Название | Engineering Acoustics |
---|---|
Автор произведения | Malcolm J. Crocker |
Жанр | Техническая литература |
Серия | |
Издательство | Техническая литература |
Год выпуска | 0 |
isbn | 9781118693827 |
where F is an n‐dimensional complex column vector of dynamic amplitude forces. We assume harmonic solutions of the form
where A is a vector of undetermined amplitudes. Substituting Eq. (2.36) into (2.35) leads to
A unique solution of Eq. (2.37) exists unless
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]
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.
Example 2.6
Let consider the two‐degree of freedom system of Example 2.5. Assume that a force F0 ejωt is applied to mass m1 and no force is applied to mass m2. Then, Eq. (2.37) becomes
Solution
Solving the system of Eq. (2.43) simultaneously, we obtain that
(2.45)
and the ratio
(2.46)
where
It is noted from Eq. (2.44) that the steady‐state amplitude of the mass m1 will become zero when r2 = 1, i.e. when the excitation frequency is
Example
Repeat the problem discussed in Example 2.6 but now assume that a force F0 ejωt is applied to the mass m2 and no force is applied to the mass m1.
Solution
Equation (2.37) becomes now
(2.47)