The vibrations which the system executes under an external periodic force are called as forced vibrations. Thus, in the case of forced vibrations, there is an impressed force on the system which keeps it vibrating. This force is called as external excitation. The excitation may be periodic, impulsive or random in nature. Again, the periodic force may be harmonic and non-harmonic. Vibrations because of impulsive force are called as transient vibrations. Earthquake is an example of random excitation forces.

We will see the harmonic excitation's. The vibrations of air compressors, internal combustion engines, machine tools are the examples of forced vibrations due to harmonic excitation's.

(A) FORCED VIBRATIONS WITH CONSTANT HARMONIC EXCITATION.

Consider a spring mass system having viscous damping, excited by a sinusoidal forcing function, $F_o$ sin wt shown in figure (1a) and free body diagram in figure (1b).

Applying Newton’s second law of motion,

$m\ddot{x} \ = \ mg \ + \ F_0 \ sin_{wt} \ – \ k (x + \triangle_{st}) - c\dot{x}$

$m\ddot{x} \ = \ mg \ + \ F_0 \ sin_{wt} \ – \ kx \ – \ k \triangle_{st} - c\dot{x}$

$m\ddot{x} \ + \ cx \ + \ kx \ = \ F_0 \ sin_{wt}$ - - - - (1)

Equation (1) is a linear, non homogeneous, second order differential equation for the forced vibration system. The solution of this equation consists of two parts: complementary function and particular integral.

The complementary function is obtained from the differential equation:

$m\ddot{x} + c\dot{x} + kx = 0$

This equation is the same as that obtained for damped free vibration system. The solution of this equation is given by,

$X_c \ = \ X \ e^{-\xi w_n t} \ sin(w_dt \ + \ \phi)$ - - - - (2)

Here, the vector method of finding the particular solution will be used to give more insight into the behavior of the system. The particular integral (solution) is a steady harmonic oscillation having frequency equal to that of excitation and the displacement vector lags the force vector by some angle. Let us therefore, assume that the particular solution is

$X_p \ = \ X sin \ (wt - \phi)$ - - - -(3)

Where X is the amplitude of the system, $\pi$ is the angle by which the displacement vector lags the force vector and $X_p$ corresponds to the particular solution.

Differentiating equation (3) twice, we have

$\dot{X_p} = w \ X \ cos (wt - \phi) = w \ X \ sin (wt - \phi + \pi / 2)$

$\ddot{X_p} = -w^2 \ X \ sin (wt - \phi) = w^2 \ X \ sin (wt - \phi + \pi )$

Substituting the values of $X_p, \dot{X_p}$ and $\ddot{X_p}$ for x,$\dot{x}$ and $\ddot{x}$ in equation (1),

we have

$mw^2 \ X \ sin (wt \ - \ \phi \ + \ \pi ) + cw \ X \ sin (wt \ - \ \phi \ + \ \pi 2) \ + \ K \ X \ sin (wt \ - \ \phi)$

$ = F_0 \ sin \ wt \ F_0 \ sin \ wt \ – \ k x \ sin \ (wt \ - \ \phi) \ – \ cw \ X \ sin (wt \ - \ \phi \ + \ \frac{\pi }{2}) \ – \ m w^2 \ X \ sin (wt \ - \ \phi \ + \ \pi ) \ = \ 0$ - - - -(4)

The four terms in equation (4), including their signs, represents the four forces, in magnitude and direction, acting on the body.

The forces are the :

• Impressed force,

• Spring force,

• Damping force and

• Inertia force respectively and their sum equal to zero.

This is nothing but the D Alembert principle. The vector representation of equation (4) is shown below in figure (a).

The impressed force on the body is $F_0$ sin wt [from equation (4)], i.e. the force vector $F_0$ is acting at an angle wt from the reference axis and is so shown in the figure below.

The displacement vector lags the force vector by an angle $\phi$ [from equation (3)] and therefore this vector acts at an angle $(wt \ - \ \phi)$ from the reference axis.

The spring force acting on the body is $– KX \ sin (wt \ - \ \phi)$ [from equation (4)], which means that a vector – KX acting at an angle $(wt \ - \ \phi)$ or vector KX acting in opposite direction to ($wt \ - \ \phi$)

Thus the spring force on the body acts in a direction opposite to that of displacement. In a similar way the damping and inertia force vectors are represented in the vector diagrams. The actual forces acting on the body are the components of the above four forces vectors along the perpendicular axis. Since the body is in equilibrium under the four forces, these vectors must form a closed polygon.

The following points are observed from the vector diagram:

1] The displacement lags the impressed force by an angle $\pi$.

2] The spring force is always opposite in direction to the displacement.

3] The damping force lags the displacement by 90˚. Since the velocity leads the displacement by 90˚, it follows that the damping force is always opposite in direction to the velocity.

4] The inertia force is in phase with the displacement.

It may be noted that the relative positions of the vectors and their magnitudes do not change with time.

In order to find the values of X and $\phi$ in equation (3), we consider the right angled triangle OAB after dropping OB perpendicular to AB.

$\therefore$ $X = \frac{F_0}{\sqrt{ [(k-mw^2)^2 + (cw)^2]}}$ - - - -(5)

[$\because$ $F^2_0 = (kx – mw^2 X)^2 + (cwX)^2]$

[$\therefore$ $F^2_0 = x^2 [(k – mw^2)^2 + (cw)^2 ]$

And $\phi \ = \ tan^{-1} [\frac{w}{k – mw^2}]$ - - - -(6)

In order to obtain the above equations in a non dimensional form, we divide the numerator and denominator by k,

$X = \frac{F_0 / k}{\sqrt{ (1 - \frac{mw^2}{k})^2 + (\frac{ew}{k})^2}} $ - - -(6a)

And $\phi \ = \ tan^{-1} [ \frac{cw/k}{1 - \frac{mw^2}{k}}] $- - - -(6b)

Now, $\frac{mw^2}{k} \ = \ \frac{w^2}{w^2_n} $ [$\because$ $w_n \ = \ \sqrt{ \frac{k}{m}}]$ - - - -(7)

Or $\frac{cw}{k} \ = \ 2 \ \xi \ \frac{w}{w_n}$ - - - - (8)

In the above equations (7) and (8), $w_n$ is the undamped natural frequency of the system and $\xi$ is the damping factor.

Let $\frac{F_o}{k} = X_{st}$ - - - -(9)

Where $X_{st}$ may be defined as zero frequency deflection of the spring mass system under a steady force $F_0$

Substituting the expression (8) and (9) in equation [(6a) and (6b)], we have

$X = \frac{X_{st}}{\sqrt{ [ 1 – (\frac{w}{w_n})^2 + [ 2 \xi \frac{w}{w_n}]^2}}$ - - - -(10)

And $\phi \ = \ tan^{-1} [ \frac{2 \xi \frac{w}{w_n}}{1 – (\frac{w}{w_n})^2}]$ - - - - (11)

Hence the particular solution of equation (3) may be written as

$X_p \ = \ \frac{X_{st} sin (wt - \phi)}{\sqrt{ [1- (\frac{w}{w_n})^2]^2 + [ 2\xi \frac{w}{w_n}]^2}}$ - - - -(12)

Therefore the complete solution is $X = X_c + X_p$, the values of $X_c$ and $X_p$ being given by equation (2) and equation (12).

$X = Xe^{- \xi w_n t} \ sin [ \sqrt{1 - \xi^2} w_nt + \phi_2] + \frac{X_{st} sin (wt - \phi)}{\sqrt{[ 1 – (\frac{w}{w_n})^2]^2 + [2 \xi \frac{w}{w_n}]^2}}$ - - - -(13)

This is the complete solution to an under-damped system subjected to sinusoidal excitation. The first part of the complete solution, i.e. complementary function is seen to decay with force and vanishes ultimately. This part, in engineering practice, is commonly called as Transient vibrations. The second part, i.e. the particular solution, is seen to be a sinusoidal vibration with a constant amplitude and is called as Steady state vibrations. The complete solution is a super position of transient and steady vibrations and is shown in the figure 3.

Physical interpretation of complete solution.

(a) Transient state. (complimentary)

(b) Steady state (Particular solution)

(c) Complete solution.

Magnification factor.

$M.F. \ = \ \frac{x}{X_{st}} = \frac{1}{\sqrt{ 1 – (\frac{w}{w_n}^2]^2 + [ 2 \xi \frac{w}{w_n}]^2}}$ - - - (14) (from equation 10)

The equation for phase lag is rewritten again from equation (11) as:

$\phi \ = \ tan^{-1} [ \frac{2 \xi \frac{w}{w_n}}{1 – (\frac{w}{w_n})^2}]$ - - - - (15)

M.F. The ratio of steady state amplitude to the zero frequency deflection, i.e. X / X_{st} is defined as magnification factor. The dimensionless plots of magnification versus frequency ratio is shown in figure 4. It is known as frequency response curve. At zero frequency the magnification is unity and is independently of the damping.

i.e. $X = X_{st}$

which itself is deflection of the zero frequency deflection.

At very high frequency the magnification tends to zero or the amplitude of vibration becomes very small.

At resonance $(w = w_n)$ , the amplitude of vibration becomes excessive. For zero damping at resonance the amplitude is infinite theoretically. With increase in frequency the damping force vector $cwX$ grows larger. Angle $\phi$ has also to increase so that components of $F_0$ perpendicular to X direction may balance the increasing damping force. Also inertia force vector grows $mw^2 \ X$

If we increase the frequency time comes when the spring force and the inertia force vectors are equal ad oppose

$kx = mw^2X$

Or $W = \sqrt{\frac{w}{m}} = w_n$

This is resonance condition of the system and the vector diagram becomes a rectangle. The impressed force completely balances the damping force and $\phi = 90˚$

Or the amplitude at resonance is

$X_r = \frac{F_0}{cw} = \frac{F_0 / k}{cw / k}$

$= \frac{X_{st}}{2 \xi (\frac{w}{w_n})}$

$= \frac{X_{st}}{2 \xi}$ $(\therefore w = w_n)$

$\therefore$ $\frac{X_r}{X_{st}} = \frac{1}{2 \xi}$

It may be obtained by putting $w/w_n = 1$ in equation of M.F.

$M.F. = \frac{X}{X_{st}} = \frac{1}{\sqrt{ [1 – (1)^2]^2 + [2 \xi]^2}}$

$= \frac{1}{\sqrt{4 \xi^2}}$

$\frac{1}{2 \xi}$

Thus at resonant frequency, the phase angle is 90 degree and the impressed force balances to damping factor. The amplitude at resonance is inversely proportional to the damping factor. By seeing frequency response curve, it is seen that the maximum amplitude occurs not at the resonance frequency but a little towards its left.

This shift increases with increase in damping.

For zero damping, the maximum amplitude (infinite value), of course, is obtained at the resonant frequency.

The frequency at which the maximum amplitude occurs which is not resonance frequency (except for zero damping) is said to be $w_p \rightarrow$ at which peak amplitude occurs.