PHY26041/102/0/2026 iQ Level: 0793226427
UNISA
PHY2606 Assignment 2 full solutions 2026
Assignment 2
Due Date: 25th May 2026 at 08:00
Assignment Unique number: 209216
_________________________________________________________________
Instruction
- This assignment consists of five (5) questions.
Chapter 3 - The Forced Oscillator (Topic 3)
, Question 1
Equation of motion: 0.40𝑥̈ + 0.20𝑥̇ + 0.275𝑥 = 0.079cos(0.750𝑡)N
Dividing by 𝑚 = 0.40kg:
𝑥̈ + 0.5𝑥̇ + 0.6875𝑥 = 0.1975cos(0.750𝑡)
So: 𝜔0 = √0.6875 = 0.8292rad/s, 2𝛽 = 0.5 ⇒ 𝛽 = 0.25s⁻¹, 𝜔𝑑𝑟𝑖𝑣𝑒 = 0.750rad/s
(a) Amplitude resonance condition: occurs at 𝜔𝑑𝑟𝑖𝑣𝑒 = √𝜔02 − 2𝛽 2 = √0.6875 − 0.125 =
√0.5625 = 0.750rad/s ✓
Steady-state amplitude:
𝐹0 /𝑚 0.1975 0.1975 0.1975
𝐴= = = = = 0.527 m
√(𝜔02 − 𝜔 2 )2 + 4𝛽 2 𝜔 2 √0 + 4(0.0625)(0.5625) √0.140625 0.375
(b) Underdamped condition: requires 𝛽 < 𝜔0 , i.e. 0.25 < 0.8292✓
(c) Amplitude decay: Free oscillation envelope: 𝑥 = 𝐴0 𝑒 −𝛽𝑡
0.005 −ln(0.009487) 4.658
0.005 = 0.527 𝑒 −0.25𝑡 ⇒ 𝑒 −0.25𝑡 = ⇒𝑡= = ≈ 18.6 s
0.527 0.25 0.25
2
UNISA
PHY2606 Assignment 2 full solutions 2026
Assignment 2
Due Date: 25th May 2026 at 08:00
Assignment Unique number: 209216
_________________________________________________________________
Instruction
- This assignment consists of five (5) questions.
Chapter 3 - The Forced Oscillator (Topic 3)
, Question 1
Equation of motion: 0.40𝑥̈ + 0.20𝑥̇ + 0.275𝑥 = 0.079cos(0.750𝑡)N
Dividing by 𝑚 = 0.40kg:
𝑥̈ + 0.5𝑥̇ + 0.6875𝑥 = 0.1975cos(0.750𝑡)
So: 𝜔0 = √0.6875 = 0.8292rad/s, 2𝛽 = 0.5 ⇒ 𝛽 = 0.25s⁻¹, 𝜔𝑑𝑟𝑖𝑣𝑒 = 0.750rad/s
(a) Amplitude resonance condition: occurs at 𝜔𝑑𝑟𝑖𝑣𝑒 = √𝜔02 − 2𝛽 2 = √0.6875 − 0.125 =
√0.5625 = 0.750rad/s ✓
Steady-state amplitude:
𝐹0 /𝑚 0.1975 0.1975 0.1975
𝐴= = = = = 0.527 m
√(𝜔02 − 𝜔 2 )2 + 4𝛽 2 𝜔 2 √0 + 4(0.0625)(0.5625) √0.140625 0.375
(b) Underdamped condition: requires 𝛽 < 𝜔0 , i.e. 0.25 < 0.8292✓
(c) Amplitude decay: Free oscillation envelope: 𝑥 = 𝐴0 𝑒 −𝛽𝑡
0.005 −ln(0.009487) 4.658
0.005 = 0.527 𝑒 −0.25𝑡 ⇒ 𝑒 −0.25𝑡 = ⇒𝑡= = ≈ 18.6 s
0.527 0.25 0.25
2
