Physics 321 Hour 11 Simple and Damped Harmonic Oscillators

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Physics 321
Hour 11
Simple and Damped Harmonic Oscillators
Bottom Line
β€’ Equation for a damped oscillator:
π‘šπ‘₯ = βˆ’π‘˜π‘₯ βˆ’ 𝑏π‘₯
π‘₯ = βˆ’πœ”02 π‘₯ βˆ’ 2𝛽 π‘₯
β€’ Damping regimes:
β€’ Underdamped:
𝛽 < πœ”0
β€’ Critically damped: 𝛽 = πœ”0
β€’ Overdamped:
𝛽 > πœ”0
Exam #1 Reminders (Spring 2015)
See Review #1, Sample Test #1, Example
Problem
Review is Wednesday 5/13
Test is available Thurs. 5/14 – Monday 5/18
Average time will be about 1 hr 15 min
Simple Harmonic Oscillator
1) What is the equation of motion?
2) Where is your origin?
3) What is the frequency of
oscillation?
4) What are the three ways you can
write the general solution?
5) How many constants do you
have to fix to get a specific
solution?
6) How do you find those
constants?
x
Mass on a Massless Spring
1)
2)
3)
4)
5)
6)
Find U and T
𝑇+π‘ˆ =0
Equilibrium
Equation of motion
Ο‰, T
Energy plot
y
Example
VertSpringUWell.nb
A Shallow Frictionless Bowl
1) Assume 𝑧 β‰ˆ 0, 𝑧 = π›Όπ‘Ÿ 2
2) Find U and T
3) 𝑇 + π‘ˆ = 0
4) 𝐹 = βˆ’π›»π‘ˆ = π‘šπ‘Ÿ
5) Ο‰, T
y
x
z
Another Bowl
1) Assume 𝑧 β‰ˆ 0, 𝑧 = 𝛼π‘₯ 2 + 𝛽𝑦 2
2) Find U
3) 𝐹 = βˆ’π›»π‘ˆ = π‘šπ‘Ÿ
4) Two frequencies
y
x
z
Example
Lissajous.nb
Damped Oscillator
The equation: π‘šπ‘₯ = βˆ’π‘˜π‘₯ βˆ’ 𝑏π‘₯
π‘˜
βˆ’ π‘₯
π‘š
A little rearranging: π‘₯ =
π‘₯ = βˆ’πœ”02 π‘₯ βˆ’ 2𝛽 π‘₯
βˆ’
𝑏
π‘₯
π‘š
A trial solution: π‘₯(𝑑) = 𝑒 π‘Ÿπ‘‘
π‘Ÿ 2 π‘₯ 𝑑 = βˆ’ πœ”02 π‘₯ 𝑑 βˆ’ 2π›½π‘Ÿπ‘₯ 𝑑
π‘Ÿ 2 + πœ”02 + 2π›½π‘Ÿ = 0
π‘Ÿ = βˆ’π›½ ± 𝛽 2 βˆ’ πœ”02
Three Regimes
β€’ Underdamped:
𝛽 < πœ”0
β€’ Critically damped: 𝛽 = πœ”0
β€’ Overdamped:
𝛽 > πœ”0
Underdamped
𝛽 < πœ”0 ,
π‘Ÿ = βˆ’π›½ ± 𝛽 2 βˆ’ πœ”02
π‘Ÿ = βˆ’π›½ ± 𝑖 πœ”02 βˆ’ 𝛽 2
Solution:
π‘₯ 𝑑 = 𝐴𝑒
or
βˆ’π›½π‘‘ +𝑖 πœ”02 βˆ’π›½ 2 𝑑
𝑒
+ 𝐡𝑒
βˆ’π›½π‘‘ βˆ’π‘– πœ”02 βˆ’π›½ 2 𝑑
𝑒
π‘₯ 𝑑
= 𝐴𝑒 βˆ’π›½π‘‘ sin( πœ”02 βˆ’ 𝛽 2 𝑑)
+ 𝐡𝑒 βˆ’π›½π‘‘ cos( πœ”02 βˆ’ 𝛽 2 𝑑)
Note: Ο‰ =
πœ”02 βˆ’ 𝛽 2
Overdamped
π‘Ÿ = βˆ’π›½ ± 𝛽 2 βˆ’ πœ”02
𝛽 > πœ”0 ,
Solution:
π‘₯ 𝑑 = 𝐴𝑒
βˆ’π›½π‘‘ + 𝛽2 βˆ’πœ”02 𝑑
𝑒
+ 𝐡𝑒
βˆ’π›½π‘‘ βˆ’ 𝛽2 βˆ’πœ”02 𝑑
𝑒
Critically Damped
𝛽 = πœ”0 ,
π‘Ÿ = βˆ’π›½ ± 𝛽 2 βˆ’ πœ”02
π‘Ÿ = βˆ’π›½
Solution:
π‘₯ 𝑑 = 𝐴𝑒 βˆ’π›½π‘‘ + 𝐡𝑑𝑒 βˆ’π›½π‘‘
Example
DampOsc5_4.nb
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