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Home , Damping, Harmonic oscillator

Damped Oscillator Wednesday, 23 October 2013

A simple subject to linear may oscillate with , or it may decay biexponen- tially without oscillating, or it may decay most rapidly when it is critically damped. When driven sinusoidally, it resonates Physics 111 at a near the natural frequency, and with very large when the damping is slight. Because the system is linear, we can superpose solutions, leading to Green’s method and the usefulness of Laplace transforms.

Many systems exhibit mechanical stability: disturbed from an equilibrium position they move back toward that equilibrium position. This is the recipe for . Provided that the disturbance from equilibrium is small, virtually any mechanically stable system will experience a linear restoring , giving rise to simple harmonic . We will start with a single degree of freedom, which will illustrate most of the impor- tant behavior: decaying oscillation and . We will soon generalize to a system with an arbitrary number of degrees of freedom, where we will find that we can always find suitable combinations of coordinates to reduce that problem to a set of decoupled one-dimensional oscillators. The simple harmonic oscillator model, therefore, is ubiq- uitous in physics. You find it in ; in electromagnetism, where it describes electromagnetic waves, , and modes; atomic physics, where it describes coupling of an atom to the electromagnetic field. . . I suspect it arises in every subdiscipline of physics. In fact, one of my graduate physics professors quipped that it couldn’t be a physics course with the simple harmonic oscillator. Our point of departure is the general form of the lagrangian of a system near its position of stable equilibrium, from which we deduce the equation of motion. We will then con- sider both unforced and periodically forced motion before turning to general methods of solving the linear second-order differential equations that describe oscillatory systems.

1. General Form of the Lagrangian

We may Taylor expand the potential about q 0: = ¯ 2 ¯ dU ¯ 1 d U ¯ 2 U(q) U(0) ¯ q 2 ¯ q (1) = + dq ¯0 + 2 dq ¯0 + ··· The first term on the right-hand side is just a constant; it cannot influence the since the variational derivative kills all constants. The second term vanishes since we expand about a minimum. The third term has to be positive (otherwise, we expand about either a maximum or a saddle point). So,

2 1 d U 2 1 2 U(q) U(0) q U0 kq (2) = + 2 dq2 + ··· = + 2 + ···

Physics 111 1 of 9 Peter N. Saeta 2. DAMPED SIMPLE HARMONIC OSCILLATOR

We may also expand the kinetic about its equilibrium value, 0:

1 T mq˙2 (3) = 2 for some constant m. Lagrange’s equations for this system then become

d µ ∂L ¶ ∂L 0 = dt ∂q˙ − ∂q µ ∂2L ¶ µ ∂2L ¶ q¨ q = ∂q˙2 − ∂q2 k 0 q¨ q (4) = + m

2 2 If k ∂ U ∂ L is negative, then this gives exponential solutions and the equilibrium is = ∂q2 = − ∂q2 unstable. If it is positive, then the equilibrium is stable and the system oscillates about the equilibrium position q 0. In that case, Eq. (4) is the simple harmonic oscillator = (SHO) equation, with solutions of the form

¡ ¢ i(ω0t ϕ) q(t) A cos ω t ϕ A Ree− + = 0 + =

p I am using i in the exponent to be consis- where ω0 k/m is called the natural frequency of the oscillator and the coefficients − ≡ tent with . A plane wave A and ϕ may be determined from initial conditions. i(kx ωt) of the form e − moves to positive x if k is positive. Furthermore, the There is another way to interpret the complex expression. Moving the real number A of the wave is k, not k, which is possible } −} inside and rearranging gives because the component carries the nega- tive sign. h³ ´ i h i iϕ iω0t iω0t q(t) Re Ae− e− Re Ae− = =

iφ where A Ae− is the complex amplitude of oscillation. The single complex amplitude = contains both the magnitude and information of the oscillation.

2. Damped Simple Harmonic Oscillator

If the system is subject to a linear damping force, F br˙ (or more generally, b r˙ ), = − − j j such as might be supplied by a viscous fluid, then Lagrange’s equations must be modi- fied to include this force, which cannot be derived from a potential. Recall that we had developed the expression d µ ∂L ¶ ∂L ∂r Fext (5) dt ∂q˙ − ∂q = · ∂q If the equations of transformation do not depend explicitly on the time—so that ∂r /∂q ∂r˙ /∂q˙—then we can simplify the damping term: j = j

X ∂rj X ∂r˙ j ∂ X b j 2 ∂F b j r˙ j b j r˙ j (v j ) − j ∂q = − j ∂q˙ = −∂q˙ j 2 ≡ −∂q˙

Physics 111 2 of 9 Peter N. Saeta 2. DAMPED SIMPLE HARMONIC OSCILLATOR

The sum in this final expression, which can be generalized to N particles in the obvious way, is called Rayleigh’s dissipation function, F. The modified Lagrange equations are Rayleigh’s dissipation function allows a gen- then eral linear damping term to be incorporated d µ ∂L ¶ ∂L ∂F into the Lagrangian formalism. 0 (6) dt ∂q˙ − ∂q + ∂q˙ =

For a single degree of freedom with linear damping, we have F b q˙2, so = 2 mq¨ kq bq˙ 0 or q¨ 2βq˙ ω2q 0 (7) + + = + + 0 = where we have defined β b/2m, which has the dimensions of a frequency, and is called = the damping parameter. We may solve this equation with the Ansatz q est , which =iωt slightly biases us to expect exponential solutions, or with the Ansatz q e− , which is = more appropriate when we expect oscillatory solutions. Making the latter choice yields the algebraic equation q ω2q 2iβωq ω2q 0 ω iβ ω2 β2 − − + 0 = =⇒ = ± 0 − This yields two linearly independent solutions to the second-order , Eq. (11), unless β ω0. This special case is called critical damping; the second solution =βt takes the form te− when β ω . = 0 When β ω , the oscillator is underdamped and oscillates when released from rest < 0 away from its equilibrium position. Let An underdamped oscillator vibrates at angu- lar frequency ω1 with decreasing amplitude q when released away from its equilibrium posi- 2 2 tion. For light damping, this frequency is just ω1 ω β (8) ≡ 0 − slightly smaller than the natural frequency ω0. Then the solution may be expressed

βt q(t) Ae− cos(ω t ϕ) (9) = 1 +

When β ω , the oscillator is overdamped; the two solutions decay exponentially with > 0 different time constants: ³ q ´ ³ q ´ β β2 ω2 t β β2 ω2 t q(t) A e− + − 0 A e− − − 0 (10) = + + − The second term has the longer ; it tends to dominate at long .

1.0 0.8 0.6 L t H 0.4 q 0.2 0.0 Figure 1: Overdamped (olive), crit- ically damped (purple), and under- 0.0 0.5 1.0 1.5 2.0 damped (blue) harmonic oscillators t released from rest.

Physics 111 3 of 9 Peter N. Saeta 3. DRIVEN DSHO 3. Driven DSHO

A damped simple harmonic oscillator subject to a sinusoidal driving force of angular fre- quency ω will eventually achieve a steady-state motion at the same frequency ω. How long it must be driven before achieving steady state depends on the damping; for very light damping it can take a great many cycles before the transient solution to the homo- geneous differential equation decays sensibly to zero, as illustrated in Fig. 2.

0.10

0.05 Figure 2: Response of L

t a lightly damped sim- H 0.00 q ple harmonic oscillator driven from rest at its - 0.05 equilibrium position. In this case, ω0/2β 20 ≈ -0.10 and the drive frequency is 15% greater than 0 5 10 15 20 25 30 the undamped natural t frequency.

The equation of motion for the driven damped oscillator is µ ¶ 2 F0 F0 iωt q¨ 2βq˙ ω q cosωt Re e− (11) + + 0 = m = m

Rather than solving the problem for the sinusoidal forcing function, let us instead look for a complex function of time, z(t), that satisfies essentially the same equation,

2 F0 iωt z¨ 2βz˙ ω z e− (12) + + 0 = m If we can find such a function, then its real part q Rez solves Eq. (11). Knowing that the = solution will eventually oscillate at the same frequency as the drive, we make the Ansatz iωt z Ae− , thereby obtaining the particular solution = " # F0/m F0/m iωt A or q Re e− (13) = ω2 ω2 2βωi = ω2 ω2 2βωi 0 − − 0 − − where the complex amplitude A encodes both the (real) amplitude A and the phase of iϕ the oscillator with respect to the drive, A Ae− , and = Ã ! F0/m 1 2ωβ A and ϕ tan− = q = 2 2 (ω2 ω2)2 4β2ω2 ω ω0 0 − + − The complete solution is then the sum of the general solution, Eq. (9), and the particular solution Eq. (13).

Physics 111 4 of 9 Peter N. Saeta 3. DRIVEN DSHO

Exercise 1 Show that the frequency of maximum amplitude (the resonant frequency) is q ω ω2 2β2. R = 0 −

The width of the resonance may be characterized by the quality factor of the , a that is defined by Roughly speaking, the quality of a resonator is the number of it undergoes before q its initial energy is reduced by a factor of 1/e. ω2 2β2 ωR 0 Q − (14) ≡ 2β = 2β

Figure 3 shows resonance curves for damped driven harmonic oscillators of several val- ues of Q between 1 and 256. For a lightly damped oscillator, you can show that Q ω0 , ≈ ∆ω where ∆ω is the frequency interval between the points that are down to 1/p2 from the maximum amplitude. The width is really defined with respect to the energy of the oscil- lator, which is proportional to the square of the amplitude. Hence, the factor of 1/p2.

Exercise 2 If a oscillating at 440 Hz takes 4 seconds to lose half its energy, roughly what is its quality? Ans: 2500. ≈

0

1 -50 0.1 j A -100 0.01

0.001 -150

0.1 0.2 0.5 1.0 2.0 5.0 10.0 0.1 0.2 0.5 1.0 2.0 5.0 10.0 Frequency Frequency

Figure 3: Amplitude and phase of a damped driven simple harmonic oscillator for several different quality factors (Q = 1, 4, 16, 64, and 256).

Physics 111 5 of 9 Peter N. Saeta 4. GREEN’S FUNCTION 4. Green’s Function

If the damped oscillator is driven by an arbitrary function of time,

F (t) q¨ 2βq˙ ω2q (15) + + 0 = m there are a variety of ways to solve for q(t). All are based on the observation that the left-hand side of this equation is a linear operator on q, L(q). That is,

d 2 d L 2β ω2 = dt 2 + dt + 0 Linear operators satisfy

L(q q ) L(q ) L(q ) 1 + 2 = 1 + 2 L(α q α q ) α L(q ) α L(q ) 1 1 + 2 2 = 1 1 + 2 2 This means that if we have a sum of forcing functions, F (t) m f (t), then i = i à ! X X ¡ ¢ X L αi qi (t) L αi qi (t) αi fi (t) i = i = i which is the principle of superposition. It says that we are free to break apart the forcing function into pieces, solve each piece separately, and add up the resulting solutions. George Green (1793–1841) looked for a solution to Eq. (15) by breaking apart the forcing function into a series of brief impulses, short time intervals ∆t of constant force:

0 t t j∆t  < 0 + L(q(t)) f j t0 j∆t t t0 (j 1)∆t =  + < < + + 0 t t (j 1)∆t > 0 + + The complete temporal behavior of q is then obtained by adding up all the individual solutions q j (t) corresponding to all the little impulses f j : X q(t) q j (t) = j

In the limit that ∆t goes to zero, we may represent the forcing functions by f (t 0) j = f δ(t 0 t ), where δ(t) is Dirac’s delta function. Recall that the delta function δ(x) has j − j the properties • δ(x) 0 if x 0 = 6= • δ(0) → ∞ • R δ(x)dx 1, provided that the range of integration includes x 0, and vanishes = = otherwise. Think of δ(x) as an infinitely tall spike at x 0 with unit area. Using the delta function, = we can represent the forcing function via Z ∞ f (t) f (t 0)δ(t t 0)dt 0 = − −∞

Physics 111 6 of 9 Peter N. Saeta 4. GREEN’S FUNCTION where t 0 is a dummy variable of integration. The only time that the delta function is nonzero is when t t 0. Integrating over this time yields f (t) times the integral of the = delta function, which is 1 by definition. So, if we knew the solution to

2 G¨ 2βG˙ ω G δ(t t 0) (16) + + 0 = − for a function G(t,t 0), we could multiply G(t,t 0) by f (t 0) and integrate over all times t 0 t < to get the solution at t: Z t q(t) G(t,t 0)f (t 0)dt 0 (17) = −∞ where f (t 0) F (t 0)/m. =

Exercise 3 Show that ( 0 t t 0 < G(t,t 0) 1 β(t t ) (18) = e− − 0 sinω1(t t 0) t t 0 ω1 − ≥

Hint: Solve for t t 0 and then integrate the defining differential equation from t 0 ² to 6= − t 0 ² and take the limit as ² 0. Then match the solution to G(t,t 0) 0 for t t 0. + − > = <

Physics 111 7 of 9 Peter N. Saeta 5. LAPLACE TRANSFORM 5. Laplace Transform

Another important technique for solving Eq. (15) is via an integral transform called the Laplace transform. The transform converts time derivatives into polynomials, which produces an algebraic equation. These are easier to solve than differential equations. After solving the algebraic equation, we then apply the inverse Laplace transform to re- turn to a time-domain expression that gives q(t). First, we define the Laplace transform of a function of time, f (t), as Z © ª ∞ st F (s) L f (t) e− f (t)dt (19) = ≡ 0

Since we will be applying this to time derivatives of , let’s out expressions for the Laplace transform of d f /dt and d 2 f /dt 2: ½ ¾ Z Z d f ∞ st d f st ¯ ∞ st ∞ L e− dt e− f ¯0 se− f dt sF (s) f (0) (20) dt = 0 dt = + 0 = − ½ 2 ¾ Z 2 d f ∞ st d f 2 L 2 e− 2 dt s F (s) s f (0) f 0(0) (21) dt = 0 dt = − − where the prime indicates differentiation with respect to the argument of the function. We now seek to apply the Laplace transform to Eq. (15). Let us assume that we have situated the origin of time such that the system is quiescent and the forcing function vanishes for t 0. Then we may take all of the integrated terms to vanish, and get < ½ F (t) ¾ 1 (s2 2βs ω2)Q(s) L F (s) (22) + + 0 = m = m

Solving for the Laplace transform of q, Q(s), gives

F (s)/m Q(s) (23) = s2 2βs ω2 + + 0 We now apply the Laplace transform inversion integral (the Bromwich integral),

1 Z γ i f (t) + ∞ Q(s)est ds (24) = 2πi γ i − ∞ where γ is a real constant that exceeds the real part of all the singularities of F (s), to solve for q(t). Equation (24) is derived in the notes on contour integration.

Physics 111 8 of 9 Peter N. Saeta 5. LAPLACE TRANSFORM

Example 1 Suppose that the oscillator described by Eq. (11) is thumped at t 0 with a = delta-function : F (t) αδ(t). Find q(t) using the Laplace transform method. =

According to Eq. (23), we need first to calculate the Laplace transform of the forc- ing function: Z ∞ st L {αδ(t)} e− αδ(t)dt α = 0 = By Eq. (23) and the inversion integral, we have

1 Z γ i α/m q(t) + ∞ est ds 2 2 = 2πi γ i s 2βs ω − ∞ + + 0 The integrand has poles at the roots of the quadratic equation in the denomina- q tor, s β β2 ω2, both of which lie to the left of s 0. So, we may integrate ± = − ± − 0 = along the imaginary axis and close in the left half-plane, where the exponen- tial sends the integrand to zero for t 0. By the residue theorem, the integral > is therefore 2πi times the sum of the two residues. Writing the denominator as (s s )(s s ), the residues are − + − −

α/m s t α/m s t a+1 e + and a−1 e − − = s s − = s s + − − − − + Combining these results gives

βt µ q q ¶ α e− β2 ω2t β2 ω2t q(t) e − 0 e− − 0 = q − m 2 β2 ω2 − 0 q Noting that we have called ω ω2 β2, we may rewrite this results as 1 = 0 −

βt iω1t iω1t βt α e− e e− αe− q(t) − sin(ω1t) = m iω1 2i = mω1 consistent with what we found before.

Physics 111 9 of 9 Peter N. Saeta