Question 5.3: Determine the steady state response of the single degree of ...
Determine the steady state response of the single degree of freedom system shown in Fig. 5.6 to the forcing function F(t), where F(t) is the periodic function given in Example 5.1.
The equation of motion of the damped single degree of freedom system due to the periodic excitation is given by
m\ddot{x} + c\dot{x} + kx = F(t)
It was shown in Example 5.1 that F(t) can be expressed in terms of the harmonic functions as
F(t) =\frac{a_{0}}{2} + \sum\limits_{n=1}^{\infty }{b_{n}} \sin ω_{n} t
where ω_{n} = nω_{f} = 2πn/T_{f}, \ a_{0} = F_{0}, and
b_{n} = \begin{cases} 2F_{0}/nπ & \text{if n is odd }\\0 & \text{if n is even} \end{cases}
The equation of motion can then be written as
m \ddot{x} + c\dot{x} + kx = \frac{a_{0}}{2} + \sum\limits_{n=1}^{\infty }{b_{n}} \sin ω_{n} t
= \frac{F_{0}}{2}+ \frac{2F_{0}}{π} \sin ω_{f}t + \frac{2F_{0}}{3π} \sin 3 ω_{f} t + . . .
Clearly, x_{p0} is given by
x_{p0} = \frac{F_{0}}{2k}
Since, in this case, a_{n} = 0, \ n = 1, 2, . . ., it is clear that the phase angle \phi_{n} in Eq. 5.35 is zero, and as such, x_{pn} is defined as
\sum\limits_{n=1}^{\infty }{x_{pn}} = \sum\limits_{n=1}^{\infty }\frac{F_n/k}{\sqrt{(1 − r^{2}_{n})^2 + (2r_{n}ξ)^2}} \sin( ω_{n}t + \phi _n − ψ_{n}) (5.35)
x_{pn} = \frac{b_{n}/k}{\sqrt{(1 − r^{2}_{n})² + (2r_{n}ξ)²}} sin(nω_{f}t − ψ_{n})
= \frac{2F_0/nπk}{\sqrt{(1 − r^{2}_{n})² + (2r_{n}ξ)²}} \sin(nω_{f}t − ψ_{n}), n = 1, 3, 5, . . .
It follows that
x_{p} = x_{p0}+ \sum\limits_{n=1.3.5}^{\infty }{x_{pn}}
\quad = \frac{F_0}{2k} + \sum\limits_{n=1.3.5}^{\infty }\frac{2F_0/nπk}{\sqrt{(1 − r^{2}_{n})² + (2r_{n}ξ)²}} \sin(n ω_{f}t − ψ_{n})
\quad = \frac{F_0}{k} \left[\frac{1}{2} + \sum\limits_{n=1.3.5}^{\infty }\frac{2}{\sqrt{(1 – r^2_n)^2 + (2r_n ξ )^2}} \sin(nω_{f}t − ψ_{n})\right]