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Driven Oscillators Sect. 3.5. Consider a 1d Driven Oscillator with damping, & a time dependent driving force F d (t). Newton’s 2 nd Law Equation of Motion: F = ma = m(d 2 x/dt 2 ) = - kx - bv + F d (t) Can be solved in closed form for many special cases of F d (t). - PowerPoint PPT Presentation

### Transcript of Driven Oscillators Sect. 3.5

• Driven Oscillators Sect. 3.5Consider a 1d Driven Oscillator with damping, & a time dependent driving force Fd(t).Newtons 2nd Law Equation of Motion:F = ma = m(d2x/dt2) = - kx - bv + Fd(t)Can be solved in closed form for many special cases of Fd(t).The text immediately goes to the very important special case of a sinusoidal driving force (at frequency ):Fd(t) = F0 cos(t)Treating this is equivalent to treating one Fourier component of a more general t dependent force! This is because Ns 2nd Law is a linear differential equation!

• Sinusoidal Driving ForcesEqtn of Motion: F = m(d2x/dt2) = - kx -bv + Fd(t) (1)Assume Fd(t) = F0 cos(t) = Angular frequency of driving force(1) becomes:mx + bx + kx = F0 cos(t)Defining: [b/(2m)]; A (F0/m), 02 (k/m) x + 2x + 02x = A cos(t)A linear, 2nd order, inhomogeneous, time dependent differential eqtn! Differential eqtn theory shows: (App. C): The general solution has 2 parts: x(t) = xc(t) + xp(t) Where xc(t) Complimentary solutionxp(t) Particular solution

• x + 2x + 02x = A cos(t)x(t) = xc(t) + xp(t) xc(t) Complimentary solution= the solution to the homogeneous eqtn (the solution with with A = 0 or the solution for the damped oscillator just discussed!) xc(t) = e-t[A1 et + A2 e-t] with [2 - 02]xp(t) Particular solution = a specific solution to the inhomogeneous eqtn (the solution with A 0)For the particular solution, try xp(t) = D cos(t - ) Note: here is NOT the same as in the discussion of the ordinary SHO!

• x + 2x + 02x = A cos(t) x(t) = xc(t) + xp(t) For the Particular solution xp(t) try xp(t) = D cos(t - ). Substitution into the differential equation gives D & (Student exercise!):{A-D[(02 -2)cos +(2)sin]}cos(t) -{D[(02 - 2)sin (2)cos]}sin(t) = 0cos(t) & sin(t) are linearly independent Coefficients of these functions are separately = 0!

• {A-D[(02 -2)cos +(2)sin]}cos(t) -{D[(02 - 2)sin (2)cos]}sin(t) = 0Algebra: sin(t) term gives: tan = 2()/(02 - 2) sin = 2()/[(02 - 2)2 +4()2] cos =(02 - 2)/[(02 - 2)2 +4()2]Algebra: cos(t) term gives:D = (A)/[(02 - 2)cos + 2()sin]Or D = (A)/[(02 - 2)2 +4()2] xp(t) = D cos(t - ) xp(t) = [A cos(t - )]/[(02 - 2)2+4()2]And = tan-1[2()/(02 - 2)]Physics: = the phase difference between the driving force Fd(t) & the response. xp(t)

• Summary: x(t) = xc(t) + xp(t) wherexc(t) = e-t [A1 et + A2 e-t] with = [2 - 02] Clearly a transient solution! Goes to zero after times t >> 1/xp(t) = [A cos(t - )]/[(02 - 2)2+4()2]with = tan-1[2()/(02 - 2)]A steady state solution! Dominates at long times t >> 1/ x(t >> 1/) xp(t) Motion details before the transient xc(t) dies to zero (t 1/):Depend strongly on the conditions at time the force is first applied. Depend clearly also on the relative magnitudes of the driving frequency and the frequency with damping: 1 [02- 2]

• To understand this, its helpful to solve Problems 3-24 & 3-25 numerically (on a computer!) Results similar to figure: If < 1 = [02- 2] the oscillator transient xc(t) greatly distorts the sinusoidal shape of the forcing function just after t = 0

• Figures for another case. If > 1 = [02- 2] the oscillator transient xc(t) modulates the sinusoidal shape of the forcing function just after t = 0

• ResonanceFocus on the steady state solution. Write it as: xp(t) = D cos(t - ) with D = (A)/[(02 - 2)2 +4()2] A= (F0/m) = tan-1[2()/(02 - 2)]Plotting D vs. clearly gives a function with a peak! Resonance frequency R the frequency at which the amplitude D() is a maximum. (dD/d) = 0 Solving for = R gives: R = [02 - 22]Clearly the resonance frequency R < 0 where 0 = (k/m) is the natural frequency of oscillator! How much less obviously depends on the size of the damping constant !

• xp(t) = D cos(t - ), D = (A)/[(02 - 2)2 +4()2] = tan-1[2()/(02 - 2)]Consider the resonance frequency R = [02 - 22]If 02 < 22, R is imaginary. In this case, there is no resonance! D simply decreases as increases.Comparison of the fundamental oscillation frequencies for the driven oscillator:Free oscillations: 02 = (k/m)Free oscillations + damping: 12 = 02 - 2 Driven oscillations + damping: R2 = 02 - 22Clearly: 0 > 1 > R

• Quality (Q) Factor of the OscillatorFor a driven oscillator, its useful to define theQUALITY FACTOR: Q [R/(2)]Q is a measure of the damping strength of the oscillator. Also (as well see next) its a measure of how sharp the resonance is. R = [02 - 22]For very small damping 2
• D = (A)/[(02 - 2)2 + 4()2] vs for different Q values (different relative sizes of 0 & ). = full width at half max

• = tan-1[2()/(02 - 2)] vs for different Q values (different relative sizes of 0 & )

• Problem 3-19 shows: For a lightly damped oscillator (02 >> 2): Q [R/(2)] [0/()]Where = full width at half max (from D() plot). This shows that Q is definitely a measure of the quality (sharpness) of the resonance! the interval between 2 points on the D() curve on either side of the max, which have an amplitude 1/(2) 0.707 of the maximum amplitude. That is, if the maximum D(R) Dm, = the interval between 2 s whereD() = Dm/(2) 0.707 Dm A measure of the linewidth (or, simply, the width) of the resonance.

• (D/Dm) vs showing relative positions of the 3 frequencies 0 , 1 & R

• Real physical oscillators: Values of Q vary greatly!Mechanical systems (e.g., loudspeakers): Q 1 to a few 100Quartz crystal oscillators & tuning forks: Q > 104 Highly tuned electrical circuits: Q 104 - 105 Atomic systems: Electron oscillations in atoms Optical radiation. Sharpness of spectral lines limited by energy loss due to radiation. Classical minimum linewidth: 2 108 0 Q 5 107 Largest known Qs: Gas lasers: Q 1014

• Energy ResonanceWhat weve talked about up to now should technically be called Amplitude Resonance since the resonance occurs in the amplitude of xp(t) = D cos(t - ), D = (A)/[(02 - 2)2 +4()2]

A similar, related phenomenon, with a (slightly) different resonance frequency is Energy Resonance.First note that the velocity is vp(t) = (dxp(t)/dt) = D sin(t - ),

• Start with: xp(t) = D cos(t - ), vp(t) = D sin(t - ), D D() = (A)/[(02 - 2)2 +4()2] The Potential Energy is:U = ()k[xp(t)]2 = ()kD2 cos2(t - )Consider the time average of U (over one period of the driving force: 0 < t < [2/]): U (/2)Udt Note that cos2(t - ) = () U ()kD2 ()k[D()]2 That is, the resonance frequency of the potential energy = the for which (dU/d) = 0 Occurs at the same as the amplitude resonance: R= [02 - 22] . So: Potential Energy Resonance is the same as amplitude resonance!

• xp(t) = D cos(t - ), vp(t) = D sin(t - ), D D() = (A)/[(02 - 2)2 +4()2] The Kinetic Energy isT = m[vp(t)]2 = m 2D2sin2(t - )Consider the time average of T (over one period of the driving force: 0 < t < [2/]): T (/2)T dt Note that cos2(t - ) = () T ()m2D2 T () m2[D()]2 . That is the resonance frequency of the kinetic energy = the for which (dT/d) = 0 Occurs at a different than the amplitude resonance:E= 0 So: Kinetic Energy Resonance is different thanamplitude resonance! It occurs at the natural frequency!

• SUMMARYPotential energy resonance occurs at R= [02 - 22] Kinetic energy resonance occurs at E = 0 PHYSICS: They occur at different frequencies because the driven, damped oscillator is not a conservative system!Energy is continually exchanged between the (external) driving mechanism & the oscillator. Energy is also continually lost to the damping medium.

• Total Energy Resonance Consider E = T + U for the driven, damped oscillator.Student exercise (as part of Ch. 3 homework!):Take time the average of E: (over one period of the driving force: 0 < t < [2/]): E (/2)E dt Compute the resonance frequency of the total energy = the for which (dE/d) = 0

• Lorentzian Resonance Curves L() [D()]2from Energy Resonance = 0 = 20 = 30