Wave Phenomena Physics 15c. Lecture 2 Damped Oscillators Driven Oscillators

Transcription

1 Wave Phenomena Physics 15c Lecture Damped Oscillators Driven Oscillators

2 What We Did Last Time Analyzed a simple harmonic oscillator The equation of motion: The general solution: Studied the solution m d x(t) dt = kx(t) x(t) = a cosωt + b sinωt where ω = Frequency, period, energy conservation Learned to deal with complex exponentials Makes it easy to solve linear differential equations Studied how the equation of motion can be linearized for small oscillations Taylor expansion of the potential near the minimum k m

3 Goals for Today Damped oscillator Harmonic oscillator with friction = energy loss (Re)introduce quality factor Q Driven (or forced ) oscillator Resonance Good exercise of complex math Finish discussion of single oscillators

4 RLC Circuit Ideal LC circuit is a harmonic oscillator Real ones aren t Because there is always resistance Oscillation dies away Let s analyze an RLC circuit Step 1: define current I and Charge q Step : Kirchhoff L di dt RI q C = 0 I = dq dt d q dt + R L Easiest way to solve this: assume q(t) = e Xt Equation becomes X + R L X + 1 LC = 0 C +q -q dq dt + q LC = 0 I R L

5 RLC Circuit Solutions X = R L ± R X 4L 1 + R L X + q LC = 0 solutions Is this positive LC or negative? Three possible scenarios Positive real solutions X = α ± β Negative complex solutions X = α ± iω Zero 1 duplicate solution X = α α = R L, β = R 4L 1 LC, ω = 1 LC R 4L Border line is R = L C Smaller R complex solutions Larger R real solutions critical resistance

6 Weak Damping For, we find R < L C q(t) = e αt e ±iωt e ix = cos x + i sin x q(t) = e αt (cosωt ± i sinωt) Use Euler s formula: Linear combinations of the solutions are q(t) = e αt (Acosωt + B sinωt) damping oscillation C α = R L, ω = 1 LC R 4L +q -q I R L Note that R 0 recovers the simple LC oscillator

7 Strong Damping For R > L C, we find Note α ± β > 0, so both solutions are exponentially decreasing Linear combinations of the solutions are A and B are determined by the initial conditions In our problem, q(0) = q 0, I(0) = 0 (α ±β)t q(t) = e q(t) = Ae (α +β)t (α β)t + Be q(0) = A + B = q 0 q(t) = q 0 α β β C +q I(0) = dq dt t =0 = (α + β)a (α β)b = 0 e (α +β)t + α + β β α = R L, β = R -q e (α β)t I R 4L 1 LC L

8 Critical Damping For, we find only R = L C q(t) = e αt α = R L We can t satisfy the initial conditions General solution turns out to be q(t) = (A + Bt)e αt Not trivial to come up with this In our problem, q(0) = q 0, I(0) = 0 q(0) = A = q 0 C I(0) = dq dt t =0 = αa + B = 0 q(t) = q 0 (1+ αt)e αt +q -q I R L

9 How Do They Look? Q 0 Critically-damped Strongly-damped 0 Weakly-damped Q 0 For same L and C, a critically damped oscillator loses energy faster than weakly/strongly damped oscillators

10 Quality Factor How many times does a damped oscillator oscillate? For weak damping, the ratio Q ω α determines this This is called the quality factor of the oscillator Defined as the number of radians it oscillates during the time it takes for the stored energy to decrease by a factor 1/e For an RLC circuit Q = ω α 1 R Large Q value = weak damping = more oscillations Critical damping occurs at Q = 1/ L C

11 Critical Damping Critically damped oscillator stops most quickly The energy is dissipated in R as fast as possible This is useful when you are trying to control the movement Shock absorbers on automobiles Feedback control systems (e.g. thermostats) Underdamped oscillation can cause disasters Overloaded tracks. Broken shock absorbers

12 Forced Oscillation F A motor drives one end of the spring ω = Angular frequency of the motor The spring is stretched/shrunk by r cosωt on the motor side x(t) on the mass side Equation of motion: F = k(x(t) r cosωt) m d x(t) dt + kx(t) = kr cosωt

13 Equation of Motion m d x(t) + kx(t) = kr cosωt dt We are interested in the motion caused by the motor The solution should oscillate with angular frequency ω Let s work in complex exponentials cosωt replace eiωt x(t) = x 0 e iωt x(t) = kr mω + k eiωt = r The amplitude of the oscillation depends on the driving frequency ω ( mω + k)x 0 e iωt = kre iωt x 0 = ω 0 ω 0 ω eiωt where ω 0 = k m kr mω + k Natural frequency without the force

14 Friction The oscillation amplitude x 0 goes to infinity when ω = ω 0 This is does not happen in reality because of friction Include friction v = dx dt Friction F is always opposite to the velocity v Simplest model: F is proportional to v Not true in many cases Good enough approximation for today Must be OK if the velocity is small (Taylor expansion) F = fv = f dx(t) dt

15 Equation of Motion m d x(t) dt + f dx(t) dt + kx(t) = kr cosωt We now have a friction term Let s work with complex exponentials again ( mω + ifω + k)x 0 e iωt = kre iωt x 0 = kr mω + ifω + k = rω 0 ω 0 ω + iγω Γ represents the damping strength Corresponds to α in the RLC circuit k ω 0 m Γ f m Quality factor is Q = ω 0 Γ = mk f

16 Solution k rω x(t) = 0 ω 0 ω 0 ω + iγω eiωt m Γ f m The amplitude and the phase of the oscillation depends on ω We assume Γ is much smaller than ω 0 (weakly damped) At low frequency x(t) = The mass moves exactly following the motor At high frequency x(t) = rω 0 ω 0 ω + iγω eiωt rω 0 ω 0 ω + iγω eiωt re iωt Real part r ω 0 ω eiωt The oscillation becomes small due to (ω 0 /ω) Re x(t) = r cosωt x(t) = r ω 0 ω cosωt

17 Resonance Suppose the motor is running at ω 0 x(t) = rω 0 ω 0 ω + iγω eiωt = rω 0 iγ eiωt The oscillation is larger than the movement of the motor end by the quality factor Q = ω 0 /Γ The smaller the damping, the larger the oscillation Q quantifies how resonant a damped oscillator is Large Q Weakly-damped, highly-resonant Small Q Heavily-damped, muffled The phase is 90 behind the driving force Re x(t) = r ω 0 Γ sinωt

18 How They Look Like Q = ω 0 Γ = 3 ω < ω 0 ω > ω 0 ω = ω 0

19 Amplitude Near the Resonance x(t) = rω 0 ω 0 ω + iγω eiωt The amplitude is x = xx * = rω 0 ( ω ω ) 0 + Γ ω

20 Phase Near the Resonance The phase of the oscillation is arg(1/ z) = arg(z) rω arg 0 ω 0 ω + iγω = arg ( ω ω + iγω ) 0 ω ω = arccot 0 Γω

21 Initial Condition The solution we found has no free parameter What happened to the parameters in the initial condition? Imagine we didn t have the motor The equation of motion would be m d x(t) dt + f dx(t) dt We know the general solution to it x(t) = rω 0 ω 0 ω + iγω eiωt + kx(t) = 0 A damped oscillator x(t) = x 0 e Γ t e iω 0 t parameters in x 0

22 Initial Condition Add the forced and the damped solutions x(t) = rω 0 ω 0 ω + iγω eiωt Because of linearity, the sum satisfies m d x(t) dt + f dx(t) dt + x 0 e Γ t e iω 0 t + kx(t) = re iωt This solution can satisfy any initial condition Second term disappears with time We concentrated on the first term ( steady-state solution )

23 What Happens to Energy? Energy is not conserved because of the friction The energy consumed per unit time = (friction) x (speed) Motor supplies the energy The work done per unit time = (force) x (speed) at the motor-spring connection friction = f x(t) speed = x(t) force = k(x(t) r cosωt) speed = rω sinωt f { x(t) } krω sinωt(x(t) r cosωt)

24 What Happens to Energy? Energy consumed by the friction must equal to the work done by the motor This is true only after averaging over time for one cycle Because the spring and the mass store/release energy Next two slides show how to confirm E f = E m E f = energy consumed by the friction in one cycle E m = energy supplied by the motor in one cycle Unfortunately, we must do this in real numbers, with sines and cosines Energy is not a linear quantity

25 Energy Consumed by Friction E f = T 0 T 0 f x dt ( ) dt = f Re iω x 0 e iωt = f = f T 0 T 0 = ftω = ftr iω x 0 e iωt iω x 0 e iωt dt ω x 0 e iωt + ω x 0 x 0 ω x 0 e iωt 4 rω 0 ω 0 ω + iωγ ω 0 4 ω (ω 0 ω ) + Γ ω rω 0 ω 0 ω iωγ dt x 0 = x(t) = x 0 e iωt Re(z) = z + z T e ±iωt dt = 0 0 rω 0 ω 0 ω + iγω

26 Work Done by Motor E m = T 0 k(x(t) r cosωt)rω sinωt dt = krω Re( ( x 0 r )e iωt )Im(e iωt ) dt = krω = krω T 0 T 0 T 0 ( )e iωt ( x 0 r )e iωt + x 0 r ( x 0 r )e iωt x 0 r = krωt Im ( x ) = krωt 0 = ftr ω 0 4 ω ( ) + x 0 r e iωt e iωt i dt ( ) x 0 r 4i rω 0 Γω (ω 0 ω ) + Γ ω ( )e iωt (ω 0 ω ) + Γ ω Γ = f m dt k m = ω 0

27 Energy Balance Sheet E f T = E m T = fr ω 4 0 ω ( ω 0 ω ) + Γ ω So the intake equals the consumption on average Energy flow is proportional to f r No energy is consumed if the friction f is zero Low frequency limit ω << ω 0 fr ω 0 ω ω 0 0 High frequency limit ω >> ω 0 fr ω 0 ω 0 ω 0

28 Energy Balance Sheet On resonance ω = ω 0 E f T = E m T = fr ω 0 4 ω ( ω 0 ω ) + Γ ω = fr 4 ω 0 = mr ω 0 Γ The motor does large amount of work only when ω is close to the resonance frequency ω 0 The energy flow at the resonance is proportional to Q 3 Q The smaller the friction, the larger the loss? Because the mass moves larger distance, and faster Γ f m Q ω 0 Γ

29 Energy vs. Frequency Average energy consumption for the same m and k. fr ω 0 4 ω ( ω 0 ω ) + Γ ω

30 Summary Analyzed a damped oscillator Behavior depends on the damping strength Weakly, strongly, or critically-damped How many oscillation does it do? Q factor Studied forced oscillation Oscillation becomes large near the resonance frequency Phase changes from 0 π/ π as the frequency increases through the resonance Work done by the force is consumed by the friction Energy consumption is large near the resonance Resonance is taller and narrower for a larger Q value Next to come: coupled oscillators