Question 4.4: Determine an expression for steady state response of an unda...

Let us write the expression for the steady state response of SDOF system as
given in Equations 4.10(a) and 4.10(b).

m\ddot{x} (t)+c\dot{x} (t)+kx(t)=p(t)=e^{i\bar{\omega }t }             (4.10)

x(t)=H(\bar{\omega } )p(t)

or, x(t)=H(\bar{\omega } )C_{n}e^{i\bar{\omega } t}

where, H(\bar{\omega } )=\frac{1}{k(1-\beta ^{2}+i2\beta \xi )} 

For an undamped system, ξ = 0,

∴ H(\bar{\omega } )=\frac{1}{k(1-\beta ^{2})} =\frac{1}{k\left[1-\left\lgroup\frac{n\bar{\omega }_{1} }{\omega } \right\rgroup^{2} \right] }

or,    H(\bar{\omega } )=\frac{1}{k\left[1-\left\lgroup\frac{n}{4} \right\rgroup^{2} \right] }

Also, derived earlier,    C_{n}=0  for n even

C_{n}=\frac{-2p_{0}i}{n\pi }    for n odd

∴ n^{th} term of response can be written as x_{n}=H(ϖ) C_{n}

or, x_{n}=\frac{-i2p_{0}}{n\pi k\left[1-\left\lgroup\frac{n}{4} \right\rgroup^{2} \right] }     for n odd

= 0  for n even

∴   \left|x_{n}\right| =\left|\frac{2p_{0}/\pi k}{n\left[1-\left\lgroup\frac{n}{4} \right\rgroup^{2} \right] } \right|       for n odd

Alternatively, this can be re-written as follows:

\frac{x_{n}\pi /2}{p_{0}/k} =\frac{1}{n\left[1-\left\lgroup\frac{n}{4} \right\rgroup^{2} \right] } =\alpha             for n odd

This can be plotted as shown in Figure 4.7.