Oscillations. Vern Lindberg. June 10, 2010

Transcription

1 Oscillations Vern Lindberg June 10, 2010 You have discussed oscillations in Vibs and Waves: we will therefore touch lightly on Chapter 3, mainly trying to refresh your memory and extend the concepts. 1 Undamped Simple Harmonic Motion If a mass m is attached to an ideal spring of constant k, and the spring force is the only net force, then one dimensional simple harmonic motion (SHM) results. The equation of motion is The solution can be written is several forms. ẍ = k m x (1) x = A 0 sin(ωt + φ 0 ) (2) = A 1 cos ωt + A 2 sin ωt (3) = A 3 exp(iωt) + A 4 exp( iωt) (4) = A 5 exp(i(ωt + φ 5 )) (5) The angular frequency, ω = 2πf, is independent of amplitude and is k ω = m Since this is a second order differential equation, all solutions have two integration constants (such as amplitude and phase, A 0 and φ 0, or two amplitudes A 1 and A 2 ) that are determined by the initial conditions. Picking Eq. 2, the first solution, If at t = 0 we have x = x 0 and v = v 0 then (6) x(0) = A sin φ 0 = x 0 (7) ẋ(0) = ωa cos φ 0 = v 0 (8) 1

2 Solving these we get tan φ 0 = ωx 0 (9) v 0 A = x v2 0 ω 2 (10) We will choose the positive square root value for amplitude, but there remains an ambiguity in the inverse tangent that is resolved by deciding what quadrant for φ 0 is needed to produce the proper signs for initial position and velocity. We will spend some time on this in class. Example: Suppose the spring constant is 18 N/m and the mass is 2 kg. Then the angular frequency is ω = 3 rad/s. If x 0 = cm and v 0 = cm/s, then the amplitude is 5.0 cm and the inverse tangent gives phase angles of either or 2.80 radians. With a bit of thought you can decide that the correct phase is 2.80 radians. Example Using Eq. 3, find the values of A 1, A 2 Use the same values as in the previous example. The potential energy associated with this simple model SHM is V (x) = 1 2 kx2 (11) This is a parabolic potential energy, with the oscillations occurring around the minimum in the potential energy. The spring force is a restoring force, linear in displacement. 1.1 The Morse Potential for Diatomic Molecules: Approximations Any potential energy that has a minimum can have oscillations around that minimum. A natural question is to ask whether the oscillations are simple harmonic. If the system oscillates about x = x 0 with force F = k(x x 0 ) and potential energy V = 1 2 k(x x 0) 2 then the motion is SHM for any amplitude. What if we don t have forces of that form? For small oscillations about the minimum we can make suitable approximations to see if the small amplitude motion is SHM. Taylor Series and Approximations A vital skill in physics is making approximations. The Taylor series is a polynomial expansion of a function about some point. If we expand f(x) about x = a, the Taylor series is f(x) = f(a) + ( f x ) x=a (x a) + 1 2! ( 2 ) f x 2 (x x=a a)2 + (12) The terms are called the zeroeth order, first order, second order,.... We frequently want to find the first higher order term (first and above) that is non-zero. 2

3 Consider the Morse potential energy, a model of the potential energy for a vibrating diatomic molecule like O 2. [ V (x) = V 0 1 exp( x x 2 0 )] V 0 (13) δ First we find the location of the minimum by solving dv/dx = 0 and find that the minimum occurs at x = x 0. This is done as an example in Chapter 2 of the text. At this minimum we find V = V 0. Computational Hint Make appropriate substitutions of variables to make the calculus easier. For the Morse potential define z = (x x 0 )/δ. Then V = V 0 (1 exp z) 2 V 0, and we can do derivatives as dv/dx = dv/dz dz/dx. For small oscillations we expand the exponential in a Taylor series, V = V 0 [1 (1 z + z 2 /2 z 3 /6 + )] 2 V 0 = V 0 [z z 2 /2 + z 3 /6 + ] 2 V 0 = V 0 [z 2 z 3 + 7z 4 /12 + ] V 0 ( ) x 2 = V 0 z 2 x0 V 0 = V 0 V 0 (14) δ This is parabolic and so we expect SHM. The restoring force for small oscillations is linear, F (x) = dv dx = 2V 0 δ 2 (x x 0) (15) Thus for small oscillation we expect SHM with an angular frequency of 2V0 ω = mδ 2 (16) Example The following system is set up on a horizontal, frictionless table. A small block of mass m is attached to two identical springs of constant k. When the springs are unstretched they have identical lengths L 0. The equilibrium position is a straight line consisting of a wall, a spring, the block, the other spring, and a second wall. The block is pulled down a small distance y L 0 perpendicular to the equilibrium line. Determine whether or not the subsequent motion is simple harmonic. 1.2 Angular SHM Consider a simple pendulum of length l with point mass m attached at one end, in a uniform gravitational field g. The motion is along an arc of a circle, so it is natural to use 3

4 an angular variable θ for the displacement. Using the angular form of Newton s Second Law, in scalar form since the angular motion is along a fixed direction, N = I θ (17) mgl sin θ = (ml 2 ) θ (18) Providing the angle (in radians) is small, we can use the Taylor Series expansion sin θ = θ θ 3 /3! + and keep it to first order to get the equation of motion. θ + g l θ = 0 (19) This is the same form as the SHM equation of motion, Equation 1, so the solution is SHM with angular frequency ω = g/l. We will solve the general equation of motion (large angles) in Mechanics 2 using Elliptic Integrals. 1.3 Energy and SHM For ideal SHM the energy is a constant of the motion and is E = 1 2 mẋ kx2 (20) 2 Damped Simple Harmonic Motion Linear Damping Suppose we add a linear damping term to the equation of motion to get ẍ = c mẋ k m x (21) It is convenient to define a damping factor γ = so that the equation of motion becomes ẍ + 2γẋ + ω0 2 = 0 (22) Next let s introduce a differential operator, D = d/dt. The operator has no independent meaning, but once it is applied to a function its meaning is clear. The equation of motion can be written [ D 2 + 2γD + ω0 2 ] x = 0 (23) [ ] [ ] D + γ γ 2 ω0 2 D + γ + γ 2 ω0 2 x = 0 (24) There are two solutions, one for the first bracket operating on x equalling 0 and one for the second bracket operating on x being zero. c 2m 4

5 These are first order differential equations and are easy to solve resulting in a general solution x(t) = A 1 exp [ (γ q)t] + A 2 exp [ (γ + q)t] (25) where q = γ 2 ω 2 0. Case 1 When γ 2 ω0 2 > 0, q is real and positive and we have the overdamped case, where a displaced mass slowly returns to equilibrium with no oscillations. Note that (γ q) > 0. Case 3 When γ 2 ω0 2 < 0, q is imaginary and we have the underdamped case where the mass oscillates about equilibrium with a decreasing amplitude. Defining = ω0 2 γ2 (26) and doing some work (shown in the text) we can write the solution can as where A and θ 0 are determined by initial conditions. x(t) = e γt A cos( t + θ 0 ) (27) Case 2 When γ 2 ω0 2 = 0 we have critical damping. The mass quickly returns to equilibrium without overshooting. If we start with Equation 25, we run into a problem: we have only one arbitrary integration constant. To find a solution with two integration constants we must return to the differential equation, Equation 24, that becomes [D + γ] [D + γ] x = 0 (28) This is a second order differential equation and we need two arbitrary constants. Here is the trick needed to get the general solution. Define u = (D + γ)x. Then (D + γ)u = 0 and this has the solution u = Ae γt. The solution for x is found from Ae γt = (D + γ)x A = e γt (D + γ)x = D(xe γt ) xe γt = At + B x = Ate γt + Be γt (29) Example Suppose we have a mass of 3.0 kg attached to a spring of constant 12.0 N/m, and with initial conditions that at t = 0, position is 1.0 m and velocity is 0.0 m/s. (a) For the underdamped case with c = 4.0 kg/s find the values of the integration constants. [A = 1.06 m, θ 0 = rad] (b) For the critically damped case with c = 12 kg/s find the values of the integration constants. [A = 2.0 m/s, B = 1.0 m] (c) For the overdamped case of c = 20 kg/s, find the values of the integration constants. [A 1 = m, A 2 = m] 5

6 2.1 Energy and Linear Damping The instantaneous energy is still E = mẋ 2 /2 + kx 2 /2 so we can look at how this changes with time. de = mẋẍ + kẋx = (mẍ + kx)ẋ (30) dt But from Equation 21, mẍ + kx = cẋ, hence de dt = cẋ2 (31) 2.2 Quality Factor for Damped SHM For the underdamped case we can characterize the system by its Q-factor, Q = 2π E E (32) where E is the energy stored in the oscillator (think in terms of the average energy) and E is the energy lost in one period of the oscillation. Q is calculated usually for weakly damped systems where the energy does not change quickly. In this case the energy at time t can be written Using Equation 27 for x we differentiate to get Then, changing variables to z = t + θ 0 E = E = 1 2 ka2 e 2γt = 1 2 mω2 0A 2 e 2γt (33) ẋ = Ae γt (γ cos( t + θ 0 ) + sin( t + θ 0 )) (34) Td 0 = ca2 cẋ 2 dt (35) 2π+θ0 θ 0 e 2γt [ γ 2 cos 2 z + 2γ cos z sin z + ω 2 d sin2 z ] dz (36) But for weakly decaying systems the exponential changes very little during the integration, so we remove it from the integral. The integration limits can start anywhere as long as they extend over 2π. So 2π E = ca2 e 2γt 0 [ γ 2 cos 2 z + 2γ cos z sin z + ω 2 d sin2 z ] dz (37) 6

7 The integrals for cos 2 and sin 2 each yield π, and the cross term becomes 0, giving Then E = πca2 e 2γt [ γ 2 + ω 2 ] πca 2 d = e 2γt ω0 2 (38) Q = 2π 1 2 mω2 0 A2 e 2γt πca 2 e 2γt ω 2 0 = 2γ Table in the text lists some Q s for underdamped systems. The earth (responding to earthquakes) has Q = 250 to 1400 while a piano string has Q = (39) 2.3 An Example Suppose a mass of 500 kg is attached to an ideal spring of force constant 2000 N/m and a linear damping shock absorber of constant 20 N s/m. Find (a) The natural (undamped) angular frequency and frequency in Hz. (b) The frequency with damping applied. (c) The Q of the system (d) The time for the amplitude to reduce to 1/10 of its original value. (e) The energy when the amplitude has reduced to 1/10 its original value. We have m = 500 kg, k = 2000 N/m, and c = 20 N s/m. (a) Use ω 0 = k/m = rad/s. This gives a frequency of f = ω 0 /2π = Hz (and a period of s). (b) Compute γ = c/2m = 0.02 rad/s. Then = ω 2 0 γ2 = rad/s. (c) The quality factor is Q = /2γ = (d) The envelope of the oscillation varies as A(t) = Ae γt, so solve 0.1 = e 0.02t to get t = s. (37 oscillations during this time.) (e) Since E A 2, when the amplitude is reduced to 1/10 of its original value, the energy is reduced to 1/100 of its original value. 3 Phase Space Plots of position versus time and velocity versus time are useful, but often a plot of velocity (or momentum) versus position is drawn. A plot such as this is called a phase plot. 7

8 For undamped SHM, x = A sin(ω 0 t + φ 0 ) and ẋ = Aω 0 cos(ω 0 t + φ 0 ). We combine these so as to eliminate time and get the parametric equation x 2 A 2 + ẋ2 A 2 ω 2 0 = 1 (40) which is the equation of an ellipse. Figure shows several trajectories in phase space corresponding to different initial conditions. All trajectories in this situation form closed paths. For underdamped oscillations the phase plot spirals in toward a final value of v = 0. The text does some very clever changes of variables to make the phase plot more obvious: I will just demonstrate using Easy JAVA Simulation, and extend to the case of critical damping and over damping. Non linear oscillators can lead to chaotic motion, a topic left for other courses. 4 Forced Damped Harmonic Motion Consider the damped harmonic oscillator subject to a periodic driving force F 0 cos ωt. The equation of motion is mẍ = kx cẋ + F 0 cos ωt (41) First consider undamped motion and guess the solution to be x = A cos(ωt φ) with two variables, A, φ to be determined. The equation of motion becomes maω 2 cos(ωt φ) + ka cos(ωt φ) = F 0 cos ωt (42) This only works for the two values φ = 0, π, implying A = F 0/m ω 2 0 ω2 φ = 0 Low frequency (43) = F 0/m ω 2 ω 2 0 This gives a resonance at ω = ω 0. φ = π High frequency (44) When we add damping and the velocity term we have difficulty when starting with a cosine solution. Instead we assume a driving force as F 0 e iωt and consider the real part of 8

9 the complex quantity to describe the physical situation. We guess a solution x = Ae iωt φ so that the equation of motion becomes mω 2 Ae iωt φ + kae iωt φ + icωae iωt φ = F 0 e iωt (45) mω 2 A + ka + icωa = F 0 e iφ = F 0 (cos φ + i sin φ) (46) The real parts must be equal, and the imaginary parts must be equal giving These can be solved simultaneously and give A(k mω 2 ) = F 0 cos φ (47) c ω A = F 0 sin φ (48) tan φ = A = 2γω ω0 2 ω2 (49) F 0 /m (ω 2 0 ω 2 ) 2 + 4γ 2 ω 2 (50) Further details should have been discussed in Vibrations and Waves. You will probably see resonance in an electrical circuit when you take Electronic Measurements. When the resonance is fairly sharp, meaning relatively small damping, the resonant frequency is very close to the natural frequency. For these cases we define the sharpness of the resonance with another Q factor same symbol and definition as before, but a different meaning. If the resonance has an amplitude A max at resonance, ω 0, then there are frequencies above and below the resonant frequency at which A = A max / 2. These two frequencies are called the half-energy (or half-power) frequencies, and the difference in frequencies is ω = 2γ = c (51) m and Q resonance = ω 0 ω ω 0 2γ (52) 9