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By forced vibration, we mean that the system is vibrating under the influence of continuous (external) forces that do not cease. The total response of a multi‐degree of freedom system due to a force excitation is the sum of a homogeneous solution and a particular solution. The homogenous solution depends upon the system properties while the particular solution is the response due to the particular form of excitation. The homogenous solution is often ignored for a system subjected to a periodic vibration for being of lesser practical importance than the particular solution. For a general form of excitation, a closed‐form solution of a multi‐degree of freedom system can be very difficult to obtain and numerical methods are often used.

The equations of motion of an n‐degree‐of‐freedom undamped linear system excited by simple harmonic forces at some arbitrary angular forcing frequency ω (all excitation terms at the same phase) can be expressed in matrix form as

(2.35)

where F is an n‐dimensional complex column vector of dynamic amplitude forces. We assume harmonic solutions of the form

(2.36)

where A is a vector of undetermined amplitudes. Substituting Eq. (2.36) into (2.35) leads to

(2.37)

A unique solution of Eq. (2.37) exists unless

(2.38)

which has the same form as Eq. (2.26). Equation (2.38) is satisfied only when the forcing frequency coincides with one of the system's natural frequencies. In this condition, called resonance, the response of the system grows linearly with time and thus use of the solution Eq. (2.36) is unsuitable. When a solution of Eq. (2.37) exists, the amplitudes can be determined as [13]

(2.39)

If we consider the two‐degree of freedom system discussed in Example 2.5 but now harmonic force excitations of frequency ω and amplitude F₁ and F₂ are applied to the masses m₁ and m₂, respectively (see Figure 2.13), the equations of motion are

(2.40a)

and

(2.40b)

Figure 2.13 Harmonically forced two‐degree‐of‐freedom system.

The particular solution is given by Eq. (2.36) as

(2.41)

Therefore, Eq. (2.37) becomes

(2.42)

which has to be simultaneously solved to find the displacement amplitudes A₁ and A₂.