Resoluções Moyses volume 2 cap. 4

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𝑇𝑒𝑛𝑑𝑜 𝑞𝑢𝑒: 
 
 
 
𝑄 =
𝜔0
𝛾
= 10 
𝐸 𝑎 𝑎𝑛𝑒𝑟𝑔𝑖𝑎 𝑡𝑜𝑡𝑎𝑙 𝑑𝑎𝑑𝑎 𝑝𝑜𝑟: 
 
𝑠𝑢𝑎 𝑣𝑎𝑟𝑖𝑎çã𝑜 𝑡𝑒𝑚𝑝𝑜𝑟𝑎𝑙 𝑠𝑒𝑟á: 
 
𝑂𝑛𝑑𝑒, 𝑢𝑚 𝑜𝑠𝑐𝑖𝑙𝑎𝑑𝑜𝑟 ℎ𝑎𝑟𝑚ô𝑛𝑖𝑐𝑜 𝑎𝑚𝑜𝑟𝑡𝑒𝑐𝑖𝑑𝑜 é 𝑑𝑒𝑠𝑐𝑟𝑖𝑡𝑜 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑎 𝑒𝑞𝑢𝑎çã𝑜: 
 
𝑚�̈� + 𝑘𝑥 = −𝜌�̇� 
𝐴𝑔𝑜𝑟𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑒𝑠𝑡á ú𝑙𝑡𝑖𝑚𝑎 𝑒𝑞𝑢𝑎çã𝑜, 𝑛𝑎 𝑣𝑎𝑟𝑖𝑎çã𝑜 𝑑𝑎 𝑒𝑛𝑒𝑟𝑔𝑖𝑎: 
𝑑𝐸
𝑑𝑡
= �̇�(𝑚�̈� + 𝑘𝑥) = �̇�(−𝜌�̇�) 
𝑑𝐸
𝑑𝑡
= −�̇�2𝜌 
 
 
𝐴𝑔𝑜𝑟𝑎 𝑣𝑎𝑚𝑜𝑠 𝑣𝑒𝑟 𝑜 𝑞𝑢𝑒 𝑜 𝑒𝑥𝑒𝑟𝑐í𝑐𝑖𝑜 𝑝𝑒𝑑𝑒: 
" … 𝑎 𝑒𝑛𝑒𝑟𝑔𝑖𝑎 𝑡𝑜𝑡𝑎𝑙 𝑑𝑜 𝑜𝑠𝑐𝑖𝑙𝑎𝑑𝑜𝑟 𝑑𝑖𝑚𝑖𝑛𝑢𝑖 𝑛𝑢𝑚𝑎 𝑡𝑎𝑥𝑎, 𝑝𝑜𝑟 𝑠𝑒𝑔𝑢𝑛𝑑𝑜, 𝑖𝑔𝑢𝑎𝑙 𝑎 4 𝑣𝑒𝑧𝑒𝑠 𝑠𝑢𝑎 
𝑒𝑛𝑒𝑟𝑔𝑖𝑎 𝑐𝑖𝑛é𝑡𝑖𝑐𝑎 𝑖𝑛𝑠𝑡𝑎𝑛𝑡â𝑛𝑒𝑎. " 
 
𝑆𝑒 𝑒𝑙𝑒 𝑑𝑖𝑚𝑖𝑛𝑢𝑖, 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎 𝑞𝑢𝑒: 
𝑑𝐸
𝑑𝑡
< 0, 𝑒 𝑜 𝑟𝑒𝑠𝑡𝑜 𝑑𝑜 𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜 𝑠𝑒𝑟á: 
 
−
𝑑𝐸
𝑑𝑡
= 4𝐾 
−
𝑑𝐸
𝑑𝑡
= 4
1
2
𝑚�̇�2 
�̇�2𝜌 = 2𝑚�̇�2 
𝜌 = 2𝑚 
𝜌
𝑚
= 2 
 
 
𝑙𝑜𝑔𝑜: 
𝛾 = 2 𝑠−1 
 
𝑉𝑜𝑙𝑡𝑎𝑛𝑑𝑜 𝑎 𝑛𝑜𝑠𝑠𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑒𝑞𝑢𝑎çã𝑜: 
𝑄 =
𝜔0
𝛾
= 10 
𝜔0
2
= 10 
𝜔0 = 20 𝑠
−1 
𝑆𝑎𝑏𝑒𝑚𝑜𝑠, 𝑒𝑛𝑡ã𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑞𝑢𝑒 𝑑𝑒𝑠𝑐𝑟𝑒𝑣𝑒 𝑢𝑚 𝑜𝑠𝑐𝑖𝑙𝑎𝑑𝑜𝑟 ℎ𝑎𝑟𝑚ô𝑛𝑖𝑐𝑜 𝑎𝑚𝑜𝑟𝑡𝑒𝑐𝑖𝑑𝑜 (𝑢𝑚𝑎 𝐸𝐷𝑂) 
𝑒 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑠𝑢𝑎 𝑠𝑜𝑙𝑢çã𝑜, 𝑛𝑜 𝑐𝑎𝑠𝑜 𝑑𝑒 𝜔0 > 𝛾, é: 
 
�̇�(𝑡) = −
𝛾
2
𝑒−
𝛾
2𝑡(−𝑎𝜔 sin 𝜔𝑡 + 𝑏𝜔 cos 𝜔𝑡) 
𝑜𝑛𝑑𝑒: 
 
𝐴 𝑞𝑢𝑒𝑠𝑡ã𝑜 𝑛𝑜𝑠 𝑓𝑜𝑟𝑛𝑒𝑐𝑒 𝑎𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑖𝑠: 
𝑥(0) = 0 
�̇�(0) = 5 𝑚/𝑠 
 
𝐴𝑠 𝑢𝑠𝑎𝑚𝑜𝑠 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 (4.1.10): 
𝑥(0) = 𝑒0(𝑎 cos 0 + 𝑏 sin 0) = 0 
𝑥(0) = 𝑎 = 0 
 
�̇�(0) = −
𝛾
2
𝑒0(−𝑎𝜔 sin 0 + 𝑏𝜔 cos 0) = 5 
�̇�(0) = −
𝛾
2
𝑏𝜔 = 5 
𝑏 = −
10
𝛾𝜔
 
𝑠𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒𝑚 𝑠ã𝑜 𝛾 𝑒 𝜔0: 
𝜔 = √𝜔02 −
𝛾2
4
 
𝜔 = √202 −
22
4
 
𝜔 = √400 − 1 
𝜔 ≅ 20 𝑠−1 
𝑙𝑜𝑔𝑜 𝑏 𝑠𝑒𝑟á: 
𝑏 = −
10
2 × 20
= −
1
4
 
 
𝐶𝑜𝑚 𝑖𝑠𝑠𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 (4.1.10)𝑓𝑖𝑐𝑎 𝑎𝑠𝑠𝑖𝑚: 
 
𝑥(𝑡) = 𝑒−𝑡(−
1
4
sin 20𝑡) 
 
 
 
𝑈𝑚 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑚á𝑥𝑖𝑚𝑜 𝑜𝑐𝑜𝑟𝑟𝑒𝑟á 𝑢𝑚𝑎 𝑜𝑠𝑐𝑖𝑙𝑎çã𝑜 𝑖𝑛𝑡𝑒𝑖𝑟𝑎 (𝑡 = 𝑇)𝑎𝑝ó𝑠 𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑜 𝑚á𝑥𝑖𝑚𝑜: 
𝑟 =
𝑥𝑚(𝑡)
𝑥𝑚(𝑡 + 𝑇)
 
𝑟 =
𝑥𝑚(𝑡)
𝑥𝑚(𝑡 + 𝑇)
=
𝐴𝑒−
𝛾
2𝑡(cos(𝜔𝑡 + 𝜑))
𝐴𝑒−
𝛾
2(𝑡+𝑇)(cos(𝜔(𝑡 + 𝑇) + 𝜑))
 
𝑛𝑢𝑚𝑎 𝑜𝑠𝑐𝑖𝑙𝑎çã𝑜, 𝑜 𝑑𝑒𝑠𝑙𝑜𝑐𝑎𝑚𝑒𝑛𝑡𝑜 𝑚á𝑥𝑖𝑚𝑜 𝑠𝑒 𝑑á 𝑞𝑢𝑎𝑛𝑑𝑜 𝑜 𝑐𝑜𝑠𝑠𝑒𝑛𝑜 é 𝑖𝑔𝑢𝑎𝑙 𝑎 1: 
𝑟 =
𝐴𝑒−
𝛾
2𝑡
𝐴𝑒−
𝛾
2(𝑡+𝑇)
=
𝑒−
𝛾
2𝑡
𝑒−
𝛾
2(𝑡+𝑇)
=
𝑒−
𝛾
2𝑡
𝑒−
𝛾
2𝑡𝑒−
𝛾
2𝑇
=
1
𝑒−
𝛾
2𝑇
= 𝑒
𝛾
2𝑇 
𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 ln 𝑑𝑒 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑙𝑎𝑑𝑜𝑠: 
ln 𝑟 =
𝛾
2
𝑇 
𝑜 𝑞𝑢𝑒 é 𝑐ℎ𝑎𝑚𝑎𝑑𝑜, 𝑝𝑒𝑙𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜, 𝑑𝑒 𝛿: 
𝛿 = ln 𝑟 =
𝛾
2
𝑇 
𝛿 =
𝛾
2
𝑇 
 
 
 
 
 
 
 
 
 
 
 
 
𝐴 𝑞𝑢𝑒𝑠𝑡ã𝑜 𝑛𝑜𝑠 𝑓𝑜𝑟𝑛𝑒𝑐𝑒 𝑎𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑖𝑠: 
 
{
𝑥(0) = 0
�̇�(0) = 𝑣0
𝑥(1) = 𝑥𝑀Á𝑋 = 3,68
 
 
𝑢𝑚 𝑜𝑠𝑐𝑖𝑙𝑎𝑑𝑜𝑟 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑚𝑒𝑛𝑡𝑒 𝑎𝑚𝑜𝑟𝑡𝑒𝑐𝑖𝑑𝑜 é 𝑑𝑒𝑠𝑐𝑟𝑖𝑡𝑜 𝑝𝑒𝑙𝑎 𝑒𝑞𝑢𝑎çã𝑜: 
 
�̇�(𝑡) = 𝑒−
𝛾
2
𝑡 (−
𝛾
2
(𝑎 + 𝑏𝑡) + 𝑏) 
𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜, 𝑛𝑒𝑙𝑎, 𝑎𝑠 𝑑𝑢𝑎𝑠 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑖𝑠: 
 
𝑥(0) = 𝑎 = 0 
�̇�(0) = 𝑏 = 𝑣0 
 
𝑙𝑜𝑔𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 4.2.18 𝑓𝑖𝑐𝑎 𝑎𝑠𝑠𝑖𝑚: 
𝑥(𝑡) = 𝑒−
𝛾
2𝑡𝑣0𝑡 
 
𝐴𝑔𝑜𝑟𝑎, 𝑢𝑠𝑎𝑚𝑜𝑠 𝑎 𝑡𝑒𝑟𝑐𝑒𝑖𝑟𝑎 𝑐𝑜𝑛𝑑𝑖çã𝑜: 
𝑥(1) = 𝑒−
𝛾
2𝑣0 = 3,68 (∗) 
 
𝑃𝑒𝑟𝑐𝑒𝑏𝑎 𝑞𝑢𝑒 𝑡𝑒𝑚𝑜𝑠 𝑑𝑢𝑎𝑠 𝑖𝑛𝑐ó𝑔𝑛𝑖𝑡𝑎𝑠, 𝑙𝑜𝑔𝑜 𝑝𝑟𝑒𝑐𝑖𝑠𝑎𝑚𝑜𝑠 𝑑𝑒 𝑚𝑎𝑖𝑠 𝑢𝑚𝑎 𝑒𝑞𝑢𝑎çã𝑜. 
𝑉𝑎𝑚𝑜𝑠 𝑝𝑒𝑛𝑠𝑎𝑟, 𝑠𝑒 𝑒𝑚 𝑡 = 1𝑠 𝑜 𝑑𝑒𝑠𝑙𝑜𝑐𝑎𝑚𝑒𝑛𝑡𝑜 é 𝑚á𝑥𝑖𝑚𝑜, 𝑎 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑𝑒 é 𝑧𝑒𝑟𝑜! 
�̇�(1) = 𝑒−
𝛾
2 (−
𝛾
2
𝑣0 + 𝑣0) = 0 
�̇�(1) = 𝑒−
𝛾
2𝑣0 (−
𝛾
2
+ 1) = 0 
 
𝑐𝑜𝑚𝑜 𝑒−
𝛾
2𝑣0 𝑑𝑒𝑣𝑒 𝑠𝑒𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒 𝑑𝑒 𝑧𝑒𝑟𝑜: 
(−
𝛾
2
+ 1) = 0 
𝛾 = 2 𝑠−1 
 
𝑆𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒𝑚 é 𝛾, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑣𝑜𝑙𝑡𝑎𝑟 𝑎 𝑒𝑞𝑢𝑎çã𝑜 (∗): 
 
𝑥(1) = 𝑒−
2
2𝑣0 = 3,68 
𝑣0 = 3,68 × 𝑒 
𝑣0 ≅ 10𝑚/𝑠 
 
 
𝑆𝑒 𝑎 𝑓𝑜𝑟ç𝑎 𝑝𝑒𝑠𝑜 é 𝑑𝑒𝑠𝑝𝑟𝑒𝑧í𝑣𝑒𝑙, 𝑎 𝑝𝑎𝑟𝑡í𝑐𝑖𝑙𝑎 𝑑𝑒𝑠𝑐𝑒 𝑑𝑒𝑠𝑎𝑐𝑒𝑙𝑒𝑟𝑎𝑛𝑑𝑜: 
𝜌�̇� = −𝑚�̈� 
𝛾�̇� + �̈� = 0 
𝑜𝑛𝑑𝑒 𝛾 =
𝜌
𝑚
 
𝑃𝑎𝑟𝑎 𝑡𝑜𝑟𝑛𝑎𝑟 𝑒𝑠𝑡𝑎 𝑐𝑜𝑛𝑡𝑎 𝑚𝑎𝑖𝑠 𝑠𝑖𝑚𝑝𝑙𝑒𝑠, 𝑐ℎ𝑎𝑚𝑜 �̇�(𝑡) = 𝑓(𝑡); 𝑜 𝑞𝑢𝑒 𝑡𝑜𝑟𝑛𝑎 �̈�(𝑡) = 𝑓̇(𝑡): 
 
𝛾𝑓 + 𝑓̇ = 0 
 
𝑞𝑢𝑒 é 𝑢𝑚𝑎 𝐸𝐷𝑂 𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑒 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑜𝑟𝑑𝑒𝑚 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎. 
𝑄𝑢𝑒 𝑝𝑜𝑑𝑒 𝑠𝑒𝑟 𝑟𝑒𝑠𝑜𝑙𝑣𝑖𝑑𝑎 𝑝𝑒𝑙𝑜 𝑚é𝑡𝑜𝑑𝑜 𝑑𝑎 𝑠𝑒𝑝𝑎𝑟𝑎çã𝑜 𝑑𝑒 𝑣𝑎𝑟𝑖á𝑣𝑒𝑖𝑠: 
 
−𝛾𝑓 = 𝑓̇ 
−𝛾𝑓 =
𝑑𝑓
𝑑𝑡
 
−𝛾𝑓 𝑑𝑡 =
𝑑𝑓
𝑑𝑡
 𝑑𝑡 
−𝛾𝑓 𝑑𝑡 = 𝑑𝑓 
−𝛾 𝑑𝑡 =
𝑑𝑓
𝑓
 
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑚𝑜𝑠 𝑑𝑒 0 𝑎 𝑡: 
−𝛾 ∫ 𝑑𝑡
𝑡
0
= ∫
𝑑𝑓
𝑓
𝑓(𝑡)
𝑓(0)
 
−𝛾𝑡 = ln 𝑓(𝑡) − ln 𝑓(0) 
𝑝𝑒𝑙𝑎 𝑝𝑟𝑜𝑝𝑟𝑖𝑒𝑑𝑎𝑑𝑒 𝑑𝑒 𝑙𝑛: 
−𝛾𝑡 = ln (
𝑓(𝑡)
𝑓(0)
) 
𝑎𝑝𝑙𝑖𝑐𝑎𝑚𝑜𝑠 exp 𝑑𝑒 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑙𝑎𝑑𝑜𝑠: 
𝑒−𝛾𝑡 =
𝑓(𝑡)
𝑓(0)
 
𝑐𝑜𝑚𝑜 𝑓(0) = �̇�(0) = 𝑣0, 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜: 
 
𝑣0𝑒
−𝛾𝑡 = 𝑓(𝑡) 
 
𝑎𝑔𝑜𝑟𝑎 𝑑𝑒𝑠𝑓𝑎𝑧𝑒𝑚𝑜𝑠 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖çã𝑜 �̇�(𝑡) = 𝑓(𝑡): 
𝑣0𝑒
−𝛾𝑡 =
𝑑𝑧
𝑑𝑡
 
𝐴𝑔𝑜𝑟𝑎 𝑜 𝑝𝑟𝑜𝑐𝑒𝑠𝑠𝑜 é 𝑠𝑒𝑚𝑒𝑙ℎ𝑎𝑛𝑡𝑒 𝑝𝑎𝑟𝑎 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑚𝑜𝑠 𝑧(𝑡): 
𝑣0𝑒
−𝛾𝑡𝑑𝑡 = 𝑑𝑧 
𝑣0 ∫ 𝑒
−𝛾𝑡𝑑𝑡
𝑡
0
= ∫ 𝑑𝑧
𝑧(𝑡)
𝑧(0)
 
𝑣0 (
𝑒−𝛾𝑡
−𝛾
)|
0
𝑡
= 𝑧(𝑡) − 𝑧(0) 
−
𝑣0
𝛾
(𝑒−𝛾𝑡 − 1) + 𝑧(0) = 𝑧(𝑡) 
𝑆𝑒𝑔𝑢𝑛𝑑𝑜 𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜: 𝑧(0) = 𝑧0 
𝑧(𝑡) = 𝑧0 +
𝑣0
𝛾
(1 − 𝑒−𝛾𝑡) 
 
 
 
𝑆𝑒𝑔𝑢𝑛𝑑𝑜 𝑜 𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜, é 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟: 
𝑚𝑔 − 𝜌�̇� = 𝑚�̈� 
𝛾�̇� + �̈� = 𝑔 (∗) 
𝑜𝑛𝑑𝑒: 𝛾 =
𝜌
𝑚
 
𝑉𝑒𝑚𝑜𝑠, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑞𝑢𝑒 𝑎 𝑔𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑎𝑝𝑎𝑟𝑒𝑐𝑒 𝑐𝑜𝑚𝑜 𝑢𝑚𝑎 𝑝𝑎𝑟𝑡𝑒 forçada 𝑛𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜. 
𝑆𝑒𝑛𝑑𝑜, 𝑝𝑜𝑟é𝑚, 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒. 𝐿𝑜𝑔𝑜, 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑔𝑒𝑟𝑎𝑙 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑡𝑒𝑟á 𝑢𝑚𝑎 𝑠𝑜𝑙𝑢çã𝑜 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎 
𝑒 𝑢𝑚𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟: 
𝑧(𝑡) = 𝑧ℎ(𝑡) + 𝑧𝑝(𝑡) 
𝑁𝑜 𝑒𝑥𝑒𝑟𝑐í𝑐𝑖𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟 (𝑞𝑢𝑒𝑠𝑡ã𝑜 5), 𝑎𝑐ℎ𝑎𝑚𝑜𝑠 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑎 𝑜 𝑐𝑎𝑠𝑜 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑜: 
 
𝑧ℎ(𝑡) = 𝑧0 +
𝑣0
𝛾
(1 − 𝑒−𝛾𝑡) 
 
𝑃𝑜𝑟é𝑚, 𝑝𝑒𝑟𝑐𝑒𝑏𝑎 𝑞𝑢𝑒, 𝑧0 𝑒 𝑣0 𝑓𝑜𝑟𝑎𝑚 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑑𝑒 𝑐𝑜𝑛𝑡𝑜𝑟𝑛𝑜, (�̇�(0) = 𝑣0 𝑒 𝑧(0) = 𝑧0) 
𝑞𝑢𝑒, 𝑠𝑒 𝑎𝑝𝑙𝑖𝑐𝑎𝑑𝑎𝑠 𝑛𝑒𝑠𝑡𝑒 𝑒𝑥𝑒𝑟𝑐í𝑐𝑖𝑜, 𝑛ã𝑜 𝑑𝑎𝑟𝑖𝑎 𝑜 𝑚𝑒𝑠𝑚𝑜 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜. 
𝐸𝑛𝑡ã𝑜, 𝑝𝑎𝑟𝑎 𝑡𝑜𝑟𝑛𝑎𝑟 𝑒𝑠𝑠𝑎 𝑠𝑜𝑙𝑢çã𝑜 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑖𝑎 𝑢𝑚 𝑝𝑜𝑢𝑐𝑜 𝑚𝑎𝑖𝑠 𝑔𝑒𝑟𝑎𝑙, 𝑐ℎ𝑎𝑚𝑎𝑟𝑒𝑚𝑜𝑠: 
�̇�ℎ(0) = 𝑎 𝑒 𝑧ℎ(0) = 𝑏, 𝑙𝑜𝑔𝑜: 
𝑧ℎ(𝑡) = 𝑏 +
𝑎
𝛾
(1 − 𝑒−𝛾𝑡) 
𝐴𝑠𝑠𝑖𝑚, 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑔𝑒𝑟𝑎𝑙 𝑠𝑒𝑟á: 
𝑧(𝑡) = 𝑏 +
𝑎
𝛾
(1 − 𝑒−𝛾𝑡) + 𝑧𝑝(𝑡) 
 
𝑃𝑟𝑒𝑐𝑖𝑠𝑎𝑚𝑜𝑠 𝑠𝑎𝑏𝑒𝑟 𝑞𝑢𝑒𝑚 é 𝑧𝑝(𝑡). 𝑁𝑎 𝑝á𝑔𝑖𝑛𝑎 78, 𝑑𝑜 𝑚𝑒𝑠𝑚𝑜 𝑙𝑖𝑣𝑟𝑜, 𝑑𝑖𝑧: 
 
𝑠𝑜𝑙𝑢çã𝑜 𝑒𝑠𝑡𝑎𝑐𝑖𝑜𝑛á𝑟𝑖𝑎 é 𝑎𝑞𝑢𝑒𝑙𝑎 𝑜𝑛𝑑𝑒 𝐹𝑟 = 0: 
 
𝑚𝑔 − 𝜌�̇� = 0 
�̇� =
𝑔
𝛾
 
𝐶𝑜𝑚𝑜 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑒𝑠𝑡𝑎𝑐𝑖𝑜𝑛á𝑟𝑖𝑎 é 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟: 
�̇�𝑝(𝑡)=
𝑔
𝛾
 
𝑑𝑧𝑝
𝑑𝑡
=
𝑔
𝛾
 
∫ 𝑑𝑧𝑝 = ∫
𝑔
𝛾
𝑑𝑡 
𝑧𝑝(𝑡) =
𝑔
𝛾
𝑡 
𝐿𝑜𝑔𝑜, 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑔𝑒𝑟𝑎𝑙 𝑠𝑒𝑟á: 
𝑧(𝑡) = 𝑏 +
𝑎
𝛾
(1 − 𝑒−𝛾𝑡) +
𝑔
𝛾
𝑡 
𝑞𝑢𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎: 
�̇�(𝑡) = 𝑎𝑒−𝛾𝑡 +
𝑔
𝛾
 
 
𝐴𝑔𝑜𝑟𝑎, 𝑐𝑜𝑚 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑚𝑎𝑖𝑠 𝑔𝑒𝑟𝑎𝑙 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑝𝑎𝑟𝑎 𝑜 𝑐𝑎𝑠𝑜 𝑑𝑒 𝑞𝑢𝑒𝑑𝑎 𝑙𝑖𝑣𝑟𝑒 𝑐𝑜𝑚 𝑟𝑒𝑠𝑖𝑠𝑡ê𝑛𝑐𝑖𝑎 
𝑣𝑎𝑚𝑜𝑠 𝑎𝑝𝑙𝑖𝑐𝑎𝑟 𝑎𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑑𝑒 𝑐𝑜𝑛𝑡𝑜𝑟𝑛𝑜: 
𝑧(0) = 𝑧0 
�̇�(0) = 𝑣0 
 
𝑧(0) = 𝑏 = 𝑧0 
�̇�(0) = 𝑎 +
𝑔
𝛾
= 𝑣0 ∴ 𝑎 = 𝑣0 −
𝑔
𝛾
 
 
𝑙𝑜𝑔𝑜, 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑠: 
 
𝑧(𝑡) = 𝑧0 +
𝑣0 −
𝑔
𝛾
𝛾
(1 − 𝑒−𝛾𝑡) +
𝑔
𝛾
𝑡 
 
 
𝑆𝑒𝑔𝑢𝑛𝑑𝑜 𝑜 𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜, é 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟: 
𝑚�̈� + 𝑘𝑥 = 𝐹0 sin 𝜔𝑡 
�̈� + 𝜔0
2𝑥 =
𝐹0
𝑚
sin 𝜔𝑡 
É 𝑝𝑜𝑠𝑠í𝑣𝑒𝑙 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑎𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑎 𝑛𝑜𝑡𝑎çã𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑥𝑎 (𝑐𝑜𝑚𝑜 𝑛𝑎 𝑝. 78), 𝑚𝑎𝑠 𝑛ã𝑜 𝑜 𝑓𝑎𝑟𝑒𝑚𝑜𝑠 𝑎𝑞𝑢𝑖. 
𝑈𝑠𝑎𝑟𝑒𝑚𝑜𝑠, 𝑛𝑜 𝑒𝑛𝑡𝑎𝑛𝑡𝑜, 𝑜𝑠 𝑚𝑒𝑠𝑚𝑜𝑠 𝑝𝑎𝑠𝑠𝑜𝑠: 
𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑔𝑒𝑟𝑎𝑙 é: 
𝑥(𝑡) = 𝑥ℎ(𝑡) + 𝑥𝑝(𝑡) 
𝐴 𝑝𝑎𝑟𝑡𝑒 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎 𝑑𝑎 𝑠𝑜𝑙𝑢çã𝑜 é 𝑐𝑜𝑛ℎ𝑒𝑐𝑖𝑑𝑎: 
𝑥ℎ(𝑡) = 𝑎 cos 𝜔0𝑡 + 𝑏 sin 𝜔0𝑡 
𝑙𝑜𝑔𝑜: 
𝑥(𝑡) = 𝑎 cos 𝜔0𝑡 + 𝑏 sin 𝜔0𝑡 + 𝑥𝑝(𝑡) 
𝑃𝑒𝑙𝑎 𝑓𝑜𝑟𝑚𝑎 𝑑𝑎 𝑝𝑎𝑟𝑡𝑒 𝑓𝑜𝑟ç𝑎𝑑𝑎, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑠𝑢𝑝𝑜𝑟 𝑞𝑢𝑒 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑒𝑟á: 
𝑥𝑝(𝑡) = 𝑀 sin 𝜔𝑡 
𝑜𝑛𝑑𝑒 𝑀 é 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑎 𝑠𝑒𝑟 𝑑𝑒𝑓𝑖𝑛𝑖𝑑𝑎. 
𝐸 𝑑𝑒𝑟𝑖𝑣𝑎𝑚𝑜𝑠: 
𝑥𝑝(𝑡) = 𝑀 sin 𝜔𝑡 
�̇�𝑝(𝑡) = 𝑀𝜔 cos 𝜔𝑡 
�̈�𝑝(𝑡) = −𝑀𝜔
2 sin 𝜔𝑡 
 
𝑜𝑛𝑑𝑒 𝑛𝑜𝑠𝑠𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑑𝑒𝑣𝑒 𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑒𝑟 𝑎 𝑝𝑎𝑟𝑡𝑒 𝑓𝑜𝑟ç𝑎𝑑𝑎: 
�̈�𝑝 + 𝜔0
2𝑥𝑝 =
𝐹0
𝑚
sin 𝜔𝑡 
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠, 𝑎𝑞𝑢𝑖, 𝑎𝑠 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎𝑠 𝑎𝑐𝑖𝑚𝑎: 
−𝑀𝜔2 sin 𝜔𝑡 + 𝜔0
2𝑀 sin 𝜔𝑡 =
𝐹0
𝑚
sin 𝜔𝑡 
(𝜔0
2 − 𝜔2)𝑀 sin 𝜔𝑡 =
𝐹0
𝑚
sin 𝜔𝑡 
(𝜔0
2 − 𝜔2)𝑀 =
𝐹0
𝑚
 
𝑀 =
𝐹0
𝑚(𝜔0
2 − 𝜔2)
 
 
𝐶𝑜𝑛ℎ𝑒𝑐𝑒𝑛𝑑𝑜 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟 𝑎 𝑔𝑒𝑟𝑎𝑙: 
𝑥(𝑡) = 𝑎 cos 𝜔0𝑡 + 𝑏 sin 𝜔0𝑡 +
𝐹0
𝑚(𝜔0
2 − 𝜔2)
sin 𝜔𝑡 
 
�̇�(𝑡) = −𝑎𝜔0 sin 𝜔0𝑡 + 𝑏𝜔0 cos 𝜔0𝑡 +
𝜔𝐹0
𝑚(𝜔0
2 − 𝜔2)
cos 𝜔𝑡 
 
𝐴𝑔𝑜𝑟𝑎, 𝑎𝑝𝑙𝑖𝑐𝑎𝑟𝑒𝑚𝑜𝑠 𝑎𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑑𝑒 𝑐𝑜𝑛𝑡𝑜𝑟𝑛𝑜 𝑑𝑎𝑑𝑎𝑠 𝑝𝑒𝑙𝑜 𝑒𝑥𝑒𝑟𝑐í𝑐𝑖𝑜: 
 
𝑥(0) = 0 
�̇�(0) = 0 
 
𝑥(0) = 𝑎 = 0 
�̇�(0) = 𝑏𝜔0 +
𝜔𝐹0
𝑚(𝜔0
2 − 𝜔2)
= 0 
𝑏 = −
𝜔𝐹0
𝑚𝜔0(𝜔0
2 − 𝜔2)
 
 
𝐴𝑔𝑜𝑟𝑎, 𝑠𝑎𝑏𝑒𝑛𝑑𝑜 𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑠, 𝑎𝑠 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑛𝑎 𝑠𝑜𝑙𝑢çã𝑜: 
 
𝑥(𝑡) = −
𝜔𝐹0
𝑚𝜔0(𝜔0
2 − 𝜔2)
sin 𝜔0𝑡 +
𝐹0
𝑚(𝜔0
2 − 𝜔2)
sin 𝜔𝑡 
𝑥(𝑡) =
𝐹0
𝑚(𝜔0
2 − 𝜔2)
(sin 𝜔𝑡 −
𝜔
𝜔0
sin 𝜔0𝑡) 
 
 
𝐴𝑡𝑒𝑛çã𝑜 𝑝𝑎𝑟𝑎 𝑛ã𝑜 𝑐𝑜𝑛𝑓𝑢𝑛𝑑𝑖𝑟 𝜌 𝑑𝑒 𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑𝑒 𝑐𝑜𝑚 𝜌 𝑑𝑒 𝑐𝑜𝑒𝑓. 𝑑𝑒 𝑝𝑟𝑜𝑝.! 
𝐴𝑛𝑡𝑒𝑠 𝑑𝑒 𝑐𝑜𝑚𝑒ç𝑎𝑟, 𝑝𝑎𝑟𝑎 𝑓𝑎𝑐𝑖𝑙𝑖𝑡𝑎𝑟 𝑣𝑎𝑚𝑜𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑡𝑒𝑟 𝑎 𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑𝑒 𝑑𝑜 𝑓𝑙𝑢𝑖𝑑𝑜: 
𝜌𝑓 = 1,25
𝑔
𝑐𝑚3
= 1250
𝑘𝑔
𝑚3
 
𝑂𝑠 𝑎𝑔𝑒𝑛𝑡𝑒𝑠 𝑞𝑢𝑒 𝑖𝑛𝑡𝑒𝑟𝑓𝑒𝑟𝑒𝑚 𝑛𝑒𝑠𝑠𝑎 𝑜𝑠𝑐𝑖𝑙𝑎çã𝑜: 
𝐺𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒, 𝑚𝑜𝑙𝑎 𝑒 𝑓𝑙𝑢𝑖𝑑𝑜 
 
𝐶𝑜𝑚𝑜 𝑎 𝑔𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑠ó 𝑖𝑛𝑡𝑒𝑟𝑓𝑒𝑟𝑒 𝑎𝑙𝑡𝑒𝑟𝑎𝑛𝑑𝑜 𝑎 𝑝𝑜𝑠𝑖çã𝑜 𝑑𝑒 𝑒𝑞𝑢𝑖𝑙í𝑏𝑟𝑖𝑜 𝑧0, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟 
𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑠𝑒𝑚 𝑒𝑙𝑎 𝑒 𝑎𝑝𝑒𝑛𝑎𝑠 𝑎𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑎𝑟 𝑧0 𝑛𝑎 𝑠𝑜𝑙𝑢çã𝑜. 
𝐴𝑐ℎ𝑎𝑛𝑑𝑜 𝑧0: 
𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑚𝑎𝑠𝑠𝑎 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 é 𝑜 𝑝𝑟𝑜𝑑𝑢𝑡𝑜 𝑑𝑒 𝑠𝑢𝑎 𝑑𝑒𝑛𝑠𝑖𝑑𝑎𝑑𝑒 𝑐𝑜𝑚 𝑠𝑒𝑢 𝑣𝑜𝑙𝑢𝑚𝑒: 
𝑚 = 8 × 103 = 8000𝑔 = 8𝑘𝑔 
𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑛𝑑𝑜 𝑎𝑠 𝑓𝑜𝑟ç𝑎𝑠 𝑞𝑢𝑒 𝑎𝑔𝑒𝑚 𝑛𝑜 𝑏𝑙𝑜𝑐𝑜: 
𝑃 − 𝐸 − 𝐹𝑒𝑙 = 0 
𝑚𝑔 − 𝜌𝑓𝑣𝑔 − 𝑘𝑧 = 0 
𝑢𝑠𝑎𝑛𝑑𝑜 𝑜𝑠 𝑑𝑎𝑑𝑜𝑠 𝑓𝑜𝑟𝑛𝑒𝑐𝑖𝑑𝑜𝑠 𝑝𝑒𝑙𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜: 
8 × 9,8 − 1250 × 10−3 × 9,8 − 40𝑧 = 0 
𝑧 = 1,65 𝑚 
 
𝑆𝑒𝑛𝑑𝑜 𝑒𝑠𝑡𝑒 ú𝑙𝑡𝑖𝑚𝑜, 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑑𝑒𝑓𝑜𝑟𝑚𝑎çã𝑜 𝑑𝑎 𝑚𝑜𝑙𝑎 𝑐𝑎𝑢𝑠𝑎𝑑𝑎 𝑝𝑒𝑙𝑎 𝑔𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒, 𝑜 𝑠𝑜𝑚𝑎𝑚𝑜𝑠 𝑎𝑜 
𝑣𝑎𝑙𝑜𝑟 𝑑𝑜 𝑐𝑜𝑚𝑝𝑟𝑖𝑚𝑒𝑛𝑡𝑜 𝑟𝑒𝑙𝑎𝑥𝑎𝑑𝑜, 𝑝𝑎𝑟𝑎 𝑜𝑏𝑡𝑒𝑟 𝑎 𝑝𝑜𝑠𝑖çã𝑜 𝑑𝑒 𝑒𝑞𝑢𝑖𝑙í𝑏𝑟𝑖𝑜 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑜𝑠𝑐𝑖𝑙𝑎𝑡ó𝑟𝑖𝑜: 
 
𝑧0 = 𝑧 + 𝑧𝑟𝑒𝑙𝑎𝑥 
𝑧0 = 1,65 + 0,5 = 2,15 𝑚 
 
𝐶𝑜𝑚𝑜 𝑑𝑖𝑡𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟𝑚𝑒𝑛𝑡𝑒, 𝑣𝑎𝑚𝑜𝑠 𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑠𝑒𝑚 𝑎 𝑔𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒: 
−𝜌�̇� − 𝑘𝑧 = 𝑚�̈� 
𝜌�̇� + 𝑘𝑧 + 𝑚�̈� = 0 
𝛾�̇� + 𝜔0
2𝑧 + �̈� = 0 
𝑜𝑛𝑑𝑒: 𝜔0
2 =
𝑘
𝑚
 𝑒 𝛾 =
𝜌
𝑚
 
𝐶𝑜𝑛ℎ𝑒𝑐𝑒𝑚𝑜𝑠 𝜌, 𝑚 𝑒 𝑘: 
𝜔0
2 =
40
8
= 5 𝑠−1 
𝛾 =
2
8
= 0,25 𝑠−1 
𝑙𝑜𝑔𝑜: 𝜔0
2 >
𝛾
2
 
𝑒𝑛𝑡ã𝑜, 𝑡𝑟𝑎𝑡𝑎 − 𝑠𝑒 𝑑𝑒 𝑢𝑚 𝑎𝑚𝑜𝑟𝑡𝑒𝑐𝑖𝑚𝑒𝑛𝑡𝑜 𝑠𝑢𝑏𝑐𝑟í𝑡𝑖𝑐𝑜, 𝑐𝑢𝑗𝑎 𝑠𝑜𝑙𝑢çã𝑜 é: 
𝑧(𝑡) = 𝑒−
𝛾
2𝑡(𝑎 cos 𝜔𝑡 + 𝑏 sin 𝜔𝑡) 
𝑜𝑛𝑑𝑒: 𝜔 = √𝜔0
2 −
𝛾2
4
 
𝑙𝑜𝑔𝑜: 
𝜔 = √5 −
0,0625
4
≅ 2,23 
𝑒: 
𝛾
2
= 0,125 
𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑓𝑖𝑐𝑎: 
𝑧(𝑡) = 𝑒−0,125𝑡(𝑎 cos 2,23𝑡 + 𝑏 sin 2,23𝑡) 
�̇�(𝑡) = −0,125𝑒−0,125𝑡(𝑎 cos 2,23𝑡 + 𝑏 sin 2,23𝑡)
+ 𝑒−0,125𝑡(−2,23𝑎 sin 2,23𝑡 + 2,23𝑏 cos 2,23𝑡) 
 
𝑎𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑑𝑒 𝑐𝑜𝑛𝑡𝑜𝑟𝑛𝑜 𝑓𝑜𝑟𝑛𝑒𝑐𝑖𝑑𝑎𝑠 𝑝𝑒𝑙𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜 𝑠ã𝑜: 
𝑧(0) = 0,01 𝑚 
�̇�(0) = 0 𝑚/𝑠 
 
𝑧(0) = 𝑒0(𝑎) = 0,01 → 𝑎 = 0,01 
 
𝑧(𝑡) = −0,125𝑒0(𝑎) + 𝑒0(2,23𝑏) = 0 = −0,125 × 0,01 + 2,23𝑏 
0,125 × 0,01 = 2,23𝑏 
𝑏 = 5,6 × 10−4 
 
𝑐𝑜𝑚 𝑎𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒𝑠, 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑓𝑖𝑐𝑎: 
𝑧(𝑡) = 𝑒−0,125𝑡(0,01 cos 2,23𝑡 + 5,6 × 10−4 sin 2,23𝑡) 
𝑧(𝑡) = 10−2𝑒−0,125𝑡(cos 2,23𝑡 + 0,056 sin 2,23𝑡) 
 
𝑎𝑔𝑜𝑟𝑎, 𝑎𝑑𝑖𝑐𝑖𝑜𝑛𝑎𝑚𝑜𝑠 𝑎 𝑛𝑜𝑣𝑎 𝑝𝑜𝑠𝑖çã𝑜 𝑑𝑒 𝑒𝑞𝑢𝑖𝑙í𝑏𝑟𝑖𝑜 𝑐𝑎𝑢𝑠𝑎𝑑𝑎 𝑝𝑒𝑙𝑎 𝑔𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑧0: 
 
𝑧(𝑡) = 2,15 + 10−2𝑒−0,125𝑡(cos 2,23𝑡 + 0,056 sin 2,23𝑡) 
 
 
 
 
 
 𝑧𝐴 
 
 𝑧𝐵 
 
 
𝑉𝑎𝑚𝑜𝑠 𝑝𝑟𝑒𝑠𝑡𝑎𝑟 𝑎𝑡𝑒𝑛çã𝑜 𝑒𝑚 𝑡𝑢𝑑𝑜 𝑞𝑢𝑒 𝑑𝑒𝑓𝑜𝑟𝑚𝑎 𝑎 𝑚𝑜𝑙𝑎. 
𝑃𝑟𝑖𝑚𝑒𝑖𝑟𝑜, 𝑎 𝑎çã𝑜 𝑑𝑎 𝑔𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒, 𝑝𝑎𝑟𝑎 𝑣𝑒𝑟 𝑖𝑠𝑠𝑜, 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒 𝑜 𝑠𝑖𝑠𝑡𝑒𝑚𝑎 𝑝𝑎𝑟𝑎𝑑𝑜: 
𝑘𝑧𝑒𝑞 = 𝑚𝑔 → 𝑧𝑒𝑞 =
𝑚𝑔
𝑘
 
𝐷𝑒𝑝𝑜𝑖𝑠, 𝑒𝑚 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜, 𝑡𝑎𝑛𝑡𝑜 𝑎 𝑚ã𝑜 𝑧𝐴 𝑞𝑢𝑎𝑛𝑡𝑜 𝑜 𝑏𝑙𝑜𝑐𝑜 𝑧𝐵 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑚 𝑎 𝑚𝑜𝑙𝑎. 
𝑀𝑎𝑠, 𝑝𝑒𝑟𝑐𝑒𝑏𝑎 𝑞𝑢𝑒, 𝑎 𝑑𝑒𝑓𝑜𝑟𝑚𝑎çã𝑜 𝑧1 𝑠𝑒𝑟á 𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛ç𝑎 𝑑𝑒 𝑧𝐴 𝑒 𝑧𝐵. 
 𝑧1 = 𝑧𝐵 − 𝑧𝐴 
É 𝑑𝑖𝑓í𝑐𝑖𝑙 𝑣𝑒𝑟 𝑞𝑢𝑒 𝑖𝑠𝑠𝑜 𝑎𝑐𝑜𝑛𝑡𝑒𝑐𝑒, 𝑒𝑛𝑡ã𝑜 𝑣𝑜𝑙𝑡𝑒 𝑛𝑜 𝑑𝑒𝑠𝑒𝑛ℎ𝑜 𝑎𝑐𝑖𝑚𝑎, 𝑙á 𝑎𝑝𝑒𝑠𝑎𝑟 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 𝑒𝑠𝑡𝑎𝑟 
𝑑𝑒𝑠𝑐𝑒𝑛𝑑𝑜 (𝑒 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑛𝑑𝑜 𝑎 𝑚𝑜𝑙𝑎), 𝑎 𝑚ã𝑜 𝑡𝑎𝑚𝑏é𝑚 𝑑𝑒𝑠𝑐𝑒. 
 
𝐸𝑛𝑡ã𝑜, 𝑎 𝑑𝑒𝑓𝑜𝑟𝑚𝑎çã𝑜 𝑡𝑜𝑡𝑎𝑙 𝑠𝑒𝑟á 𝑎 𝑠𝑜𝑚𝑎 𝑑𝑎 𝑞𝑢𝑒 𝑎 𝑔𝑟𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑝𝑟𝑜𝑣𝑜𝑐𝑎 𝑐𝑜𝑚 𝑎 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝑧 = 𝑧𝑒𝑞 + 𝑧1 
𝑧 = 
𝑚𝑔
𝑘
+ 𝑧𝐵 − 𝑧𝐴 
 
𝐴𝑝ó𝑠 𝑒𝑠𝑠𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟 𝑎 𝑒𝑞𝑢çã𝑜 𝑑𝑒 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝑚𝑔 − 𝑘𝑧 = 𝑚�̈�𝐵 
𝑚𝑔 − 𝑘 (
𝑚𝑔
𝑘
+ 𝑧𝐵 − 𝑧𝐴) = 𝑚�̈�𝐵 
−𝑘( 𝑧𝐵 − 𝑧𝐴) = 𝑚�̈�𝐵 
𝑘( 𝑧𝐵 − 𝑧𝐴) + 𝑚�̈�𝐵 = 0 
𝜔0
2( 𝑧𝐵 − 𝑧𝐴) + �̈�𝐵 = 0 
𝑜𝑛𝑑𝑒: 𝜔0
2 =
𝑘
𝑚
 
𝜔0
2𝑧𝐵 + �̈�𝐵 = 𝑧𝐴𝜔0
2𝑇𝑒𝑚𝑜𝑠 𝑞𝑢𝑒, 𝑧𝐴 é 𝑜 𝑀𝐻𝑆 𝑑𝑎 𝑚ã𝑜, 𝑓𝑜𝑟𝑛𝑒𝑐𝑖𝑑𝑜 𝑝𝑒𝑙𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜, 𝑙𝑜𝑔𝑜: 
𝑧𝐴(𝑡) = 0,05 cos(𝜔𝑡 + 𝜙) 
𝑐𝑜𝑚𝑜 𝑐𝑜𝑚𝑒ç𝑎 𝑑𝑎 𝑝𝑜𝑠𝑖çã𝑜 𝑑𝑒 𝑒𝑞𝑢𝑖𝑙í𝑏𝑟𝑖𝑜: 
𝑧𝐴(0) = 0 
𝑧𝐴(0) = 𝐴 cos(𝜙) = 0 
𝑙𝑜𝑔𝑜: 
𝜙 =
𝜋
2
 
𝑒𝑛𝑡ã𝑜: 
𝑧𝐴(𝑡) = 𝐴 cos (𝜔𝑡 +
𝜋
2
) 
𝑞𝑢𝑒 𝑝𝑜𝑑𝑒 𝑠𝑒𝑟 𝑒𝑠𝑡𝑟𝑖𝑡𝑜 𝑐𝑜𝑚𝑜: 
𝑧𝐴(𝑡) = 𝐴 sin 𝜔𝑡 
𝐶𝑜𝑚𝑜 𝑖𝑠𝑠𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑒 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑓𝑖𝑐𝑎: 
𝜔0
2𝑧𝐵 + �̈�𝐵 = 𝜔0
2𝐴 sin 𝜔𝑡 
𝑠𝑒𝑛𝑑𝑜 𝑎 𝑎𝑠𝑠𝑖𝑚, 𝑢𝑚 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 ℎ𝑎𝑟𝑚ô𝑛𝑖𝑐𝑜 𝑓𝑜𝑟ç𝑎𝑑𝑜. 
𝐽á 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑆𝑜𝑙𝑢çã𝑜 𝑔𝑒𝑟𝑎𝑙 𝑡𝑒𝑟á 𝑢𝑚𝑎 𝑝𝑎𝑟𝑡𝑒 ℎ𝑜𝑚𝑜𝑔ê𝑛𝑒𝑎 𝑒 𝑢𝑚𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟. 
𝑧𝐵(𝑡) = 𝑧ℎ(𝑡) + 𝑧𝑝(𝑡) 
𝐸 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑖𝑔𝑎𝑚𝑖𝑛𝑎𝑟 𝑎 𝑓𝑜𝑟𝑚𝑎 𝑑𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑝𝑒𝑙𝑜 𝑡𝑖𝑝𝑜 𝑑𝑒 𝑓𝑜𝑟ç𝑎: 
𝑧𝑝(𝑡) = 𝑀 sin 𝜔𝑡 
�̈�𝑝(𝑡) = −𝜔
2𝑀 sin 𝜔𝑡 
𝑒 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢í𝑚𝑜𝑠 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑒 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
𝜔0
2𝑧𝐵 + �̈�𝐵 = 𝜔0
2𝐴 sin 𝜔𝑡 
𝜔0
2𝑀 sin 𝜔𝑡 − 𝜔2𝑀 sin 𝜔𝑡 = 𝜔0
2𝐴 sin 𝜔𝑡 
𝑀(𝜔0
2 − 𝜔2) = 𝜔0
2𝐴 
𝑀 =
𝜔0
2𝐴
(𝜔0
2 − 𝜔2) 
 
𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 é: 
𝑧𝑝(𝑡) =
𝜔0
2𝐴
(𝜔0
2 − 𝜔2) 
 sin 𝜔𝑡 
𝑒 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑔𝑒𝑟𝑎𝑙: 
𝑧𝐵(𝑡) = 𝑎 𝑐𝑜𝑠𝜔0𝑡 + 𝑏 sin 𝜔0𝑡 +
𝜔0
2𝐴
(𝜔0
2 − 𝜔2) 
 sin 𝜔𝑡 
�̇�𝐵(𝑡) = −𝜔0𝑎 sin 𝜔0𝑡 + 𝜔0𝑏 cos 𝜔0𝑡 +
𝜔𝜔0
2𝐴
(𝜔0
2 − 𝜔2) 
 cos 𝜔𝑡 
 
𝐴𝑔𝑜𝑟𝑎 𝑢𝑠𝑎𝑟𝑒𝑚𝑜𝑠 𝑎𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑑𝑒 𝑐𝑜𝑛𝑡𝑜𝑟𝑛𝑜 𝑓𝑜𝑟𝑛𝑒𝑐𝑖𝑑𝑎𝑠 𝑝𝑒𝑙𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜: 
𝑧𝐵(0) = 0 𝑚 
�̇�𝐵(0) = 0 𝑚/𝑠 
 
𝑧𝐵(0) = 𝑎 = 0 
�̇�𝐵(0) = 𝜔0𝑏 +
𝜔𝜔0
2𝐴
(𝜔0
2 − 𝜔2)
= 0 
𝑏 = −
𝜔𝜔0𝐴
(𝜔0
2 − 𝜔2)
 
 
𝐿𝑜𝑔𝑜, 𝑎 𝑠𝑜𝑙𝑢çã𝑜 𝑔𝑒𝑟𝑎𝑙 𝑑𝑜 𝑏𝑙𝑜𝑐𝑜 é: 
𝑧𝐵(𝑡) = −
𝜔𝜔0𝐴
(𝜔0
2 − 𝜔2)
sin 𝜔0𝑡 +
𝜔0
2𝐴
(𝜔0
2 − 𝜔2) 
 sin 𝜔𝑡 
𝑧𝐵(𝑡) =
𝜔0
2𝐴
(𝜔0
2 − 𝜔2)
 (sin 𝜔𝑡 −
𝜔
𝜔0
sin 𝜔0𝑡) 
 
 
 
 
 
 𝜃1 𝜃2 
 
 
 𝑥1 𝑥2 
𝑉𝑎𝑚𝑜𝑠 𝑑𝑒𝑐𝑜𝑚𝑝𝑜𝑟 𝑎𝑠 𝑓𝑜𝑟ç𝑎𝑠 𝑞𝑢𝑒 𝑎𝑡𝑢𝑎𝑚 𝑒𝑚 𝑐𝑎𝑑𝑎 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎 
𝐹𝑎𝑙𝑎𝑛𝑑𝑜 𝑒𝑚 𝑡𝑒𝑟𝑚𝑜𝑠 𝑑𝑎 𝑓𝑜𝑟ç𝑎 𝑒𝑙á𝑠𝑡𝑖𝑐𝑎, 𝑎 𝑑𝑒𝑓𝑜𝑟𝑚𝑎çã𝑜 𝑑𝑎 𝑚𝑜𝑙𝑎 𝑠𝑒𝑟á 𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛ç𝑎 𝑑𝑜𝑠 
𝑑𝑒𝑠𝑙𝑜𝑐𝑎𝑚𝑒𝑛𝑡𝑜𝑠 𝑑𝑎𝑠 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎𝑠: 
𝑥 = 𝑥2 − 𝑥1 
 𝐴𝑜 𝑠𝑒 𝑑𝑖𝑠𝑡𝑒𝑛𝑑𝑒𝑟/𝑐𝑜𝑛𝑡𝑟𝑎𝑖𝑟, 𝑎 𝑚𝑜𝑙𝑎 𝑎𝑢𝑥𝑖𝑙𝑖𝑎 𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑢𝑚𝑎 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎 𝑒𝑛𝑞𝑢𝑎𝑛𝑡𝑜 
𝑎𝑡𝑟𝑎𝑝𝑎𝑙ℎ𝑎 𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 𝑑𝑎 𝑜𝑢𝑡𝑟𝑎. 𝐼𝑠𝑠𝑜 𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎 𝑞𝑢𝑒, 𝑒𝑚 𝑢𝑚𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑒 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜 
𝑎 𝑓𝑜𝑟ç𝑎 𝑒𝑙á𝑠𝑡𝑖𝑐𝑎 𝑎𝑝𝑎𝑟𝑒𝑐𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑎 𝑒 𝑒𝑚 𝑜𝑢𝑡𝑟𝑎, 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑎. 
𝐶𝑜𝑚𝑜 𝑠ó 𝑜 𝑞𝑢𝑒 𝑛𝑜𝑠 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑠𝑎 é 𝑎 𝑐𝑜𝑚𝑝𝑜𝑛𝑡𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑜 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑜: 
 
−𝑃 sin 𝜃1 + 𝑘(𝑥2 − 𝑥1) = 𝑚�̈�1 
−𝑚𝑔 sin 𝜃1 + 𝑘(𝑥2 − 𝑥1) = 𝑚�̈�1 
−𝑔
𝑥1
𝑙
+ 𝐾(𝑥2 − 𝑥1) = �̈�1 
𝜔0
2𝑥1 + �̈�1 = 𝐾(𝑥2 − 𝑥1) (∗) 
 
𝑜𝑛𝑑𝑒: 
 
 
𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑜 𝑚𝑒𝑠𝑚𝑜 𝑝𝑎𝑟𝑎 𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑝𝑎𝑟𝑡í𝑐𝑢𝑙𝑎: 
 
𝜔0
2𝑥2 + �̈�2 = 𝐾(𝑥2 − 𝑥1) (∗∗) 
 
𝑉𝑎𝑚𝑜𝑠 𝑢𝑠𝑎𝑟 𝑢𝑚 𝑚𝑒𝑖𝑜 𝑝𝑎𝑟𝑎 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑒𝑠𝑠𝑒 𝑡𝑖𝑝𝑜 𝑑𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎 𝑞𝑢𝑒 𝑗á 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓𝑢𝑛𝑐𝑖𝑜𝑛𝑎. 
𝑆𝑜𝑚𝑎𝑛𝑑𝑜 𝑎𝑠 𝑒𝑞𝑢𝑎çõ𝑒𝑠 (∗)𝑒 (∗∗): 
𝜔0
2(𝑥1 + 𝑥2) + (�̈�1 + �̈�2) = 0 
𝐷𝑖𝑣𝑖𝑑𝑖𝑚𝑜𝑠 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 𝑝𝑜𝑟 2 𝑒 𝑐ℎ𝑎𝑚𝑎𝑚𝑜𝑠: 
 
{
𝑞1(𝑡) =
1
2
(𝑥1 + 𝑥2)
�̈�1(𝑡) =
1
2
(�̈�1 + �̈�2)
 
 
𝑙𝑜𝑔𝑜: 
𝜔0
2𝑞1 + �̈�1 = 0 (∗∗∗) 
 
𝐸 𝑠𝑒, 𝑝𝑜𝑟 𝑜𝑢𝑡𝑟𝑜 𝑙𝑎𝑑𝑜, 𝑠𝑢𝑏𝑡𝑟𝑎í𝑠𝑠𝑒𝑚𝑜𝑠 𝑎𝑠 𝑒𝑞𝑢𝑎çõ𝑒𝑠 (∗)𝑒 (∗∗): 
𝜔0
2(𝑥1 − 𝑥2) + (�̈�1 − �̈�2) = −2𝐾(𝑥1 − 𝑥2) 
 
𝑁𝑜𝑣𝑎𝑚𝑒𝑛𝑡𝑒 𝑑𝑖𝑣𝑖𝑑𝑖𝑚𝑜𝑠 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑙𝑎𝑑𝑜𝑠 𝑝𝑜𝑟 2 𝑒 𝑐ℎ𝑎𝑚𝑎𝑚𝑜𝑠: 
 
{
𝑞2(𝑡) =
1
2
(𝑥1 − 𝑥2)
�̈�2(𝑡) =
1
2
(�̈�1 − �̈�2)
 
 
𝑙𝑜𝑔𝑜: 
𝜔0
2𝑞2 + �̈�2 = −2𝐾𝑞2 
𝑠𝑒 𝑐ℎ𝑎𝑚𝑜: 𝜔2
2 = 𝜔0
2 + 2𝐾: 
𝜔2
2𝑞2 + �̈�2 = 0 (∗∗∗∗) 
 
𝑃𝑒𝑟𝑐𝑒𝑏𝑎 𝑞𝑢𝑒, (∗∗∗) 𝑒 (∗∗∗∗)𝑠ã𝑜 𝑒𝑞𝑢𝑎çõ𝑒𝑠 𝑑𝑒 𝑜𝑠𝑐𝑖𝑙𝑎𝑑𝑜𝑟𝑒𝑠 ℎ𝑎𝑟𝑚ô𝑛𝑖𝑐𝑜𝑠, 𝑙𝑜𝑔𝑜: 
 
𝑞1(𝑡) = 𝐴1cos (𝜔0𝑡 + 𝜙1) 
𝑞2(𝑡) = 𝐴2cos (𝜔2𝑡 + 𝜙2) 
 
𝐴𝑠𝑠𝑖𝑚 𝑐𝑜𝑚𝑜, 𝑠𝑢𝑏𝑡𝑟𝑎í𝑚𝑜𝑠 𝑎𝑠 𝑒𝑞𝑢𝑎çõ𝑒𝑠 (∗∗) 𝑒 (∗) 𝑒 𝑐ℎ𝑒𝑔𝑎𝑚𝑜𝑠 𝑛𝑎𝑠 (∗∗∗) 𝑒 (∗∗∗∗) 
𝐹𝑎𝑟𝑒𝑚𝑜𝑠 𝑜 𝑐𝑎𝑚𝑖𝑛ℎ𝑜 𝑜𝑝𝑜𝑠𝑡𝑜, 𝑚𝑎𝑠 𝑐𝑜𝑚 𝑎𝑠 𝑠𝑜𝑙𝑢çõ𝑒𝑠, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑣𝑎𝑚𝑜𝑠 𝑠𝑜𝑚𝑎𝑟 𝑒 𝑠𝑢𝑏𝑡𝑟𝑎𝑖𝑟 𝑞1 𝑒 𝑞2 
𝑝𝑎𝑟𝑎 𝑎𝑐ℎ𝑒𝑟𝑚𝑜𝑠 𝑥1 𝑒 𝑥2: 
𝑥1(𝑡) = 𝑞1 + 𝑞2 
𝑥2(𝑡) = 𝑞1 − 𝑞2 
 
𝑥1(𝑡) = 𝐴1 cos(𝜔0𝑡 + 𝜙1) + 𝐴2cos (𝜔2𝑡 + 𝜙2) 
𝑥2(𝑡) = 𝐴1 cos(𝜔0𝑡 + 𝜙1) − 𝐴2cos (𝜔2𝑡 + 𝜙2) 
𝐸𝑠𝑡𝑎𝑠 𝑠ã𝑜 𝑎𝑠 𝑠𝑜𝑙𝑢çõ𝑒𝑠. 
 
𝑂 𝑒𝑥𝑒𝑟𝑐í𝑐𝑖𝑜 𝑛𝑜𝑠 𝑓𝑜𝑟𝑛𝑒𝑐𝑒 𝑎𝑠 𝑐𝑜𝑛𝑑𝑖çõ𝑒𝑠 𝑖𝑛𝑖𝑐𝑖𝑎𝑖𝑠: 
 
𝑥1(𝑡) = 0 
�̇�1(𝑡) = 0 
𝑥2(𝑡) = 0 
�̇�2(𝑡) = 𝑣 = 10 𝑐𝑚/𝑠 
 
𝐴𝑔𝑜𝑟𝑎 𝑎𝑠 𝑎𝑝𝑙𝑖𝑐𝑎𝑚𝑜𝑠 𝑛𝑎𝑠 𝑠𝑜𝑙𝑢çõ𝑒𝑠: 
𝑥1(0) = 𝐴1 cos(𝜙1) + 𝐴2 cos(𝜙2) = 0 
�̇�1(0) = −𝜔0𝐴1 sin(𝜙1) − 𝜔2𝐴2 sin(𝜙2) = 0 
𝑥2(0) = 𝐴1 cos(𝜙1) − 𝐴2 cos(𝜙2) = 0 
�̇�2(0) = −𝜔0𝐴1 sin(𝜙1) + 𝜔2𝐴2 sin(𝜙2) = 𝑣 
 
𝑇𝑒𝑚𝑜𝑠, 𝑒𝑛𝑡ã𝑜, 𝑢𝑚 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜 𝑑𝑒 4 𝑒𝑞𝑢𝑎çõ𝑒𝑠 𝑒 4 𝑖𝑛𝑐ó𝑔𝑛𝑖𝑡𝑎𝑠, 𝑙𝑜𝑔𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑠𝑜𝑙𝑢çã𝑜. 
𝑃𝑎𝑟𝑎 𝑟𝑒𝑠𝑜𝑙𝑣𝑒𝑟: 
𝒙𝟏(𝟎) − 𝒙𝟐(𝟎) 
𝐴1 cos(𝜙1) + 𝐴2 cos(𝜙2) − 𝐴1 cos(𝜙1) + 𝐴2 cos(𝜙2) = 0 
2 cos(𝜙2) = 0 ∴ 𝜙2 =
𝜋
2
 
 
𝒙𝟏(𝟎) + 𝒙𝟐(𝟎) 
𝐴1 cos(𝜙1) + 𝐴2 cos(𝜙2) + 𝐴1 cos(𝜙1) − 𝐴2 cos(𝜙2) = 0 
2 cos(𝜙1) = 0 ∴ 𝜙1 =
𝜋
2
 
 
�̇�𝟏(𝟎) − �̇�𝟐(𝟎) 
−𝜔0𝐴1 sin(𝜙1) − 𝜔2𝐴2 sin(𝜙2) + 𝜔0𝐴1 sin(𝜙1) − 𝜔2𝐴2 sin(𝜙2) = −𝑣 
−𝜔2𝐴2 sin(𝜙2) − 𝜔2𝐴2 sin(𝜙2) = −𝑣 
𝜔2𝐴2 sin (
𝜋
2
) + 𝜔2𝐴2 sin (
𝜋
2
) = 𝑣 
𝐴2 =
𝑣
2𝜔2
 
 
�̇�𝟏(𝟎) + �̇�𝟐(𝟎) 
−𝜔0𝐴1 sin(𝜙1) − 𝜔2𝐴2 sin(𝜙2) − 𝜔0𝐴1 sin(𝜙1) + 𝜔2𝐴2 sin(𝜙2) = 𝑣 
−𝜔0𝐴1 sin (
𝜋
2
) − 𝜔0𝐴1 sin (
𝜋
2
) = 𝑣 
𝐴1 = −
𝑣
2𝜔0
 
 
𝑆𝑎𝑏𝑒𝑛𝑑𝑜 𝑞𝑢𝑒 𝑎 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑡𝑟𝑎𝑡𝑎 𝑑𝑜𝑠 𝑚á𝑥𝑖𝑚𝑜𝑠 𝑑𝑒 𝑜𝑠𝑐𝑖𝑙𝑎çã𝑜 (±) é 𝑐𝑜𝑟𝑟𝑒𝑡𝑜 𝑑𝑖𝑧𝑒𝑟: 
 𝐴1 =
𝑣
2𝜔0
 
𝐴𝑔𝑜𝑟𝑎𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑟𝑒𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟 𝑎𝑠 𝑠𝑜𝑙𝑢çõ𝑒𝑠: 
𝑥1(𝑡) =
𝑣
2𝜔0
cos (𝜔0𝑡 +
𝜋
2
) +
𝑣
2𝜔2
cos (𝜔2𝑡 +
𝜋
2
) 
𝑥2(𝑡) =
𝑣
2𝜔0
cos (𝜔0𝑡 +
𝜋
2
) −
𝑣
2𝜔2
cos (𝜔2𝑡 +
𝜋
2
) 
𝑜𝑢 𝑝𝑜𝑠𝑠𝑜 𝑒𝑠𝑐𝑟𝑒𝑣𝑒𝑟 𝑎𝑠𝑠𝑖𝑚: 
𝑥1(𝑡) =
𝑣
2𝜔0
sin(𝜔0𝑡) +
𝑣
2𝜔2
sin(𝜔2𝑡) 
𝑥2(𝑡) =
𝑣
2𝜔0
sin(𝜔0𝑡) −
𝑣
2𝜔2
sin(𝜔2𝑡) 
 
𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑐𝑜𝑛ℎ𝑒𝑐𝑖𝑑𝑜𝑠: 
 
𝑥1(𝑡) = 0,0113 sin(4,43𝑡) + 0,003 sin(14,8𝑡) 
𝑥2(𝑡) = 0,0113 sin(4,43𝑡) − 0,003 sin(14,8𝑡)