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and<br />
⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫<br />
⎨η 1 ′ (0) ⎬<br />
⎩<br />
η 2 ′ (0) ⎭ = 0.925 −0.657 ⎨x ⎣ ⎦<br />
′ 1 (0) ⎬<br />
0.38 1.6<br />
⎩<br />
x ′ 2 (0) ⎭<br />
⎡<br />
⎤ ⎧ ⎫<br />
0.925 −0.657 ⎨1.5⎬<br />
= ⎣ ⎦<br />
0.38 1.6<br />
⎩<br />
3<br />
⎭<br />
⎧ ⎫<br />
⎨−0.584⎬<br />
=<br />
⎩<br />
5.37<br />
⎭<br />
Each of these EOM are solved using any of the standard methods. This results in solutions η 1 (t) and η 2 (t) .<br />
Hence the following EOM’s are solved<br />
and also<br />
η ′′<br />
1 (t) + 3.47η 1 (t) = −0.219 sin (5t)<br />
η 1 (0) = −0.657<br />
η ′ 1 (0) = −0.584<br />
η ′′<br />
2 (t) + 0.192η 2 (t) = 0.534 sin (5t)<br />
η 2 (0) = 1.6<br />
η ′ 2 (0) = 5.37<br />
The solutions η 1 (t) , η 2 (t) are found using basic methods shown in other parts of these notes. The last step<br />
is to transform back to normal coordinates by applying step 7<br />
⎧ ⎫ ⎡<br />
⎤ ⎧ ⎫<br />
⎨x 1 (t) ⎬<br />
⎩<br />
x 2 (t)<br />
⎭ = 0.925 0.380 ⎨η ⎣ ⎦ 1 (t) ⎬<br />
−0.219 0.534<br />
⎩<br />
η 2 (t)<br />
⎭<br />
⎧<br />
⎫<br />
⎨ 0.925 η 1 + 0.38η 2 ⎬<br />
=<br />
⎩<br />
0.534 η 2 − 0.219 η<br />
⎭<br />
1<br />
Hence<br />
and<br />
x 1 (t) = 0.925η 1 (t) + 0.38η 2 (t)<br />
x 2 (t) = 0.534η 1 (t) − 0.219η 2 (t)<br />
The above shows that the solution x 1 (t) and x 2 (t) has contributions from both nodal solutions.<br />
2 Fourier series representation of a periodic function<br />
Given a periodic function f (t) with period T then its Fourier series approximation ˜f (t) using N terms is<br />
(<br />
˜f (t) = 1 N<br />
)<br />
2 F ∑<br />
2π<br />
in<br />
0 + Re F n e T t<br />
= 1 2 F 0 + 1 2<br />
= 1 2<br />
N∑<br />
n=−N<br />
n=1<br />
N∑<br />
F n e<br />
n=1<br />
2π<br />
in<br />
F n e T t<br />
in<br />
2π<br />
T t + F ∗ ne<br />
2π<br />
−in T t<br />
14