-Damped System (​ξ < 1)

Comments Fig-1

## Underdamped system $\ (\xi < 1)$

If the damping factor $\ \xi$ is less than one or the damping coefficient $c$ is less than critical damping coefficient $c_c$, then the system is said to be an under-damped system.

$\xi < 1 \quad \text OR \quad {c \over c_c} < 1\implies c < c_c$
• We know that roots of differential equations are:
$S_1 = \big [-\xi + \sqrt{\xi^2 -1} \big] \omega_n \\ S_2 = \big [-\xi - \sqrt{\xi^2 -1} \big] \omega_n$

• But for $\ \xi < 1$; the roots for under-damped system are given by $S_1$ and $S_2$ as below:
\begin{aligned} S_1 = & \big [-\xi + i \sqrt{(1 - \xi^2)}] \omega_n \\ S_2 = & \big [-\xi - i \sqrt{(1 - \xi^2)}] \omega_n \\ \end{aligned}

Where $i = \sqrt{-1}$ is the imaginary unit of complex root

• The roots are complex and negative, so the solution of differential equation is given by  \begin{aligned} x &= Ae^{S_1t} + Be^{S_2t} \\ \therefore x &= Ae^{[-\xi + i \sqrt{(1 - \xi^2)}] \omega_nt} + Be^{[-\xi - i \sqrt{(1 - \xi^2)}] \omega_nt} \\ \therefore x &= A.e^{-\xi\omega_nt}.e^{i \sqrt{(1 - \xi^2)} \omega_nt} + B.e^{-\xi\omega_nt}.e^{- i \sqrt{(1 - \xi^2)} \omega_nt} \\ \therefore x &= e^{-\xi\omega_nt} \big [A.e^{i \sqrt{(1 - \xi^2)} \omega_nt} + B.e^{- i \sqrt{(1 - \xi^2)} \omega_nt}] \\ \therefore x &= e^{-\xi\omega_nt} \big [A.e^{i\omega_dt} + B.e^{- i\omega_dt}] \quad \big [ \because \sqrt {(1 - \xi^2) \omega_n} = \omega_d] \end{aligned}
• According to Euler’s theorem, above equation can be written as:
$x = Xe^{-\xi \omega_nt} \big [\sin (\omega_dt + \varnothing)] \\ \text Where \ \ X \text and \ \ \varnothing \text { are constants}$
• Above equation shows the equation of motion for an underdamped system, and the amplitude reduces gradually and finally becomes zero after some time.

• Amplitude decreases by $\ X.e^{-\xi \omega_nt}$

• The natural angular frequency of damped free vibrations is given by:

$\omega_d = \sqrt {(1 - \xi^2)}\omega_n$
• Time period for under-damped vibration is given by:
$t_p = {2\pi \over \omega_d} = {2\pi \over \sqrt{(1 - \xi^2)}\omega_n} sec.$

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