Damped Harmonic Motion

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1 Daped Haronic Motion PY154 Special Topics in Physics PY154 1

2 Driven Daped Haronic Motion What if we apply a haronic force?: F h Be it The total force is then: dx F Fh kx b dt d x dt Assue a solution of the for: x Ae it PY154

3 Driven Daped Haronic Motion A B i A b B The aplitude of the oscillations lations%and%waves/drivenshm.swf PY154 3

4 Driven Daped revised What if we apply a haronic force?: F h Be it The total force is then: Assue a solution of the for: dx i t iae dt d x dt i Ae t dx F Fh kx b dt x i Ae t d x dt PY154 4

5 Driven Daped revised But, the iaginary part ust be zero, so: sin cos tan 1 PY154 5

6 Driven Daped revised Take a look at the phase: φ = tan 1 γω ω ω = tan 1 γω ω ω Which fit into: i ωt+φ x = Ae We can also note that if: i ωt φ x = Ae φ = tan 1 γω ω ω Both signs possible, depending on initial assuptions PY154 6

7 Driven Daped revised The aplitude can be calculated : B = A ω ω + ω γ Thus: A B as before When: A B PY154 7

8 Driven Daped Haronic Motion What if we apply a haronic force?: F h Bsint dx d x The total force is then: F Fh kx b dt dt Assue a solution of the for: x C dx C cost D sint dt d x C sint D cost dt sint D cost Aplitude: A C D PY154 8

9 Driven Daped Haronic Motion t kc sint D cost Bsin b C cost Dsint C sint D cos t Separate into sine & cosine ters: Bsint kd cost b kc sint b k Dsint C cost b D cost C sint sine: B C D C k cosine: C D D PY154 C D 9

10 PY154 Driven Daped Haronic Motion 1 D C B C D C B B B C D C A But we need:

11 PY154 Driven Daped Haronic Motion 11 1 C D C A C B C

12 Driven Daped Haronic Motion B So, finally A = ω ω + ω γ Can also calculate phase w.r.t. B, fro C and D Soe algebra to siplify, first: B A = ω 1 ω + ω γ ω ω ω PY154 1

13 Driven Daped Haronic Motion Then set B = ω, A = B ω = ω ω + ω ω γ ω When: ω ω A ω = ω = 1 γ = ω γ ω PY154 13

14 Driven Daped Haronic Motion The Q factor Q = ω γ, γ ω = 1 Q A = 1 1 ω ω + ω ω 1 Q PY154 14

15 The aplitude vs. frequency A PY154 15

16 The Phase tan 1 where: b If: PY154 16

17 The Phase tan 1 where: b If: PY154 17

18 The phase vs. frequency PY154 18

19 Power absorbed The oscillator will absorb power fro the driving force, which results in the increased oscillations: F h = B sin ωt x = A sin ωt + φ P = F h v v = dx = ωa cos ωt + φ dt P = ABω sin ωt cos ωt + φ = ABω sin ωt cos ωt cos φ sin ωt sin φ = ABω sin ωt cos ωt cos φ sin ωt sin φ PY154 19

20 Power absorbed Consider the tie averages: T sin ωt cos ωt = 1 T න sin ωt cos ωt dt x = sin(ωt), dx dx = ω cos ωt, = cos ωt dt dt ω sin ωt cos ωt = 1 Tx T න dx = x ቚ T sin ωt T = ቚ = PY154

21 Power absorbed Consider the tie averages: sin ωt = 1 T න = 1 T Tsin ωt dt = 1 x cos ωt ω Thus the power absorbed: T T න = 1 T 1 cos ωt dt P = ABω sin ωt cos ωt cos φ sin ωt sin φ = 1 ABω sin φ Referred to in assignent PY154 1

22 The power vs. frequency Q = ω γ = ω Δω P Δω PY154

23 How long does it take? Look at the following web site, and exaine how long it takes for the oscillations to increase with sall daping: lations%and%waves/drivenshm.swf The picture on the next page gives an idea, about why Q oscillations are required, where: Q = ω γ = ω Δω PY154 3

24 The oscillator takes tie to charge PY154 4

Physics 41 HW Set 1 Chapter 15 Serway 7 th Edition

Physics 41 HW Set 1 Chapter 15 Serway 7 th Edition Physics HW Set Chapter 5 Serway 7 th Edition Conceptual Questions:, 3, 5,, 6, 9 Q53 You can take φ = π, or equally well, φ = π At t= 0, the particle is at its turning point on the negative side of equilibriu,