Question 5.4: 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. 5.7, due to the periodic forcing function F(t) given in Example 5.2. Neglect the gravity effect.
The equation of motion of this system is given by
m_{e} \ddot{θ} + c_{e} \dot{θ} + k_{e}θ = F(t) l
where
m_{e} = \frac{ml²}{3}, c_{e} = ca², k_{e} = ka²
where c and k are, respectively, the damping and spring coefficients. It was shown in Example 5.2 that the periodic forcing function F(t) can be written as
F(t) = \frac{a_{0}}{2} +\sum\limits_{n=1}^{\infty }{a_{n} cos nω_{f}t}
where
a_{0} = F_{0}
Since
cos nω_{f}t = sin (nω_{f}t + \frac{π}{2})
the periodic force function F(t) can be written as
F(t) = \frac{a_{0}}{2} + \sum\limits_{n=1}^{\infty }{a_{n}} sin (nω_{f}t + \frac{π}{2})
= \frac{F_{0}}{2} + \frac{2F_{0}}{π} sin(nω_{f}t + \frac{π}{2}) − \frac{2F_{0}}{3π}sin (3ω_{f}t + \frac{π}{2})+ . . .
where the angle Φ_{n}, n = 1, 2, . . ., of Eq. 5.29 is given by
F_0 = \frac{a_0}{2}, \quad F_n = \sqrt{a^2_n + b^2_n}, \quad \phi_n = \tan^{-1} (\frac{a_n}{b_n}) (5.29)
Φ_{n} = \frac{π}{2}, n= 1, 2, . . .
The equation of motion of this single degree of freedom system can be written as
m_{e}\ddot{θ} + c_{e} \dot{θ} + k_{e}θ = \frac{F_{0}l}{2} + \frac{2F_{0}l}{π}sin (nω_{f}t + \frac{π}{2}) − \frac{2F_{0}l}{3π}sin (3ω_{f}t + \frac{π}{2})+ . . .
Consequently,
x_{p}0 = \frac{F_{0}l}{2k_{e}}
Therefore,
x_{p} = \frac{F_{0}l}{2k_{e}}+\sum\limits_{n=1, 3, 5}^{\infty }{(−1)^{(n−1)/2}\frac{2F_{0}l/nπk_{e}}{\sqrt{ (1 − r_{n}^{2})^{2} + (2r_{n}ξ)^{2}}}}sin(nω_{f}t + \frac{π}{2}− ψ_{n})
= \frac{F_{0}l}{2k_{e}}+\sum\limits_{n=1, 3, 5}^{\infty }{(−1)^{(n−1)/2}\frac{2F_{0}l/nπk_{e}}{\sqrt{ (1 − r_{n}^{2})^{2} + (2r_{n}ξ)^{2}}}} \cos(nω_{f}t − ψ_{n})where the damping factor ξ is given by
ξ = \frac{c_{e}}{C_{c}} = \frac{ca^{2}}{2m_{e}ω} = \frac{3ca^{2}}{2ml^{2}ω}
and ω is the system natural frequency defined by
ω = \sqrt{\frac{k_{e}}{m_{e}}} =\sqrt{\frac{3ka^{2}}{ml^{2}}}