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PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Damped Harmonic Motion
PY1054
Special Topics in Physics
1
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
2
What if we apply a harmonic force?: ti
h BeF
2
2
dt
xdm
dt
dxbkxFF h The total force is then:
Assume a solution of the form: tiAex
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
3
http://www.cabrillo.edu/~jmccullough/Applets/Flash/Fluids,%20Oscil
lations%20and%20Waves/DrivenSHM.swf
The amplitude of the oscillations…
im
BA
22
0 22222
0
m
BA
m
b
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped revised
4
What if we apply a harmonic force?:ti
h BeF
2
2
dt
xdm
dt
dxbkxFF h The total force is then:
Assume a solution of the form: tiAex
tiAeidt
dx
tiAedt
xd 2
2
2
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped revised
5
But, the imaginary part must be zero, so:
cossin22
0
22
0
1tan
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped revised
6
Take a look at the phase:
𝜙 = tan−1 −𝛾𝜔
𝜔02 − 𝜔2
= −tan−1𝛾𝜔
𝜔02 − 𝜔2
Which fit into: 𝑥 = 𝐴𝑒𝑖 𝜔𝑡+𝜙
We can also note that if: 𝑥 = 𝐴𝑒𝑖 𝜔𝑡−𝜙
𝜙 = tan−1𝛾𝜔
𝜔02 − 𝜔2
Both signs possible, depending on initial assumptions
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped revised
7
Thus:
The amplitude can be calculated :
22222
0
m
BA as before
0m
BA When: 0
𝐵2 = 𝑚2𝐴2 𝜔02 − 𝜔2 2 + 𝜔2𝛾2
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
8
What if we apply a harmonic force?: tBFh sin
2
2
dt
xdm
dt
dxbkxFF h The total force is then:
Assume a solution of the form: tDtCx cossin
tDtCdt
dx sincos
tDtCdt
xd cossin 22
2
2
222 DCA
Amplitude:
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
9
tDmtCmtDtCb
tDtCktB
cossinsincos
cossinsin
22
tCmtDbtkCtB sinsinsinsin 2
tDmtCbtkD coscoscos 2
Separate into sine & cosine terms:
sine: DCmCDm
bC
m
kmB
22
0
2
cosine: DDm
kC 22
0
2
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
10
DCmB 22
0CD
22
0
CmB
22
0
2222
0
22222
0
22
0
22
0
2222
0
m
B
m
BC
222 DCA But we need:
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
11
2
222
0
22
1 C
222 DCA
2
222
0
22222
0 C
222222
0
2
222
0
22
m
BC
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
12
So, finally
Can also calculate phase w.r.t. B, from C and D
𝐴 =𝐵
𝑚 𝜔02 − 𝜔2 2 + 𝜔2𝛾2
Some algebra to simplify, first:
A =𝐵
𝑚𝜔02 1 −
𝜔𝜔0
2 2
+𝜔𝜔0
2 𝛾2
𝜔02
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
13
Then set 𝐵 = 𝑚𝜔02, →
𝐵
𝑚𝜔02 = 1
𝐴 =1
1 −𝜔𝜔0
2 2
+𝜔𝜔0
2 𝛾2
𝜔02
When: 𝜔 → 𝜔0 𝐴 𝜔 = 𝜔0 =1
𝛾2
𝜔02
=𝜔0
𝛾
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Driven Damped Harmonic Motion
14
The Q factor 𝑄 =𝜔0
𝛾, →
𝛾2
𝜔02 =
1
𝑄2
𝐴 =1
1 −𝜔𝜔0
2 2
+𝜔𝜔0
2 1𝑄2
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
The amplitude vs. frequency
15
0
A
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
The Phase
16
22
0
1tan
If: 0 2
m
b0where:
0
22
0
022
0
2
0 022
0
0
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
The Phase
17
22
0
1tan
If: 0 2
m
b0where:
0
22
0
022
0
2
0 022
0
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
The phase vs. frequency
18
0
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Power absorbed
19
The oscillator will absorb power from the driving force, which results in the increased oscillations:
𝑃 = 𝐹ℎ𝑣
𝐹ℎ = 𝐵 sin 𝜔𝑡 𝑥 = 𝐴 sin 𝜔𝑡 + 𝜙
𝑣 =𝑑𝑥
𝑑𝑡= 𝜔𝐴 cos 𝜔𝑡 + 𝜙
𝑃 = 𝐴𝐵𝜔 sin 𝜔𝑡 cos 𝜔𝑡 + 𝜙
= 𝐴𝐵𝜔 sin 𝜔𝑡 cos 𝜔𝑡 cos𝜙 − sin 𝜔𝑡 sin𝜙
= 𝐴𝐵𝜔 sin 𝜔𝑡 cos 𝜔𝑡 cos𝜙 − sin2 𝜔𝑡 sin𝜙
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Power absorbed
20
Consider the time averages:
sin 𝜔𝑡 cos 𝜔𝑡 =1
𝑇න0
𝑇
sin 𝜔𝑡 cos 𝜔𝑡 𝑑𝑡
𝑥 = sin(𝜔𝑡) ,𝑑𝑥
𝑑𝑡= 𝜔 cos 𝜔𝑡 ,→
𝑑𝑥
𝜔= cos 𝜔𝑡 𝑑𝑡
sin 𝜔𝑡 cos 𝜔𝑡 =1
𝑇න0
𝑇
𝑥 𝑑𝑥 =𝑥2
2ቚ0
𝑇=sin2 𝜔𝑡
2ቚ0
𝑇= 0
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
Power absorbed
21
Consider the time averages:
sin2 𝜔𝑡 =1
𝑇න0
𝑇
sin2 𝜔𝑡 𝑑𝑡 =1
2𝑇න0
𝑇
1 − cos 2𝜔𝑡 𝑑𝑡
=1
2𝑇𝑥 −
cos 2𝜔𝑡
2𝜔0
𝑇
=1
2
Thus the power absorbed:
𝑃 = 𝐴𝐵𝜔 sin 𝜔𝑡 cos 𝜔𝑡 cos𝜙 − sin2 𝜔𝑡 sin𝜙
=1
2𝐴𝐵𝜔 sin𝜙 Referred to in assignment
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
The power vs. frequency
22
0
P Δ𝜔0
𝑄 =𝜔0
𝛾=
𝜔0
Δ𝜔0
PY1054Coláiste na hOllscoile Corcaigh, Éire University College Cork, Ireland
ROINN NA FISICEDepartment of Physics
How long does it take?
23
Look at the following web site, and examine how long it takes for the oscillations to increase with small damping:
http://www.cabrillo.edu/~jmccullough/Applets/Flash/Fluids,%20Oscil
lations%20and%20Waves/DrivenSHM.swf
The picture on the next page gives an idea, about why Q oscillations are required, where:
𝑄 =𝜔0
𝛾=
𝜔0
Δ𝜔0
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