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Phase of response complex<br />
amplitude for underdamped<br />
and when r1<br />
Phase will be from -90 to -<br />
180 degrees<br />
Phase of response complex<br />
amplitude for<br />
underdamped and when<br />
r=1<br />
Phase will -90 degrees<br />
2r<br />
2r<br />
1 r 2<br />
1 r 2<br />
Examples. System has ζ =<br />
)<br />
0.1 and<br />
√<br />
m = 1, k = 1 subjected for force 3 cos (0.5t) find the steady state solution.<br />
Answer y (t) = Re<br />
(Ŷ e<br />
iϖt<br />
, ω n = = 1 rad/sec, hence r = 0.5 under the response is<br />
k<br />
m<br />
(∣ ∣∣ y (t) = Re Ŷ ∣ e iϖt)<br />
∣<br />
= ∣Ŷ ∣ cos (ϖt)<br />
= F k<br />
1<br />
√(1 − r 2 ) 2 + (2ζr) 2 cos<br />
(<br />
( ))<br />
2 (0.1) 0.5<br />
.5t − tan −1 1 − 0.5 2<br />
1<br />
= 3<br />
cos (.5t − 7.59 √(1 ◦ )<br />
− 0.5 2 ) 2 + (2 (0.1) 0.5) 2<br />
= 3. 964 9 cos (.5t − 7.59 ◦ )<br />
5 Summary table of solutions<br />
5.1 Harmonic loading mu ′′ + cu ′ + ku = F sin ϖt<br />
The equation of motion can also be written as u ′′ + 2ζωu ′ + ω 2 u = F m<br />
sin ϖt.<br />
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