Exam (elaborations)
SME3701 Minor Test 2 Answers Year 2024
- Course
- Solid Mechanics IV (SME3701)
- Institution
- University Of South Africa (Unisa)
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Solid Mechanics IVMinor 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)
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