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Mathematical Methods I&II Tutorial 1

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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 + m!2x = 0 ; dt2 dt 0 or equivalently 2 mx¨ + mγx_ + m!0x = 0 ; with x = x0 andx _ = 0 at t = 0, explaining the physical meaning of the parameters m, γ and !0. (b) Find and sketch solutions for (i) overdamping, (ii) critical damping, and (iii) underdamp- ing. (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 = !0/γ. We now add a driving force F cos(!t) to the harmonic oscillator so that its equation becomes 2 mx¨ + mγx_ + m!0x = F cos(!t) : (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 !. (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 ∆! as the full width at half maximum of the resonance peak calculate ∆!=!0 to leading order in γ=!0. (h) For a lightly damped driven oscillator near resonance, calculate the energy stored and the power supplied to the system. Hence confirm that Q = !0/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 = −m!0x 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 + !0 = 0 leading to eigenfrequencies r γ γ2 λ = − ± − !2 : 1;2 2 4 0 We get (i) overdamping when γ > 2!0 and hence solutions do not oscillate, (ii) critical damping for γ = 2!0 and (iii) underdamping for γ < 2!0. 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 = j!0 − γ =4j) 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 w0 t 2 4 w0 t w0 t 0 2 4 0 2 4 -1 Figure 1: Oscillator displacement for different dampings. (c) The energy stored in the harmonic oscillator is the sum of kinetic and elastic energy mx_(t)2 m!2x(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 !0 we may approximate α ≈ !0 and neglect the second term inx _(t). Then the expression for the energy simplifies to m!2 E(t) = 0 A2e−γt : 2 2 A radian corresponds to the time difference τ = 1=!0 and so we find the energy lost per radian m!2 m! γ E = E(0) − E(1=! ) = 0 A2(1 − e−γ=!0 ) ≈ 0 A2 : L 0 2 2 −γ=! by expanding e 0 ≈ 1 − γ=!0 for γ !0. Hence the result Q = E(0)=EL = !0/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) = < fz(t)g where z(t) is the particular solution to the equation 2 i!t mx¨ + mγx_ + m!0x = F e : (1) We find z(t) by the ansatz z(t) = A(!)ei!t, which leads to F −i' A(!) = 2 2 = jA(!)je ; m(!0 − ! + iγ!) where 8 γ! F < arctan !2−!2 for ! ≤ !0 jA(!)j = and ' = 0 : p 2 2 2 2 2 γ! m ((!0 − ! ) + γ ! ) : π + arctan 2 2 for ! > ! !0 −! The magnitude jA(!)j and phase ' are shown in Fig. 2 as a function of !. The displacement and velocity are given by x(t) = < fz(t)g = < jA(!)jei!t = −|A(!)j cos(!t − ') andx _(t) = < fz_(t)g = < ijA(!)j!ei!t = −|A(!)j! cos(!t − (' + π=2)), i.e. the velocity is phase-shifted by π=2 relative to the displacement. The additional factor of ! shifts the maximum amplitude ofx _(t) compared to that of x(t). (e) Amplitude and phase ofx _(t) are shown in Fig. 2. f A A0 p 10 È È p 5 2 w w0 w w0 0 1 2 0 1 2 f ê w A A0w0 ê 10 È ÈH L p 5 w w0 w w0 1 2 0 1 2 Figure 2: Displacement and velocity response to periodic driving for γ = !0=10 and γ = !0=4. (f) The maximum of the displacement amplitude is found by solving djA(!)j=d! = 0 giv- 2 2 2 ing a resonance frequency !x = !0 − γ =2. For the maximum velocity amplitude we solve dj!A(!)j=d! = 0 and find the resonance frequency !x_ = !0. 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 ∆! = !2 − !1 with A(!i=1;2) = A(!x)=2. We take the square of this expression and find 1 1 1 2 2 2 2 2 = 2 2 2 2 2 (!0 − !i ) + γ !i 4 (!0 − !x) + γ !x 4 2 2 2 2 2 2 This can be re-written as γ + γ (!i − 4!0) + (!0 − !i ) = 0. This can in principle be solved for !i but since we have assumed the oscillator to be lightly damped and have worked out quantities like Q only to lowest order in γ=!0 we instead only look for a solution valid to this 4 2 2 2 2 2 order. We thus substitute !i ≈ !0(1 + βγ) obtaining γ + γ (βγ!0 − 3!0) + β γ !0 = 0. Wep 3 4 now ignore any termsp of O(γ ) and O(γ ) and thus get the approximate solution β = ± 3 2 2 and thus !0 − !i ≈ ± 3γ!0 to lowest order. A Taylor series expansion in γ=!0 thus yields q p q p p ∆! = !2 − !1 = !0 1 + 3γ=!0 − 1 − 3γ=!0 ≈ 3γ : p Hence ∆!=!0 = 3γ=!0. (h) Near resonance ! ≈ !0 and we thus find for the energy of the oscillator m m!2 F 2 E = x_(t)2 + 0 x(t)2 ≈ : 2 2 2mγ2 The average supplied power is given by P = F cos(!t)_x(t) = −F jA(!)j!cos(!t) sin(!t − ') = −F jA(!)j!cos(!t)(sin(!t) cos(') − cos(!t) sin(')) : (2) Near resonance we have ' ≈ π=2 and ! ≈ !0 so that F jA(! )j! 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=!0. Thus p E ! ∆! 3 Q = = 0 and = : EL γ !0 Q 4.
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