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

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1 Background Material

1.1 Electrical Oscillations of an LCR Circuit

Figure 1: Schematic of LCR circuit where R is the resistance, C is thecapacitance, 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 displace-ment for the above driven spring-mass system are an exact parallel of theequations of charge on the capacitor in a driven LCR circuit. The differentialequation which describes the charge on the capacitor is

Ld2q

dt2+R

dq

dt+

1

Cq = 0, (1)

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 ofcharacteristic times

q + 2αq + ω20q = 0 (2)

where α = R/2L and ω0 = 1/√LC, which is the resonance frequency of the

circuit.A useful parameter is the damping ratio, ξ = α/ω0. For a damped oscil-

lation, 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

2

√C

L(3)

Depending on the value of ξ, the solution of Eq. (2) breaks down to thefollowing three cases:

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1. under-damped case when ξ < 1.

q = q0e−αt cos(ωt+ ϕ) (4)

where ω =√ω2

0 − α2. q0 and ϕ depends on the initial conditions, q(0)and q(0).

Note the amplitude of a damped oscillation in n Eq. (4) decays expo-nentially as a function of time as e−αt.

2. over-damped case for ξ > 1.

q = q1e−α−√α2−ω2

0 + q2e−α+√α2−ω2

0 (5)

3. critically damped when ξ = 1. This is the famous degenerate case,where the solution takes the form

q(t) = (q1 + q2t) e−αt (6)

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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, InductorL, Capacitor C, and resistor R.

3 Procedure

The essential equipment for this lab consists of

• Digital oscilloscope, Tektronix 312B. Familiarize yourself with the os-cilloscope’s capabilities. You will find out that it has tremendous func-tionality, and it is easy to get lost. In that case, select the factorydefault settings. The oscilloscope has an ethernet port to export data.When turning on the oscilloscope, it will prompt an IP address. Recordthat IP address in your notebook and type it in the web browser of yourcomputer. You will be able to see the oscilloscope output in the webbrowser. The lab TA will show you how to download data to yourcomputer.

• Function generator, HP 8116A

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Figure 3: Four different ways to form an LCR cirucuit.

• High-precision multimeter with four-wire resistance measurement ca-pability, 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 itexhibits 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 circuitfed into the oscilloscope.. As shown in Fig. 3, the components can beconnected in several different ways to form an LCR circuit. You areencouraged to explore a few configurations and find the one that youfeel comfortable to work with. Please note that Eq. (3) is only validfor 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 ofresistances R = 0, 10, 20, . . . , 120 Ω. Be sure that the range includes

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the resistance for critical damping. Because the actual resistance likelydiffers from the dial on the potentiometer, use the four-wire techniquemeasure the resistance of the circuit across the capacitor (disconnectthe driving portion of the circuit).

5. Show the under-damped, critical damped and over-damped cases foryour 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 com-pare these with the theory given by Eq. (4). You may find the agree-ment inadequate - consider adding an ac component to the measuredresistance. This ac component arises to dissipation in the fluid-filledcapacitor, because charging the capacitor causes alignment and motionof the dielectric. Your four-lead measurements were in DC mode, sothey 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 theohmmeter.

• Measurement of effective resistance of the variable resistor at differentdials 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 sig-nals.

• For under-damped oscillations: a) determine the oscillation period andcompare the measured oscillation frequency with the theoretical value;b) Plot of peak voltage vs. time in a semi-log plot to determine thedamping ratio ξ; c) Analyze the oscillation period for heavily dampedoscillations.

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5 Appendix: An Introduction to Resonance

Behavior

5.1 The Quality of Oscillations

Almost everything twangs[1]. Nuclei, atoms, molecules, crystals, bells, violinstrings, pot covers, electrical circuits, bridges, yea the globe itself. Disturbany one of these and it will oscillate around some equilibrium position, grad-ually dissipating the energy that distributed it. If you drive the system at itsresonant frequency, the oscillations may grow to alarming size. The durationof the vibrations of a distributed system, and the sharpness of its responseto a driving force, can be characterized by a dimensionless constant – Q.

Oscillations are the most common form of motion because they are causedby one of the simplest force laws: F = −kx, where F is the restoring forceon an object that is displaced a distance, x, from its equilibrium position. Tofirst approximation, for small displacement, Hooke’s law does indeed describethe response of most bound systems. This is true of the displacement ofatoms bound in their crystalline of molecular positions, and so is true onthe macroscopic scale of solids. A Hooke’s law force generates sinusoidaloscillations.

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 fairapproximation for many cases, then the friction force term is bv = bdx

dt, where

b is the friction proportionality constant. The oscillator equation becomesmd2x

dt2= −bdx

dt−kx, which yields the familiar damped oscillation curve shown

in Fig. 4For ease of analysis, many texts (as well as common practice) define the

following quantities:

ω0 =√k/m, the “natural” resonant frequency

γ = b/m, the damping factor, with dimensions of frequency

Q = ω0

γ=√mkb

= ω0mb

, the “quality” factor, which is dimensionless.

Lets investigate the nature of the “Q” of a system. Evidently, the largerthe friction proportionality constant, b, the lower the Q and the faster the

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Figure 4: Dependence of amplitude A(ω) and phase δ(ω) for a forced oscil-lation of amplitude F0, and an oscillator of spring constant k and naturalfrequency ω0. Figure taken from Reference [2].

oscillations die down. For very large Q, the damping is small and ωm ≈ω0, where ωm is the “peak” frequency at which the amplitude response ismaximum. Even if Q = 1, ω = 0.87ω0, which is low by only 13 percent. Youcan 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. Thisis typical of a piano or violin string that sings for a second or so after it hasbeen plucked. With a fundamental frequency of several hundred, the Q ofsuch a string must be of the order of 103. A pot cover in our kitchen rings forabout ten seconds until until the intensity is down to about 1

3. Its frequency

is 660 Hz. Since it takes about 6600 oscillations to reduce its energy by e, itsQ must be 40,000! Seismic vibrations lose intensity very slowly, consideringthat it’s the Earth that’s vibrating, and have they have Q values of severalhundred. An atomic transition producing visible light has a duration (1/etime) of about 10−8 s. The period of visible light is about 10−15 s, and so theQ must be about 108. A gamma ray from the nucleus of Fe57 (when boundin a crystal - the Mossbauer effect) has a Q of over 1012 Relatively, it ringsforever. Note that signal source with high Q generates a very long durationsine wave and therefore produces a very monochromatic signal.

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Figure 5: (a) Amplitude as functions of driving frequency for different valuesof Q, assuming driving force of constant magnitude but variable frequency.(b) Phase difference δ as functions of driving frequency for different valuesof Q. Figure taken from Reference [2].

5.2 Derivations of Resonance Behavior

In this section we derive the equations of motion and examine the behaviorof spring-mass systems. In the following section, the ideas presented herewill be applied to the case of an LCR circuit. For an undamped spring-masssystem (no friction of other dissipative forces), Newton’s Second Law givesthe usual second order differential equation of motion:

F −ma = md2x

dx2= −kx (7)

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which yields

x = A sin(√k/m+ α) = A sin(ω0t+ α) (8)

where ω0 is the angular frequency,√k/m, and α is the starting phase. The

larger the spring constant, k, the higher the resonant frequency. The largerthe mass of the object, m, the smaller the frequency ω0. The frequency of thesystem in hertz is f0 = ω0/2π, and the period is T = 1/f0. (Note that thedifferential equation requires that the second derivative of the displacement,x, with respect to time, t, is proportional to minus the displacement. Thesine function has this property. The derivative of sine is cosine, and takingthe derivative of cosine yields negative sine.)

A damped oscillation is described by

x = (x0e(−b/2m)t) sin(

√k

m− b2

4m2t+ α) (9)

The amplitude of the sinusoidal term is (x0e−(b/2m)t). It is equal to x0 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, ωm is related to the

“natural” frequency by

ω2m = ω2

0 −γ2

4= ω0(1− 1

4Q2). (10)

The equation for a damped harmonic oscillation becomes,

d2x

dt2+ γ

dx

dt+ ω2

0x = 0, (11)

ord2x

dt2+ω0

Q

dx

dt+ ω2

0x = 0. (12)

The solution is

x = x0e−γt/2 sin(ωt+ α) = x0e

−ω0t/Q sin(ω0

√1− 1

4Q2t+ α) (13)

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Another way to see the significance of Q is to calculate the amplitudeof energy loss per cycle in a damped oscillation. Instead of expressing tin seconds, express time in terms of number of periods of oscillations, n:t = nT0 = n2π

ω0. In these terms, x = x0e

−nπ/Q sin(ωt + α). The amplitudefalls 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 onecycle: ∆x

x= π

Qand ∆E

E= 2π

Q. (Since e−nπ/Q ≈ 1 − nπ/Q, then x0 = x0,

x1 ≈ x0 − π/Q, and x2 ≈ x0 − 2π/Q ≈ x1 − π/Q, etc...) We can get rid ofthe 2π factor by observing that the energy falls by e in Q radians and thatthe fractional energy lost per radian is ∆E

E= 1

Q. (For instance, if Q = 12,

the energy decreases by a factor of roughly 2.7 in 2 cycles, or in 12 radiansor roughly 12.5 radians. the fractional loss of energy is 1

2in one cycle of 1

12

in one radian.)

References

[1] The text here is borrowed almost directly from an old PH2651 materialsprovided by Stephen Jasperson, Richard Quimby, and Stephan Kholer.

[2] Vibrations and Waves, by A. P. French, W. W. Norton and Company(1971).

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