Fig-1
Underdamped system ( ξ < 1 ) \ (\xi < 1) ( ξ < 1 )
If the damping factor ξ \ \xi ξ is less than one or the damping coefficient c c c is less than critical damping coefficient c c c_c c c , then the system is said to be an under-damped system.
ξ < 1 O R c c c < 1 ⟹ c < c c \xi < 1 \quad \text OR \quad {c \over c_c} < 1\implies c < c_c ξ < 1 O R c c c < 1 ⟹ c < c c
We know that roots of differential equations are:
S 1 = [ − ξ + ξ 2 − 1 ] ω n S 2 = [ − ξ − ξ 2 − 1 ] ω n S_1 = \big [-\xi + \sqrt{\xi^2 -1} \big] \omega_n \\
S_2 = \big [-\xi - \sqrt{\xi^2 -1} \big] \omega_n S 1 = [ − ξ + ξ 2 − 1 ] ω n S 2 = [ − ξ − ξ 2 − 1 ] ω n
But for ξ < 1 \ \xi < 1 ξ < 1 ; the roots for under-damped system are given by S 1 S_1 S 1 and S 2 S_2 S 2 as below:
S 1 = [ − ξ + i ( 1 − ξ 2 ) ] ω n S 2 = [ − ξ − i ( 1 − ξ 2 ) ] ω n \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} S 1 = S 2 = [ − ξ + i ( 1 − ξ 2 ) ] ω n [ − ξ − i ( 1 − ξ 2 ) ] ω n
Where i = − 1 i = \sqrt{-1} i = − 1 is the imaginary unit of complex root
The roots are complex and negative, so the solution of differential equation is given by
x = A e S 1 t + B e S 2 t ∴ x = A e [ − ξ + i ( 1 − ξ 2 ) ] ω n t + B e [ − ξ − i ( 1 − ξ 2 ) ] ω n t ∴ x = A . e − ξ ω n t . e i ( 1 − ξ 2 ) ω n t + B . e − ξ ω n t . e − i ( 1 − ξ 2 ) ω n t ∴ x = e − ξ ω n t [ A . e i ( 1 − ξ 2 ) ω n t + B . e − i ( 1 − ξ 2 ) ω n t ] ∴ x = e − ξ ω n t [ A . e i ω d t + B . e − i ω d t ] [ ∵ ( 1 − ξ 2 ) ω n = ω d ] \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} x ∴ x ∴ x ∴ x ∴ x = A e S 1 t + B e S 2 t = A e [ − ξ + i ( 1 − ξ 2 ) ] ω n t + B e [ − ξ − i ( 1 − ξ 2 ) ] ω n t = A . e − ξ ω n t . e i ( 1 − ξ 2 ) ω n t + B . e − ξ ω n t . e − i ( 1 − ξ 2 ) ω n t = e − ξ ω n t [ A . e i ( 1 − ξ 2 ) ω n t + B . e − i ( 1 − ξ 2 ) ω n t ] = e − ξ ω n t [ A . e i ω d t + B . e − i ω d t ] [ ∵ ( 1 − ξ 2 ) ω n = ω d ]
According to Euler’s theorem, above equation can be written as:
x = X e − ξ ω n t [ sin ( ω d t + ∅ ) ] W h e r e X a n d ∅ are constants x = Xe^{-\xi \omega_nt} \big [\sin (\omega_dt + \varnothing)] \\
\text Where \ \ X \text and \ \ \varnothing \text { are constants} x = X e − ξ ω n t [ sin ( ω d t + ∅ )] W h ere X a n d ∅ 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 − ξ ω n t \ X.e^{-\xi \omega_nt} X . e − ξ ω n t
The natural angular frequency of damped free vibrations is given by:
ω d = ( 1 − ξ 2 ) ω n \omega_d = \sqrt {(1 - \xi^2)}\omega_n ω d = ( 1 − ξ 2 ) ω n
Time period for under-damped vibration is given by:
t p = 2 π ω d = 2 π ( 1 − ξ 2 ) ω n s e c . t_p = {2\pi \over \omega_d} = {2\pi \over \sqrt{(1 - \xi^2)}\omega_n} sec. t p = ω d 2 π = ( 1 − ξ 2 ) ω n 2 π sec .
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