Transcript Physics 321 Hour 12 Driven Harmonic Oscillators

Physics 321
Hour 12
Driven Harmonic Oscillators
Bottom Line
• Starting from rest, the system tries to oscillate at the
natural frequency
• In time, it oscillates at the driving frequency
• The equation for a driven, damped oscillator:
𝑥 + 2𝛽𝑥 + 𝜔0
2𝑥
𝐹(𝑡)
=
≡ 𝑓0 cos 𝜔𝑡
𝑚
• The steady state solution is:
𝑥 𝑡 = 𝐴 cos(𝜔𝑡 − 𝛿)
• FWHM of the resonance curve is approximately 2β
Driven Oscillator
2
The equation: 𝑥 + 2𝛽𝑥 + 𝜔0 𝑥 =
𝐹(𝑡)
𝑚
≡ 𝑓(𝑡)
Let 𝑓 𝑡 = 𝑓0 cos 𝜔𝑡
The oscillator wants to oscillate at 𝜔0 but the
driver forces it to oscillate at 𝜔. This leads to
transient vs steady state behavior!
Example
Driven_Osc.nb
Driven Oscillator
𝑥 + 2𝛽𝑥 + 𝜔02 𝑥 = 𝑓0 cos 𝜔𝑡
We assume a solution something like
𝑥 𝑡 = 𝐴 cos(𝜔𝑡 − 𝛿)
But 𝑥(𝑡) = −𝐴𝜔 sin(𝜔𝑡 − 𝛿)
So we employ a trick…
The driving force is the real part of
𝑓0 𝑒 𝑖𝜔𝑡
Driven Oscillator
𝑧 + 2𝛽 𝑧 + 𝜔02 𝑧 = 𝑓0 𝑒 𝑖𝜔𝑡
We assume a solution of the form
𝑧 𝑡 = 𝐴𝑒 −𝑖𝛿 𝑒 𝑖𝜔𝑡 = 𝐶𝑒 𝑖𝜔𝑡
This gives:
(−𝜔2 + 2𝑖𝛽𝜔 + 𝜔02 )𝐶𝑒 𝑖𝜔𝑡 = 𝑓0 𝑒 𝑖𝜔𝑡
𝑓0
𝐶= 2
𝜔0 − 𝜔 2 + 2𝑖𝛽𝜔
Driven Oscillator
Conclusion 1:
2
𝑓0
|𝐶| = 𝐴 =
(𝜔02 − 𝜔 2 )2 +4𝛽 2 𝜔 2
2
2
Driven Oscillator
(−𝜔2 + 2𝑖𝛽𝜔 + 𝜔02 )𝐴𝑒 −𝑖𝛿 𝑒 𝑖𝜔𝑡 = 𝑓0 𝑒 𝑖𝜔𝑡
(−𝜔2 + 2𝑖𝛽𝜔 + 𝜔02 )𝐴 = 𝑓0 𝑒 𝑖𝛿
𝜔02 − 𝜔2 𝐴 + 2𝑖𝛽𝜔𝐴 = 𝑓0 cos 𝛿 + 𝑖𝑓0 sin 𝛿
Real parts: 𝜔02 − 𝜔2 𝐴 = 𝑓0 cos 𝛿
Imaginary parts: 2𝛽𝜔𝐴 = 𝑓0 sin 𝛿
𝑓0 sin 𝛿
2𝛽𝜔
tan 𝛿 =
= 2
𝑓0 cos 𝛿 𝜔0 − 𝜔 2
Driven Oscillator
Conclusion 2:
2𝛽𝜔
tan 𝛿 = 2
𝜔0 − 𝜔 2
And finally the steady state solution is:
𝑥 𝑡 = 𝐴 cos(𝜔𝑡 − 𝛿)