SME3701 Minor Test 2 Answers Year 2024

Solid Mechanics IV

Minor Test 2 Answers

Year 2024
20 SEPTEMBER 2024
DUE DATE

,
, QUESTION 1




k = 4 000 N/m m = 10 kg c = 30 N.s/m F(t) = 200 cos 12t


𝐹0 = 200 𝑁 𝜔 = 12 𝑟𝑎𝑑/𝑠 𝑥0 = 0.1 𝑥̇ 0 = 0


𝑘 4 000
𝜔𝑛 = √ = √ = 20 𝑟𝑎𝑑/𝑠
𝑚 10


𝐹0 200
𝛿𝑠𝑡 = = = 0.05 𝑚
𝑘 4 000

𝑐 𝑐 30
𝜁= = = = 0.075
𝑐𝑐 2√𝑘𝑚 2√(4000)(10)


𝜔𝑑 = √1 − 𝜁 2 . 𝜔𝑛 = √1 − 0.0752 × 20 = 19.943671 𝑟𝑎𝑑/𝑠

𝜔 12
𝑟= = = 0.6
𝜔𝑛 20

𝛿𝑠𝑡 0.05
𝑋= =
√(1 − 𝑟 2 )2 + (2𝜁𝑟)2 √(1 − 0.62 )2 + ((2)(0.075)(0.6))
2




= 0.077364

2𝜁𝑟 2 × 0.075 × 0.6
𝜙 = 𝑎𝑟𝑐𝑡𝑎𝑛 ( 2
) = arctan ( )
1−𝑟 1 − 0.62

= 0.139709 𝑟𝑎𝑑


Steady state response:


𝑥𝑝 (𝑡) = 𝑋 cos(𝜔𝑡 − 𝜙)


𝒙𝒑 (𝒕) = 𝟎. 𝟎𝟕𝟕𝟑𝟔𝟒 𝐜𝐨𝐬(𝟏𝟐𝒕 − 𝟎. 𝟏𝟑𝟗𝟕𝟎𝟗)