WPI Physics Dept. Intermediate Lab 2651 Free and Damped Electrical Oscillations Qi Wen March 11, 2017 Index: 1. Background Material, p. 2 2. Pre-lab Exercises, p. 4 3. Report Checklist, p. 6 4. Appendix, p.7 1
1 Background Material 1.1 Electrical Oscillations of an LCR Circuit Figure 1: Schematic of LCR circuit where R is the resistance, C is the capacitance, L is the inductance, and E is the AC emf or driving voltage. Figure 1 shows a schematic of an LCR circuit. The equations of displacement for the above driven spring-mass system are an exact parallel of the equations of charge on the capacitor in a driven LCR circuit. The differential equation which describes the charge on the capacitor is L d2 q dt + Rdq 2 dt + 1 q = 0, (1) C where L is the inductance in Henry (H), C is the capacitance in Farad (F ) and R is the resistance in Ohm (Ω). Eq. (1) can be rewritten in terms of characteristic times q + 2α q + ω 2 0q = 0 (2) where α = R/2L and ω 0 = 1/ LC, which is the resonance frequency of the circuit. A useful parameter is the damping ratio, ξ = α/ω 0. For a damped oscillation, ξ is related to the quality factor Q as: ξ = 1/2Q. In the case of serial RLC circuit, the damping ratio is ξ = R C (3) 2 L Depending on the value of ξ, the solution of Eq. (2) breaks down to the following three cases: 2
1. under-damped case when ξ < 1. where ω = and q(0). q = q 0 e αt cos(ωt + ϕ) (4) ω 2 0 α 2. q 0 and ϕ depends on the initial conditions, q(0) Note the amplitude of a damped oscillation in n Eq. (4) decays exponentially as a function of time as e αt. 2. over-damped case for ξ > 1. q = q 1 e α α 2 ω 2 0 + q2 e α+ α 2 ω 2 0 (5) 3. critically damped when ξ = 1. This is the famous degenerate case, where the solution takes the form q(t) = (q 1 + q 2 t) e αt (6) 3
2 Pre-lab Exercises For the resonant circuit shown below (Fig 2), where C = 4.0µF, L = 10mH, and R is in the range 0 5Ω, calculate the following: (a) the frequency, the angular frequency, and the period of undamped (R = 0) oscillations; (b) the frequency, the angular frequency, and the period of damped (R = 5Ω) oscillations; (c) the decay constant for oscillations when R = 5Ω; (d) the time for the amplitude envelope to decrease to 1/2 its initial value. Figure 2: Schematic of LCR circuit showing AC voltage source E, Inductor L, Capacitor C, and resistor R. 3 Procedure The essential equipment for this lab consists of Digital oscilloscope, Tektronix 312B. Familiarize yourself with the oscilloscope s capabilities. You will find out that it has tremendous functionality, and it is easy to get lost. In that case, select the factory default settings. The oscilloscope has an ethernet port to export data. When turning on the oscilloscope, it will prompt an IP address. Record that IP address in your notebook and type it in the web browser of your computer. You will be able to see the oscilloscope output in the web browser. The lab TA will show you how to download data to your computer. Function generator, HP 8116A 4
Figure 3: Four different ways to form an LCR cirucuit. High-precision multimeter with four-wire resistance measurement capability, HP3468A A few resistors, circuit boards with built-in inductors and capacitors Cables and connectors The main portion of this lab deals with the analysis and making figures. In general, an entire course can be devoted to the RCL circuit because it exhibits such rich behavior that is classical for electronics and mechanics. The following is a minimal list of suggested steps for this lab. 1. Use RL, RC circuits to find out the L and C. 2. Measure the actual resistance of the variable resistor. 3. Set up the RLC circuit and get the proper signals from the LCR circuit fed into the oscilloscope.. As shown in Fig. 3, the components can be connected in several different ways to form an LCR circuit. You are encouraged to explore a few configurations and find the one that you feel comfortable to work with. Please note that Eq. (3) is only valid for serial RLC circuit. The relationship between ξ and the values of R, L, and C should be different for other types of RLC connections. 4. Transfer the wave forms via internet to your computer, for a range of resistances R = 0, 10, 20,..., 120 Ω. Be sure that the range includes 5
the resistance for critical damping. Because the actual resistance likely differs from the dial on the potentiometer, use the four-wire technique measure the resistance of the circuit across the capacitor (disconnect the driving portion of the circuit). 5. Show the under-damped, critical damped and over-damped cases for your data. Each of these cases are best plotted using different axes: linear-linear (regular), semilogx, and semilogy. 6. Fit the data for these three cases using the analytical solution. 7. For under-damped data sets, make plots of your fitted ω vs R, and compare these with the theory given by Eq. (4). You may find the agreement inadequate - consider adding an ac component to the measured resistance. This ac component arises to dissipation in the fluid-filled capacitor, because charging the capacitor causes alignment and motion of the dielectric. Your four-lead measurements were in DC mode, so they do not account for the AC component of the resistance. 4 Report Checklist In your report, you should (as a minimum) include the following: Measurement of the resistances of resistors in the circuit using the ohmmeter. Measurement of effective resistance of the variable resistor at different dials settings. Measurement of the L of the inductor and C of capacitor. Schematic diagram showing your electrical connections. Show plots for under-damped, critically damped, and over-damped signals. For under-damped oscillations: a) determine the oscillation period and compare the measured oscillation frequency with the theoretical value; b) Plot of peak voltage vs. time in a semi-log plot to determine the damping ratio ξ; c) Analyze the oscillation period for heavily damped oscillations. 6
5 Appendix: An Introduction to Resonance Behavior 5.1 The Quality of Oscillations Almost everything twangs[1]. Nuclei, atoms, molecules, crystals, bells, violin strings, pot covers, electrical circuits, bridges, yea the globe itself. Disturb any one of these and it will oscillate around some equilibrium position, gradually dissipating the energy that distributed it. If you drive the system at its resonant frequency, the oscillations may grow to alarming size. The duration of the vibrations of a distributed system, and the sharpness of its response to a driving force, can be characterized by a dimensionless constant Q. Oscillations are the most common form of motion because they are caused by one of the simplest force laws: F = kx, where F is the restoring force on an object that is displaced a distance, x, from its equilibrium position. To first approximation, for small displacement, Hooke s law does indeed describe the response of most bound systems. This is true of the displacement of atoms bound in their crystalline of molecular positions, and so is true on the macroscopic scale of solids. A Hooke s law force generates sinusoidal oscillations. Oscillations die down. The energy is dissipated by friction or radiation. If we assume that the friction is proportional to the velocity, which is a fair approximation for many cases, then the friction force term is bv = b dx, where dt b is the friction proportionality constant. The oscillator equation becomes m d2 x = b dx kx, which yields the familiar damped oscillation curve shown dt 2 dt in Fig. 4 For ease of analysis, many texts (as well as common practice) define the following quantities: ω 0 = k/m, the natural resonant frequency γ = b/m, the damping factor, with dimensions of frequency Q = ω 0 γ = mk b = ω 0m, the quality factor, which is dimensionless. b Lets investigate the nature of the Q of a system. Evidently, the larger the friction proportionality constant, b, the lower the Q and the faster the 7
Figure 4: Dependence of amplitude A(ω) and phase δ(ω) for a forced oscillation of amplitude F 0, and an oscillator of spring constant k and natural frequency ω 0. Figure taken from Reference [2]. oscillations die down. For very large Q, the damping is small and ω m ω 0, where ω m is the peak frequency at which the amplitude response is maximum. Even if Q = 1, ω = 0.87ω 0, which is low by only 13 percent. You can see the difference between oscillations with large Q and small Q in Fig 5 For a Q of 600, about one percent of the energy is lost in each cycle. This is typical of a piano or violin string that sings for a second or so after it has been plucked. With a fundamental frequency of several hundred, the Q of such a string must be of the order of 10 3. A pot cover in our kitchen rings for about ten seconds until until the intensity is down to about 1. Its frequency 3 is 660 Hz. Since it takes about 6600 oscillations to reduce its energy by e, its Q must be 40,000! Seismic vibrations lose intensity very slowly, considering that it s the Earth that s vibrating, and have they have Q values of several hundred. An atomic transition producing visible light has a duration (1/e time) of about 10 8 s. The period of visible light is about 10 15 s, and so the Q must be about 10 8. A gamma ray from the nucleus of Fe 57 (when bound in a crystal - the Mossbauer effect) has a Q of over 10 12 Relatively, it rings forever. Note that signal source with high Q generates a very long duration sine wave and therefore produces a very monochromatic signal. 8
Figure 5: (a) Amplitude as functions of driving frequency for different values of Q, assuming driving force of constant magnitude but variable frequency. (b) Phase difference δ as functions of driving frequency for different values of Q. Figure taken from Reference [2]. 5.2 Derivations of Resonance Behavior In this section we derive the equations of motion and examine the behavior of spring-mass systems. In the following section, the ideas presented here will be applied to the case of an LCR circuit. For an undamped spring-mass system (no friction of other dissipative forces), Newton s Second Law gives the usual second order differential equation of motion: F ma = m 9 d2 x = kx dx2 (7)
which yields x = A sin( k/m + α) = A sin(ω 0 t + α) (8) where ω 0 is the angular frequency, k/m, and α is the starting phase. The larger the spring constant, k, the higher the resonant frequency. The larger the mass of the object, m, the smaller the frequency ω 0. The frequency of the system in hertz is f 0 = ω 0 /2π, and the period is T = 1/f 0. (Note that the differential equation requires that the second derivative of the displacement, x, with respect to time, t, is proportional to minus the displacement. The sine function has this property. The derivative of sine is cosine, and taking the derivative of cosine yields negative sine.) A damped oscillation is described by k x = (x 0 e ( b/2m)t ) sin( m b2 t + α) (9) 4m2 The amplitude of the sinusoidal term is (x 0 e (b/2m)t ). It is equal to x 0 at t = 0, but then decays to 1/e of its starting amplitude at t = (2m/b). As t, x 0. Without friction ω = k/m. Note that the presence of friction decreases the frequency. In terms of γ and Q, the actual resonant frequency, ω m is related to the natural frequency by or ωm 2 = ω0 2 γ2 4 = ω 0(1 1 ). (10) 4Q2 The equation for a damped harmonic oscillation becomes, The solution is d 2 x dt 2 + γ dx dt + ω2 0x = 0, (11) d 2 x dt + ω 0 dx 2 Q dt + ω2 0x = 0. (12) x = x 0 e γt/2 sin(ωt + α) = x 0 e ω 0t/Q sin(ω 0 10 1 1 t + α) (13) 4Q2
Another way to see the significance of Q is to calculate the amplitude of energy loss per cycle in a damped oscillation. Instead of expressing t in seconds, express time in terms of number of periods of oscillations, n: t = nt 0 = n 2π ω 0. In these terms, x = x 0 e nπ/q sin(ωt + α). The amplitude falls be e ( 2.7) in Q/π cycles, and the energy falls by e in Q/2π cycles (since the energy is proportional to the square of the amplitude). In one x cycle: = π E and = 2π. (Since x Q E Q e nπ/q 1 nπ/q, then x 0 = x 0, x 1 x 0 π/q, and x 2 x 0 2π/Q x 1 π/q, etc...) We can get rid of the 2π factor by observing that the energy falls by e in Q radians and that the fractional energy lost per radian is E = 1. (For instance, if Q = 12, E Q the energy decreases by a factor of roughly 2.7 in 2 cycles, or in 12 radians or roughly 12.5 radians. the fractional loss of energy is 1 in one cycle of 1 2 12 in one radian.) References [1] The text here is borrowed almost directly from an old PH2651 materials provided by Stephen Jasperson, Richard Quimby, and Stephan Kholer. [2] Vibrations and Waves, by A. P. French, W. W. Norton and Company (1971). 11