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## Critically Damped System (ξ = 1)

Fig.1(Critically Damped System)

• Critically damped system(ξ=1): If the damping factor ξ is equal to one, or the damping coefficient c is equal to critical damping coefficient "cc", then the system is said to be a critically damped system.
$\xi=1 \quad \text OR \quad {c \over c_c} = 1\implies c = c_c$
• Two roots for critically damped system are given by S1 and S2 as below:
$S_1 = \big [-\xi + \sqrt{\xi^2 -1} \big] \omega_n \\ S_2 = \big [-\xi - \sqrt{\xi^2 -1} \big] \omega_n$
• For $ξ=1$; $S_1 = S_2 = -\omega_n$
Here both the roots are real and equal, so the solution to the differential equation can be given by
$x = (A + Bt)e^{-\omega_n t} \quad \quad ...(1)$
• Now differentiating equation (1) with respect to ‘t’, we get:
$\mathring x = Be^{-\omega_nt} - \omega_n(A + Bt)e^{-\omega_nt} \quad \quad ...(2)$
• Now, let at
$t = 0$ : $x = X_0$
$t = 0$ : $\mathring x =0$

• Substituting these values in equation (1):

$X_0 = A \quad ...(3)$
• Same way, from equation (2), we get
\begin{aligned} 0&= B - \omega_n(A + 0) \\ 0&= B - \omega_nA \\ B&= \omega_nA \\ B&= \omega_nX_0 \qquad ...(4) \end{aligned}
• Now putting the values of A and B in equation (1), we get:
\begin{aligned} x&= (X_0 + \omega_nX_0t)e^{-\omega_nt} \\ x&= X_0(1 + \omega_nt)e^{-\omega_nt} \qquad ...(5) \end{aligned}

### Conclusion

• From above equation (5), it is seen that as time t increases, the displacement x decreases exponentially.

• The motion of a critically damped system is aperiodic (aperiodic motion motions are those motions in which the motion does not repeat after a regular interval of time i.e non periodic motion) and so the system does not shows vibrations.

• For critically damped systems, if a system is displaced from its initial position, it will try to reach its mean position in a very short time.

• Critically damped systems are generally seen in hydraulic doors closer as it is necessary for the door to come to its initial position in a very short time.

## Damped free Vibration-Numerical 1

#### A mass of 85 kg is supported on a spring which deflects 18 mm under the weight of the mass. The vibrations of the mass are constrained to be linear and vertical. A dashpot is provided which reduces the amplitude to one-quarter of its initial value in two complete oscillations. Calculate magnitude of the damping force at unit velocity and periodic time of damped vibration.

Solution:

Fig-1

Given data:

mass $\ m = 85$ kg
Static deflection, $\ x = 18\ \text{mm} = 0.018\ \text{m}$

Let,
$x_0 =$ initial amplitude
$x_2=$ final amplitude after two complete cycle $= {1\over4}$

Taking ratio,

${x_0 \over x_2}={x_0 \over {1 \over 4} x_0}=4$

#### Damping force:

$\text F = c.\mathring x= c$

Now,

\begin{aligned} \xi= {c \over c_c} \quad \text {OR} \quad &\xi= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\ \therefore \ &c=(2m\omega_n)\xi \ \dots\dots\dots\dots \ (1) \end{aligned}

#### 1. The Logarithmic decrement:

\begin{aligned} \delta &= {1 \over n} log_e\bigg ({x_0 \over x_n}\bigg)\\ &= {1 \over 2} log_e\bigg ({x_0 \over x_2}\bigg)\\ &= {1 \over 2} log_e(4) \end{aligned}

#### 2. Damping factor:

\begin{aligned} \xi &= {\delta \over \sqrt{4 \pi^2 + {\delta}^2} } \\ &={\delta \over \sqrt{4 \pi^2 + {\delta}^2} }\\ &= 0.1091 \end{aligned}

#### 3. Frequency of undamped free vibration:

The natural circular frequency of vibration is,

\begin{aligned} \omega_n &= {\sqrt{k \over \ m}} \\ &= {\sqrt{m \ g \over \ x \ m}}\\ &= {\sqrt{g \over x}} \\ &= {\sqrt{9.81 \over 0.018}}\\ &= 23.34 \ \text {rad/s} \end{aligned}

From equation (1); we get,
Damping force at unit velocity;

\begin{aligned} \text F = c &= (2m\omega_n)\xi \\ &= (2 \ \text x \ 85\ \text x \ 23.34)\ \text x \ (0.1091) \\ &= 91.67 \ \text N \end{aligned}

The time period of damped vibration is,

\begin{aligned} t_p &= {2\pi \over \omega_d} \\ &= {2\pi \over \omega_n \ \sqrt {1 -\xi^2}} \\ &= {2\pi \over (23.34) \ \sqrt {1 -(0.1091)^2}} \\ &=0.2708 \ \text {sec} \end{aligned}

## V-Engine

A V-engine is a two cylinder engine, which has a common crank and the axis of cylinder makes a "V" shape.

Since V-engines have a common crank and the crank revolves in one plane, there is no primary or secondary couple acting on the engine.

Consider a V-engine as shown in fig.1 having common crank OC and two connecting rods CP and CQ. The lines of stroke OP and OQ are inclined to vertical axis OY at an angle ‘α’.

Fig-1

Let,

\begin{aligned} m &= \text{mass of reciprocating parts per cylinder, kg}\\ l &= \text{length of connecting rod, m}\\ r &= \text{radius of crank, m}\\ n &= \text{Obliquity ratio }= l/r\\ θ &= \text{Crank angle, measured from vertical axis OY, at any instant}\\ ω &= \text{Angular velocity of crank, rad/s}\\ 2 \alpha &= \text{V-angle i.e. angle between lines of strokes of two cylinders}\\ \end{aligned}

We know that,

• Primary unbalanced force in single cylinder engine is,
$F_P = mr\omega^2 \cos \theta$
• Secondary unbalanced force in single cylinder engine is,
$F_S = mr\omega^2 \bigg({\cos2\theta \over n} \bigg)$

### 1. Primary forces ($F_P$)

#### (i) Primary forces in individual cylinders:

$F_{P1} = mr\omega^2 \cos (\alpha -\theta)\\ F_{P2} = mr\omega^2 \cos (\alpha +\theta)$

#### (ii) Total primary force along Y axis ($F_{PV}$):

As both components $(F_{P1}\sdot \cos\alpha)$ and $(F_{P2} \sdot \cos\alpha)$ are acting in same direction;

\begin{aligned} F_{PV} &= F_{P1} \cos\alpha + F_{P2} \cos\alpha\\ &= mr\omega^2 \cos (\alpha -\theta)\cos\alpha \ - mr\omega^2 \cos (\alpha +\theta)\cos\alpha\\ &= mr\omega^2 \cos\alpha [\cos (\alpha -\theta) + \cos (\alpha+\theta)]\\ &= mr\omega^2 \cos\alpha \sdot (2 \cos\theta \cos\alpha)\\ &= 2mr\omega^2 \cos^2\alpha \cos\theta \end{aligned}

#### (iii) Total primary force along X-axis ($F_{PH}$):

As both components $(F_{P1} \sdot \sin \alpha)$ and $(F_{P2}\sdot \sin \alpha)$ are acting opposite to each other;

\begin{aligned} F_{PH} &= F_{P1} \sin\alpha + F_{P2} \sin\alpha\\ &= mr\omega^2 \cos (\alpha -\theta)\sin\alpha \ - mr\omega^2 \cos (\alpha +\theta)\sin\alpha\\ &= mr\omega^2\sin\alpha [\cos (\alpha -\theta)-\cos (\alpha +\theta)]\\ &= mr\omega^2\sin\alpha \ \sdot (2\sin \theta \sin \alpha)\\ &= 2mr\omega^2\sin^2\alpha\sin \theta \end{aligned}

#### (iv) Resultant primary force ($F_p$):

\begin{aligned} F_p &= \sqrt{(F_{PV})^2 +(F_{PH})^2 }\\ &= \sqrt{(2mr\omega^2 \cos^2\alpha \cos\theta)^2 + (2mr\omega^2\sin^2\alpha\sin \theta)^2}\\ &= 2mr\omega^2\sqrt{(\cos^2\alpha \cos\theta)^2 + (\sin^2\alpha\sin \theta)^2} \end{aligned}

The angle made by resultant primary force with vertical axis is;

\begin{aligned} \beta_P &= \tan^{-1} \bigg ( {F_{PH} \over F_{PV}}\bigg)\\ &= \tan^{-1} \bigg ( {2mr\omega^2\sin^2\alpha\sin \theta \over 2mr\omega^2 \cos^2\alpha \cos\theta} \bigg)\\ &= \tan^{-1} (\tan^2\alpha \ \sdot \ \tan\theta) \end{aligned}

### 2. Secondary forces ($F_S$)

#### (i) Secondary forces in individual cylinder:

$F_{S1} = mr \omega^2 {\cos2(\alpha-\theta) \over n}\\ F_{S2} = mr \omega^2 {\cos2(\alpha+\theta) \over n}\\$

#### (ii) Total secondary force along Y axis ($F_{SV}$):

As both components $(F_{S1}\sdot\cos \alpha)$ and $(F_{S2}\sdot \cos \alpha)$ are acting in same direction;

\begin{aligned} F_{SV} &= F_{S1} \cos \alpha + F_{S2} \cos \alpha\\ &= mr \omega^2 {\cos2(\alpha-\theta) \over n}\cos\alpha + mr \omega^2 {\cos2(\alpha+\theta) \over n}\cos\alpha\\ &= mr \omega^2\cos\alpha \bigg[{\cos2(\alpha-\theta) \over n} + {\cos2(\alpha+\theta) \over n}\bigg]\\ &={1\over n} mr \omega^2\cos\alpha \sdot[2\sdot\cos2\theta\sdot\cos2\alpha]\\ &= {2 \over n}mr \omega^2\cos\alpha\sdot\cos2\alpha\sdot\cos2\theta \end{aligned}

#### (iii) Total secondary force along X-axis ($F_{SH}$):

As both components $(F_{S1}\sdot\sin\alpha)$ and $(F_{S2}\sdot\sin\alpha)$ are acting opposite to each other;

\begin{aligned} F_{SH} &= F_{S1} \sin \alpha - F_{S2} \sin \alpha\\ &= mr \omega^2 {\cos2(\alpha-\theta) \over n}\sin\alpha - mr \omega^2 {\cos2(\alpha+\theta) \over n}\sin\alpha\\ &= mr \omega^2\sin\alpha \bigg[{\cos2(\alpha-\theta) \over n} - {\cos2(\alpha+\theta) \over n}\bigg]\\ &= {1\over n} mr \omega^2\sin\alpha \sdot[2\sdot\sin2\theta\sdot\sin2\alpha]\\ &= {2 \over n}mr \omega^2\sin\alpha\sdot\sin2\alpha\sdot\sin2\theta \end{aligned}

#### (iv) Resultant secondary force ($F_S$):

\begin{aligned} F_s &= \sqrt{(F_{SV})^2 + (F_{SH})^2}\\ &= \sqrt{( {2 \over n}mr \omega^2\cos\alpha\sdot\cos2\alpha\sdot\cos2\theta)^ + ({2 \over n}mr \omega^2\sin\alpha\sdot\sin2\alpha\sdot\sin2\theta)^2}\\ &= {2 \over n} mr\omega^2 \sqrt{(\cos\alpha\sdot\cos2\alpha\sdot\cos2\theta)^2 + (\sin\alpha\sdot\sin2\alpha\sdot\sin2\theta)^2} \end{aligned}

The angle made by resultant secondary force with vertical axis is;

\begin{aligned} \beta_S &= \tan^{-1} \bigg( {F_{SH} \over F_{SV}}\bigg)\\ &= \tan^{-1} \bigg( {{2 \over n}mr \omega^2\sin\alpha\sdot\sin2\alpha\sdot\sin2\theta \over {2 \over n}mr \omega^2\cos\alpha\sdot\cos2\alpha\sdot\cos2\theta} \bigg)\\ &= \tan^{-1}\ (\tan\alpha \sdot \tan2\alpha \sdot \tan2\theta) \end{aligned}

• The design of V-engine is compact compared to multi-cylinder inline engines. So they consume less space.
• Power output of V-engine is more compared to single cylinder engines because the crank receives power from both the cylinders.
• The operation of V-engine is smoother for high speed performance.

## Applications of V-engines

• Because of its compactness, smoother operations and high power output, V-engines are used in sports cars.

## Logarithmic Decrement (​δ)

fig-1: displacement V/s time curve for under damped system

### Logarithmic decrement

Logarithmic decrement is defined as the natural logarithm of the ratio of successive amplitude on the same side of mean position.

The rate of decay in the amplitudes of under-damped system is measured by the parameter known as logarithmic decrement.

Rate of decay in amplitudes depends on the amount of damping present in the system. So if the damping is more, then the rate of decay will also be more.

Let A and B are the two points on the successive cycles which shows maximum deflection as shown in figure.

The periodic time:

\begin{aligned} t_p &= t_2 − t_1 \\ &= {2 \pi \over \omega_d} \\ &= {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \end{aligned}

The amplitude at time $t_1$ and $t_2$ are:

$x_1 = Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)]$

And,

\begin{aligned} x_2 &= Xe^{-\xi \omega_n t_2} [sin(\omega_d t_2 + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ \omega_d (t_1 + t_p) + \varnothing \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d t_p + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + \omega_d \bigg({2 \pi \over \omega_d}\bigg) + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin ( \omega_d t_1 + 2 \pi + \varnothing)] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin \{ 2 \pi + (\omega_d t_1 + \varnothing) \}] \\ \therefore \quad x_2 &= Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] \\ \end{aligned}

Now,
Taking ratio, we get;

\begin{aligned} {x_1 \over x_2} &= {Xe^{-\xi \omega_n t_1} [sin(\omega_d t_1 + \varnothing)] \over Xe^{-\xi \omega_n (t_1 + t_p)} [sin (\omega_d t_1 + \varnothing)] }\\ \therefore \quad \quad {x_1 \over x_2} &= e^{-\xi \omega_n (t_1-t_1-t_p)} \\ \therefore \quad \quad {x_1 \over x_2} &= e^{\xi \omega_n t_p} \end{aligned}

Now,
The logarithmic decrement is given by;

\begin{aligned} \delta &= log_e \bigg({x_1 \over x_2}\bigg) \\ \therefore \quad \delta &= log_e (e^{\xi \omega_n t_p}) \\ \therefore \quad \delta &= \xi \omega_n t_p \\ \therefore \quad \delta &= \xi \omega_n {2 \pi \over \big(\sqrt {1-\xi^2}\big) \ \omega_n} \\ \therefore \quad \delta &= {2 \pi \xi\over \big(\sqrt {1-\xi^2}\big)} \end{aligned}

The logarithmic decrement can also be determined as follows;

\begin{aligned} \delta &= log_e \bigg({x_0 \over x_1}\bigg)= log_e \bigg({x_1 \over x_2}\bigg)= log_e \bigg({x_2 \over x_3}\bigg) = \dotso = log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \text {Adding upto 'n' terms}\\ n\delta &= log_e \bigg({x_0 \over x_1} \bigg) + log_e \bigg({x_1 \over x_2}\bigg) + log_e \bigg({x_2 \over x_3}\bigg) + \dotso + log_e \bigg({x_{n-1} \over x_n}\bigg) \\ \therefore \ \ n\delta &= log_e \bigg({x_0 \over x_1} \ . {x_1 \over x_2} \ . {x_2 \over x_3} \ . \dots \ . {x_{n-1} \over x_n}\bigg) \\ \text {Or} \qquad \\ n\delta &= log_e \bigg({x_0 \over x_n}\bigg) \\ \therefore \quad \delta &= {1 \over n} log_e \bigg({x_0 \over x_n}\bigg) \end{aligned}

where,
$x_0$ = amplitude at the starting position
$x_n$ = amplitude after ‘n’ cycles

## Static and Dynamic Balancing

When machine is in working condition different forces are acting on it which may cause machine to vibrate and cause damage to machine parts. The different forces acting on machine parts are static forces and dynamic forces.

### Static force

The force which depends on weight of a body, is known as static force. (Generally, a static force acts when vibrations occurs in same plane.)

### Dynamic or Inertia force

The force which depends on acceleration of a body, is known as dynamic force. (Generally, dynamic force acts when vibrations occurs in different planes.)

Due to these forces, the efficiency of the system decreases and life span of the system also decreases.
Due to these forces, the machine starts vibrating and sometimes when the vibrations increases, the machine would lift from it’s position and cause damage to other machine or human. So to avoid this, foundation is made below the machine (as you can see in the Fig-1 below), which absorbs the vibration and protects the machine against causing damage. Thus, balancing of the machine is required.

Fig-1

### Balancing

Balancing is the process of eliminating, the effect of static forces and dynamic forces acting on machine components.

## When is system said to be unbalanced?

In any system with one or more rotating masses, if the centre of mass of the system does not lie on the axis of rotation, then the system is said to be unbalanced.

As you can see in the Fig-2 we have a rotor which is mounted on a shaft and the shaft has its own axis of rotation. Now you can see that the C.G(centre of gravity) of the rotor is at a distance r from the axis of rotation of the shaft, so when the rotor will start to rotate, a centrifugal force will act on it which is in the outward direction as you can see in Fig-3. Due to this force our system will become unbalanced and it will start to vibrate.

Fig-2

Fig-3

### Static balancing

A system is said to be statically balanced, if the centre of masses(C.G) of the system lies on the axis of rotation.

#### Condition for Static Balancing

The resultant of all the centrifugal forces (dynamic forces) acting on the system during rotation must be zero.

\begin{aligned} \small ∑ \text {Centrifugal forces acting on the system} &= \text {zero}\\ i.e. \ +mrω^2 – mrω^2 &= 0 \end{aligned}

### Dynamic balancing

A system is said to be dynamically balanced, if it satisfies following two conditions:-

1. The resultant of all the dynamic forces acting on the system during rotation must be zero.
$∑ \text {Dynamic forces acting on the system} = \text {zero}$
1. The resultant couple due to all the dynamic forces acting on the system during rotation, about any plane, must be zero.
$∑ \text {couple} = \text {zero}$

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:
$t_p = {2\pi \over \omega_d} = {2\pi \over \sqrt{(1 - \xi^2)}\omega_n} sec.$

## Over-Damped System (​ξ>1)

Fig-1 (Over damped system)

We know that the characteristic equation of the damped free vibration system is,

$mS^2 + cS + K = 0$

This is a quadratic equation having two roots $S_1$ and $S_2$;

$S_{1,2} = {-c \over 2m} \pm \sqrt{\bigg({-c \over 2m}\bigg)^2 - {K \over m}}$

In order to convert the whole equation in the form of $\xi$ , we will use two parameters, critical damping coefficient '$c_c$' and damping factor '$\xi$'. So the roots $S_1$ and $S_2$ can be written as follows;

\begin{aligned} \xi= {c \over c_c} \quad \text {OR} \quad \xi &= {c \over 2m\omega_0} \quad (\text {as} \quad c_c=2m\omega_n)\\ \therefore {c \over 2m} &= \xi \omega_n \end{aligned}

where, $\omega_n$ = natural frequency of undamped free vibration = $\sqrt{K \over m}$ rad/s

$\therefore \omega_n^2 = {K \over m}$

So we can write roots $S_1$ and $S_2$ and as;

\begin{aligned} S_{1,2} &= -\xi \omega_n \pm \sqrt{(\xi \omega_n)^2 - \omega_n^2}\\ \therefore S_{1,2} &= [-\xi \pm \sqrt{\xi^2 - 1}]\omega_n\\ \therefore \ \ S_{1} &= [-\xi + \sqrt{\xi^2 - 1}]\omega_n\\ \text{And, }\\ S_{2} &= [-\xi - \sqrt{\xi^2 - 1}]\omega_n \end{aligned}

## Overdamped system (ξ>1)

If the damping factor ‘$\xi$’ is greater than one or the damping coefficient ‘$c$’ is greater than critical damping coefficient ‘$c_c$’, then the system is said to be over-damped.

$\xi > 1 \quad \text{Or} \quad {c \over c_c}\quad \text{Or} \quad c > c_c$

In overdamped system, the roots are given by;

\begin{aligned} S_{1} &= [-\xi + \sqrt{\xi^2 - 1}]\omega_n\\ \text{And, }\\ S_{2} &= [-\xi - \sqrt{\xi^2 - 1}]\omega_n \end{aligned}

For $\xi > 1$ , we get $S_1$ and $S_2$ as real and negative so we get,

\begin{aligned} x &= Ae^{S_1t} + Be^{S_2t}\\ \therefore x &= Ae^{[-\xi + \sqrt{\xi^2-1}]\omega_nt} + Be^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} \quad \dots \dots \text{(1)} \end{aligned}

Now differentiating equation (1) with respect to ‘t’, we get;

\begin{aligned} \mathring x = &Ae^{[-\xi + \sqrt{\xi^2-1}]\omega_nt}[-\xi + \sqrt{\xi^2-1}]\omega_n \ + \\ &Be^{[-\xi - \sqrt{\xi^2-1}]\omega_nt} [-\xi - \sqrt{\xi^2-1}]\omega_n\quad \dots \dots \dots \text{(2)} \end{aligned}

Now, let at
$t = 0$ : $x = X_0$
$t = 0$ : $\mathring x =0$

Substituting this value in equation (1) we get;

$X_0 = A+B\quad \dots \dots \dots \text{(3)}$

Substituting this value in equation (2) we get;

$0= A[-\xi + \sqrt{\xi^2-1}]\omega_n + B[-\xi - \sqrt{\xi^2-1}]\omega_n \quad \dots \dots \text{(4)}$

From equation (3), $B= X_0 - A$ and putting value of B in (4);

\begin{alignedat}{2} &\therefore \qquad 0 &=& \ A[-\xi + \sqrt{\xi^2-1}]\omega_n + (X_0 - A)[-\xi - \sqrt{\xi^2-1}]\omega_n\\ &\therefore \qquad 0 &=& \ -A\xi + A\sqrt{\xi^2-1} + X_0 [-\xi - \sqrt{\xi^2-1}]+A\xi+A\sqrt{\xi^2-1}\\ &\therefore \qquad 0 &=& \ 2A\sqrt{\xi^2-1} + X_0 [-\xi - \sqrt{\xi^2-1}]\\ &\therefore 2A\sqrt{\xi^2-1} \ &=& \ X_0 [\ \xi + \sqrt{\xi^2-1}]\\ &\therefore \qquad A &=& \ {X_0 [\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1} }\quad \dots \dots \text{(5)} \end{alignedat}

Now,
Putting $A=X_0-B$, in equation (4);

\begin{alignedat}{2} &\therefore \qquad 0 &=& \ (X_0-B)[-\xi + \sqrt{\xi^2-1}]\omega_n + B[-\xi - \sqrt{\xi^2-1}]\omega_n\\ &\therefore \qquad 0 &=& \ X_0[-\xi + \sqrt{\xi^2-1}] + B\xi - B\sqrt{\xi^2-1} - B\xi - B\sqrt{\xi^2-1}\\ &\therefore \qquad 0 &=& \ X_0[-\xi + \sqrt{\xi^2-1}] - 2B\sqrt{\xi^2-1}\\ &\therefore 2B\sqrt{\xi^2-1} &=& \ X_0[-\xi + \sqrt{\xi^2-1}]\\ &\therefore \qquad B &=& \ {X_0[-\xi + \sqrt{\xi^2-1}] \over 2\sqrt{\xi^2-1}} \quad \dots \dots \text{(6)} \end{alignedat}