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.

Question

Accepted Answer

The equation of motion of this system is given by
[latex]m_{e} \ddot{θ} + c_{e} \dot{θ} + k_{e}[/latex]θ = F(t) l
where
[latex]m_{e} = \frac{ml²}{3}[/latex], [latex]c_{e}[/latex] = ca², [latex]k_{e}[/latex] = 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) = [latex]\frac{a_{0}}{2} +\sum\limits_{n=1}^{\infty }{a_{n} cos nω_{f}t}[/latex]
where
[latex]a_{0} = F_{0}[/latex]

[latex]a_{n}=\begin{cases} 0 & \quad if n is even\ (−1)^{(n−1)/2} (2F_{0}/nπ) & if n is odd \end{cases}[/latex]

Since
cos n[latex]ω_{f}[/latex]t = sin (n[latex]ω_{f}t + \frac{π}{2}[/latex])

the periodic force function F(t) can be written as

F(t) = [latex]\frac{a_{0}}{2} + \sum\limits_{n=1}^{\infty }{a_{n}}[/latex] sin (n[latex]ω_{f}t + \frac{π}{2})[/latex]

= [latex]\frac{F_{0}}{2} + \frac{2F_{0}}{π}[/latex] sin(n[latex]ω_{f}t + \frac{π}{2}[/latex]) − [latex]\frac{2F_{0}}{3π}[/latex]sin (3[latex]ω_{f}t + \frac{π}{2})[/latex]+ . . .

where the angle [latex]Φ_{n}[/latex], n = 1, 2, . . ., of Eq. 5.29 is given by

[latex]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})[/latex] (5.29)

[latex]Φ_{n} = \frac{π}{2}[/latex], n= 1, 2, . . .

The equation of motion of this single degree of freedom system can be written as

[latex]m_{e}\ddot{θ} + c_{e} \dot{θ} + k_{e}[/latex]θ = [latex]\frac{F_{0}l}{2} + \frac{2F_{0}l}{π}[/latex]sin (n[latex]ω_{f}[/latex]t + [latex]\frac{π}{2}) − \frac{2F_{0}l}{3π}[/latex]sin (3[latex]ω_{f}t + \frac{π}{2}[/latex])+ . . .

Consequently,
[latex]x_{p}0 = \frac{F_{0}l}{2k_{e}}[/latex]

[latex]x_{pn}=\begin{cases}0 & if n is even \(−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}) & if n is odd\end{cases}[/latex]

Therefore,

[latex]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}}}}[/latex]sin(n[latex]ω_{f}t + \frac{π}{2}− ψ_{n})[/latex]

[latex]= \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})[/latex]

where the damping factor ξ is given by
ξ = [latex]\frac{c_{e}}{C_{c}} = \frac{ca^{2}}{2m_{e}ω} = \frac{3ca^{2}}{2ml^{2}ω}[/latex]

and ω is the system natural frequency defined by
ω = [latex]\sqrt{\frac{k_{e}}{m_{e}}} =\sqrt{\frac{3ka^{2}}{ml^{2}}}[/latex]

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