Question 4.2: Find the steady state response of an undamped SDOF system to...
Find the steady state response of an undamped SDOF system to a square pulse. Given ω = 4 ϖ_{1}.
The steady state response of an undamped SDOF subjected to a harmonic
excitation p_{0} sin ϖ t is given as follows:
x(t)=\frac{p_{0}}{k} \frac{1}{(1-\beta ^{2})} \sin \bar{\omega } t
where β = ϖ/ω = n ϖ_{1}/ω = β_{n}
The load term can be written as follows:
p(t)=\frac{4p_{0}}{\pi } \sum\limits_{n=1,3,5,}^{\infty }\frac{1}{n} \sin (n\bar{\omega }_{1}t ) for n = 1, 3, 5, etc.
= 0 for n = 0, 2, 4,etc
The n^{th} term of the steady state response is of the form:
x_{n}=\frac{4p_{0}}{n\pi } \left\lgroup\frac{1}{k} \right\rgroup \frac{\sin (n\bar{\omega }_{1}t )}{1-\beta ^{2}_{n} }
Hence, the steady state response is given by
x(t)=\sum\limits_{n=1,3,5}^{\infty }\frac{4p_{0}}{k\pi } \frac{1}{n} \frac{\sin (n\bar{\omega }_{1}t )}{\left\lgroup1-\left\lgroup\frac{n\bar{\omega }_{1} }{\omega } \right\rgroup ^{2} \right\rgroup }
Let us plot the amplitude of each harmonic term of load as well as displacement versus frequency.
p_{n}=\frac{4p_{0}}{n\pi } or \frac{p_{n}}{p_{0}} =\frac{4}{n\pi }
or \frac{p_{n}}{p_{0}}\frac{\pi}{4 }=\frac{1}{n} where n = 1, 3, 5, etc.
= 0 n = 0, 2, 4, etc.
Its plot is shown in Figure 4.5(a).
Similarly, the amplitude of each harmonic of the response term can be plotted as follows:
\frac{x_{n}}{\frac{p_{0}}{k} } =\frac{4}{n\pi \left\lgroup1-\left\lgroup\frac{n\bar{\omega }_{1} }{\omega } \right\rgroup^{2} \right\rgroup } for n = 1, 3, 5
= 0 for n = 0, 2, 4, 6
Given ϖ_{1}/ω = ¼
∴ \frac{x_{n}}{\frac{p_{0}}{k} } =\frac{4}{n\pi \left\lgroup1-\left\lgroup\frac{n}{4} \right\rgroup^{2} \right\rgroup } for n = odd
Its plot is shown in Figure 4.5(b).
