Mathematical Methods I&II Tutorial 1

Tomi Johnson1 Keble College - Hilary 2014 CP3&4: Mathematical methods I&II Tutorial 1 - Damped harmonic oscillator

Prepare full solutions to the ‘problem’ with a self assessment of your progress on a cover page. Leave these at Keble lodge by 5pm on Monday of 1st week. Suggested reading: FR 4.

Goals

• Understand the behaviour of this paradigm exactly solvable physics model that appears in numerous applications.

• Understand the connection between the response to a sinusoidal driving force and intrinsic oscillator properties.

• Understand the connection between the Q factor, width of this response and energy dissipation.

Problem

1. A damped harmonic oscillator is displaced by a distance x0 and released at time t = 0. (a) Show that the subsequent motion is described by the diﬀerential equation

d2x dx m + mγ + mω2x = 0 , dt2 dt 0 or equivalently 2 mx¨ + mγx˙ + mω0x = 0 ,

with x = x0 andx ˙ = 0 at t = 0, explaining the physical meaning of the parameters m, γ and ω0. (b) Find and sketch solutions for (i) overdamping, (ii) critical damping, and (iii) underdamping. (iv) What happens for γ = 0? (c) For a lightly damped oscillator the quality factor, or Q-factor may be deﬁned as energy stored Q = . energy lost per radian of oscillation

Show that Q = ω0/γ. We now add a driving force F cos(ωt) to the harmonic oscillator so that its equation becomes

2 mx¨ + mγx˙ + mω0x = F cos(ωt) .

(d) Explain what is meant by the steady state solution of this equation, and calculate the steady state solution for the displacement x(t) and the velocityx ˙(t). (e) Sketch the amplitude and phase of x(t) andx ˙(t) as a function of ω. (f) Determine the resonant frequency for both the displacement and the velocity.

1These problems were compiled by Prof. D. Jaksch based on problem sets by Prof. J. Yeomans and past Oxford Prelims exam questions.

1 (g) Deﬁning ∆ω as the full width at half maximum of the resonance peak calculate ∆ω/ω0 to leading order in γ/ω0. (h) For a lightly damped driven oscillator near resonance, calculate the energy stored and the power supplied to the system. Hence conﬁrm that Q = ω0/γ as shown above. How is Q related to the width of the resonance peak?

2 Solution: (a) The forces on the mass m are Fs = −kx = −mω0x due to the spring and Ff = −mγx˙ due to friction γ. The equation follows from Newton’s law mx¨ = Fs + Ff . λt 2 2 (b) The characteristic polynomial for ansatz x(t) = e is λ + γλ + ω0 = 0 leading to eigenfrequencies r γ γ2 λ = − ± − ω2 . 1,2 2 4 0

We get (i) overdamping when γ > 2ω0 and hence solutions do not oscillate, (ii) critical damping for γ = 2ω0 and (iii) underdamping for γ < 2ω0. Diﬀerent solutions are shown in Fig. 1. The general solution is given by n o λ1t λ2t x(t) = < A1e + A2e .

p 2 2 and can be simpliﬁed for the diﬀerent situations (writing α = |ω0 − γ /4|) for the three cases

(i) x(t) = e−γt/2[A cosh(αt) + B sinh(αt)], or equivalently x(t) = e−γt/2(Ceαt + De−αt), (ii) x(t) = e−γt/2(A + Bt), (iii) x(t) = e−γt/2[A cos(αt) + B sin(αt)], using the standard procedure for degenerate roots of the characteristic polynomial in (ii). By matching the initial conditions we ﬁnd for the diﬀerent cases

(i) A = x0 and B = x0γ/(2α), or equivalently C = x0(α + γ/2)/(2α) and D = x0(α − γ/2)/(2α),

(ii) A = x0 and B = x0γ/2,

(iii) A = x0 and B = x0γ/(2α),

overdamped critical underdamped x t x t x t 1 1 1

H L H L H L 0 t 2 4 0 t 0 t 0 2 4 0 2 4 1

Figure 1: Oscillator displacement for diﬀerent dampings.

mx˙(t)2 mω2x(t)2 E(t) = + 0 . 2 2 In order to proceed for the lightly damped case it is easiest to write x(t) = A cos(αt−φ)e−γt/2 −γt/2 and thusx ˙(t) = −Aα sin(αt − φ)e − γx(t)/2. Since lightly damped means γ ω0 we may approximate α ≈ ω0 and neglect the second term inx ˙(t). Then the expression for the energy simpliﬁes to mω2 E(t) = 0 A2e−γt . 2

2 A radian corresponds to the time difference τ = 1/ω0 and so we find the energy lost per radian mω2 mω γ E = E(0) − E(1/ω ) = 0 A2(1 − e−γ/ω0 ) ≈ 0 A2 . L 0 2 2 −γ/ω by expanding e 0 ≈ 1 − γ/ω0 for γ ω0. Hence the result Q = E(0)/EL = ω0/γ follows as required. (d) We now turn to the forced damped harmonic oscillator. The solutions to the homogeneous equation will damp out on a time scale 1/γ. At times t 1/γ only terms arising from the particular solution will remain. These terms describe the stationary state2. We simplify our route to finding a particular solution using the ansatz x(t) = < {z(t)} where z(t) is the particular solution to the equation

2 iωt mx¨ + mγx˙ + mω0x = F e . (1) We ﬁnd z(t) by the ansatz z(t) = A(ω)eiωt, which leads to

F −iϕ A(ω) = 2 2 = |A(ω)|e , m(ω0 − ω + iγω) where

 γω F  arctan ω2−ω2 for ω ≤ ω0 |A(ω)| = and ϕ = 0 . p 2 2 2 2 2 γω m ((ω0 − ω ) + γ ω )  π + arctan 2 2 for ω > ω ω0 −ω The magnitude |A(ω)| and phase ϕ are shown in Fig. 2 as a function of ω. The displacement and velocity are given by x(t) = < {z(t)} = < |A(ω)|eiωt = −|A(ω)| cos(ωt − ϕ) andx ˙(t) = < {z˙(t)} = < i|A(ω)|ωeiωt = −|A(ω)|ω cos(ωt − (ϕ + π/2)), i.e. the velocity is phase-shifted by π/2 relative to the displacement. The additional factor of ω shifts the maximum amplitude ofx ˙(t) compared to that of x(t). (e) Amplitude and phase ofx ˙(t) are shown in Fig. 2.

 A A0  10

» »ê

 5 2

 0  0 0 1 2 0 1 2

 ê  A A00 ê 10

» »êH L

 5

 0  0 1 2 0 1 2

ê ê

Figure 2: Displacement and velocity response to periodic driving for γ = ω0/10 and γ = ω0/4.

(f) The maximum of the displacement amplitude is found by solving d|A(ω)|/dω = 0 giv- 2 2 2 ing a resonance frequency ωx = ω0 − γ /2. For the maximum velocity amplitude we solve d|ωA(ω)|/dω = 0 and ﬁnd the resonance frequency ωx˙ = ω0.

2Since we can always move a term from the homogeneous solution to the particular solution it is strictly speaking not accurate to say that the particular solution is the stationary state.

3 (g) We write the full width half maximum as ∆ω = ω2 − ω1 with A(ωi=1,2) = A(ωx)/2. We take the square of this expression and ﬁnd 1 1 1 2 2 2 2 2 = 2 2 2 2 2 (ω0 − ωi ) + γ ωi 4 (ω0 − ωx) + γ ωx

4 2 2 2 2 2 2 This can be re-written as γ + γ (ωi − 4ω0) + (ω0 − ωi ) = 0. This can in principle be solved for ωi but since we have assumed the oscillator to be lightly damped and have worked out quantities like Q only to lowest order in γ/ω0 we instead only look for a solution valid to this 4 2 2 2 2 2 order. We thus substitute ωi ≈ ω0(1 + βγ) obtaining γ + γ (βγω0 − 3ω0) + β γ ω0 = 0. We√ 3 4 now ignore any terms√ of O(γ ) and O(γ ) and thus get the approximate solution β = ± 3 2 2 and thus ω0 − ωi ≈ ± 3γω0 to lowest order. A Taylor series expansion in γ/ω0 thus yields q √ q √ √ ∆ω = ω2 − ω1 = ω0 1 + 3γ/ω0 − 1 − 3γ/ω0 ≈ 3γ .

√ Hence ∆ω/ω0 = 3γ/ω0.

(h) Near resonance ω ≈ ω0 and we thus ﬁnd for the energy of the oscillator

m mω2 F 2 E = x˙(t)2 + 0 x(t)2 ≈ . 2 2 2mγ2 The average supplied power is given by

P = F cos(ωt)x ˙(t) = −F |A(ω)|ωcos(ωt) sin(ωt − ϕ) = −F |A(ω)|ωcos(ωt)(sin(ωt) cos(ϕ) − cos(ωt) sin(ϕ)) . (2)

Near resonance we have ϕ ≈ π/2 and ω ≈ ω0 so that

F |A(ω )|ω F 2 P = 0 0 = . 2 2mγ In the steady state the energy dissipated per radian must be equal to the energy supplied by the external force per radian EL = P τ = P/ω0. Thus √ E ω ∆ω 3 Q = = 0 and = . EL γ ω0 Q