Mathematical Physics

Transcription

1 Mathematical Physics MP205 Vibrations and Waves Lecturer: Office: Lecture 9-10 Dr. Jiří Vala Room 1.9, Mathema<cal Physics, Science Building, North Campus Phone: (1) E- Mail:

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3 t = 0: the driving force is turned on time transient (both natural & driving vibrations) steady-state (no natural vibrations)

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6 C(ω) F 0 /k ω 0 ω

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10 Example: A (nearly) undamped pendulum of length 38.8 cm, i.e. of a natural frequency ω 0 = 8π/5 rad.s -1, with the forcing done by moving the suspension point back and forth 2 millimetres each way, but with different frequencies: ω = ω 0 /2 ω = ω 0 ω = 2ω 0 resonance

11 Complex representation of the sinusoidal driving force and displacement vector in the forced oscillations:

12 (4) solution mω 2 A + ka e i(ωt+α) = F0e iωt ω 2 0 ω2 A = F 0 m e iα = F 0 m cos α if 0 m sin α the equation has to be satisfied for both the real and imaginary parts: ω 2 0 ω 2 A = F 0 m cos α and 0 = F 0 m sin α from these equations we get the final expressions for the amplitude and phase: A = F 0 /m ω 2 0 ω2 α = 0 for ω < ω 0 α = π for ω > ω 0

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14 Forced oscillations with damping The EOM is obtained by adding both the damping and the driving force to the Hook s law: m d2 x = kx bdx dt2 dt + F 0 cos ωt d 2 x dt 2 + γdx dt + ω2 0 x = F 0 cos ωt (6) m where ω 2 = k/m and γ = b/m. 0

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19 Barton s pendulum in which several pendula of differing lengths are all driven by an oscillation of the same frequency: Time: t t + Δt

20 Barton s pendulum in which several pendula of differing lengths are all driven by an oscillation of the same frequency: Time: t t + Δt

21 Barton s pendulum: The driving frequency is ω = 6π/5 rad.s -1, the damping constant is the same, i.e. Q = 1. The ten blue pendula which would have the above behaviour in real life are of lengths from 6.9 cm to cm in increments of 13.8 cm. The red pendulum is at resonance, i.e. its natural frequency is identical to the driving frequency, and would be 69.0 cm long. Note that all pendula shorter than this red one lead it, while the longer ones lag behind.

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23 It will be convenient to use the ratio ω/ω 0, rather than ω itself, as a variable: or A(ω) = F 0 mω 2 0 ω 0 /ω ω0 ω ω ω 2 1/ Q 2 A(ω) = F 0 k ω 0 /ω ω0 ω ω ω 2 1/ Q 2 (12) tan δ(ω) = 1/Q ω 0 ω ω ω 0 (13)

24 Example: the effect of damping on kicked pendula: pendula are kicked by a half-second long oscillation of amplitude 3.9 cm, with the damping constants such that Q = 2 underdamped Q = 1/2 critically damped Q = 2/5 overdamped Note that, in both the under- and overdamped cases, the pendulum slighly overshoots its initial position, whereas in the critically damped case, it gradually approaches the vertical without overshooting. This is why the indicator needles in instruments like ammeters and voltmeters are critically damped, so that they quickly approach a final reading rather than wobble about it.

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28 Sharpness of tuning of a resonant system with Q: example: Barton s pendulum Q 1 < Q 2 < Q 3

29 t = 0: the driving force is turned on time transient (both natural & driving vibrations) steady-state (no natural vibrations)

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33 Let the equation of free vibrations contain two adjustable constants - an amplitude and an initial phase (B and β), then the complete solution of the forced motion is x = B cos ω 0 t + β + C cos (ωt) where C = F 0/m ω 2 We can now tailor the equation to fit the initial conditions (in this ω2. 0 case) that x = 0 and dx/dt = 0 at t = 0: which requires that x = 0: 0 = B cos β + C dx dt : dx = ω dt 0 B sin ω 0 t + β ωc sin ωt 0 = ω 0 B sin (β) β = 0 or β = π

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36 Damping To include damping, we postulate the solution in the form of the combination of free and steady-state motions: x = Be γt/2 cos (ω 1 t + β) + A cos (ωt δ) where B and β are obtained by fitting the solution to x and dx/dt at t = 0, and ω 1 = ω 2 0 γ2 1/2 4 and A and δ are given as A(ω) = F 0 /m ω 2 0 ω (γω) 2 1/2 γω tan δ(ω) = ω 2 0 ω2

37 Response to a periodic driving force Undamped harmonic oscillator the beat pattern continues indefinitely Transient behavior of damped harmonic oscillator driven off-resonance Transient behavior of damped harmonic oscillator driven at resonance

38 The power absorbed by a driven oscillator Problem: At what rate energy must be supplied to a driven oscillator to maintain its oscillations at a fixed amplitude? the instantaneous power input P = dw dt = F dx dt = Fv where F is the driving force and v = dx/dt is the velocity.

39 (1) The undamped oscillator since in this case there is no dissipation, the mean power input is zero F = F 0 cos ωt x = F 0 /m cos ωt = C cos ωt ω2 ω 2 0 v = ωc sin ωt P = ωcf 0 sin ωt cos ωt sin (2ωt) therefore P is positive half the time and is negative the other half, averaging out to zero over any integral number of half-periods of oscillations. That is the energy is pumped into the system during one quarter-cycle and is pumped out again during the next quarter-cycle.

40 (2) The forced oscillator with damping x = A cos (ωt δ) v = ωa sin (ωt δ) = v 0 sin (ωt δ) where v 0 is the maximal velocity for any given value of F 0 and ω: v 0 (ω) = F 0 ω 0 /k ω0 ω ω ω 2 1/ Q 2 the value of v 0 passes through a maximum at ω = ω 0 exactly; we call this velocity resonance.

41 P = F 0 v 0 cos ωt sin (ωt δ) = F 0 v 0 cos ωt (sin ωt cos δ cos ωt sin δ) = F 0 v 0 cos δ sin ωt cos ωt + F 0 v 0 sin δ cos 2 ωt If we average over many cycles, the first term gives zero, and the second term contributes as P = 1 2 F 0v 0 sin δ = 1 2 ωaf 0 sin δ P = F2 0 ω 0 2kQ The P reaches its maximal value for ω = ω 0 : P m = F2 0 ω 0Q 2k 1 ω0 ω ω ω (15) Q 2 = QF2 0 2mω 0

42 _ P(ω) Mean power absorbed by a forced oscillator Q = 30 in units of F 02 /2mω 0 _ P = QF max 0 2 /2mω 0 Q = 10 Q = 3 Q = 1 ω/ω 0

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44 Sharpness of resonance curve determined in terms of power curve P max P max /2 Width of power resonance curve at half-height = γ or ω 0 /Q very nearly ω 0 ω 0 γ/2 ω 0 + γ/2

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46 The damping constant γ = ω 0 /Q characterizes the rate at which the energy of a damped oscillator decays in the absence of the driving force E = E 0 e (ω 0/Q)t = E 0 e γt using the damping constant we get P approx (ω) = γf2 0 2m 1 4 ω 0 ω 2 + γ 2 The frequencies ω 0 ± ω at which P(ω) falls to half of the maximum P(ω 0 ) are thus defined as 4 ( ω) 2 = γ 2 or 2 ω = γ = ω 0 Q (16)

47 Sharpness of resonance curve determined in terms of power curve P max P max /2 Width of power resonance curve at half-height = γ or ω 0 /Q very nearly ω 0 ω 0 γ/2 ω 0 + γ/2

48 This implies that the width of the resonance curve for the driven oscillator (measured by the power input) is equal to the reciprocal of the time needed for the free oscillations to decay to 1/e of their initial energy. This leads to the following conclusions (i) narrow resonance = slow decay of oscillations narrow 2 ω ω 0 << 1 slow: E E γ t-asmall fraction of energy is lost during one period of oscillation. Since t = 2π/ω 0, we get E E 2πγ ω 0 = 2π Q << 1 slow (17) (ii) broad resonance = fast decay

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