Dynamics cheat sheet

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