Section 24–2 Damped oscillations

(Guessing the answer / Solving under-damped oscillation / Rationalizing the answer)

In this section, Feynman discusses an under-damped oscillation from the perspective of guessing an answer, solving a damped transient motion, and rationalizing the answer. This section could be titled as “under-damped oscillations” because the next section is about over-damped oscillations.

1. Guessing the answer:

“So, as a guess, for a relatively high Q system, we would suppose that the following equation might be roughly right (we will later do it exactly, and it will turn out that it was right!) … (Feynman et al., 1963, section 24–2 Damped oscillations).”

Feynman guesses that the answer for the damped transient motion will be an oscillation of frequency that is close to the resonant frequency ω₀. In addition, the amplitude of the sine-wave motion will diminish as e^−γt/2 and thus, he expresses the displacement as x = A₀e^−γt/2cos ω₀t. However, this is not really a good guess because we should not ignore the presence of a resistive force or friction. Intuitively speaking, one would expect the period of the oscillation to be slightly longer as a result of friction that reduces the average speed of the oscillator. Feynman later clarifies that the frequency is ω_γ instead of ω₀, and says that this is the only error, but we have the right idea.

Feynman provides a better guess for a relatively high Q system, that is, we would have dE/dt = −ωE/Q. He mentions that he will later do this equation exactly and it will turn out that the equation was right! As a suggestion, we may explain that this equation can be directly proved by using the equation (24.19) that is x = e^−γt/2(Ae^iω_γt + Be^−iω_γt) in which ω_γ = Ö(ω₀² − γ²). Thus, the total energy stored in a forced oscillator is proportional to e^−γt (or x²) because its potential energy is equal to ½kx² and its kinetic energy is equal to ½mω²(x)². If we let E₀ to be the maximum energy of the oscillator, we would have E = E₀e^−γt and dE/dt = −γE₀e^−γt = −γE. Using the definition of Q as ω₀/γ (or γ = ω₀/Q), we can show that dE/dt = −ω₀E/Q (instead of dE/dt = −ωE/Q).

2. Solving under-damped oscillation:

“So, starting with Eq. (24.1), with no outside force, how do we solve it? Being physicists, we do not have to worry about the method as much as we do about what the solution is (Feynman et al., 1963, section 24–2 Damped oscillations).”

Feynman solves the under-damped oscillation by using the equation md²x/dt²+ γmdx/dt + mω₀²x = 0. Then, he simplifies it to −α² + iαγ + ω₀² = 0 that has two solutions: α₁ = iγ/2 + Ö(ω₀² − γ²/4) (24.14) and α₂ = iγ/2 −Ö(ω₀² − γ²/4) (24.15). However, one may prefer using the equation md²x/dt² + 2γmdx/dt + mω₀²x = 0 in which 2γmdx/dt is the resistive force. In this approach, the two solutions can be re-expressed as α₁ = iγ + Ö(ω₀² − γ²) and α₂ = iγ – Ö(ω₀² − γ²). Thus, we have x = (e^−γt)(Ae^iω_γt + Be^−iω_γt) or equivalently, x = Ce^−γt cos (ω_γt + q). One advantage of this approach is to allow the two solutions to look simpler without the need of remembering ½ and ¼. Another advantage is that we can relate the Q factor to the real part of the solution, ω₀² − γ², because Q may also be defined as a ratio of ω₀ to γ.

According to Feynman, we assume that γ is fairly small compared with ω₀ and this implies that the resonance frequency is close to ω₀ and Ö(ω₀² − γ²) is greater than zero. One may clarify that this transient motion is commonly known as an under-damped harmonic oscillation. It is different from an over-damped harmonic oscillation in which γ is greater than ω₀ and Ö(ω₀² − γ²) becomes an imaginary number. Furthermore, Feynman did not completely solve this problem by determining the two arbitrary constants A and B. Using x = (e^−γt)(Ae^iω_γt + Be^−iω_γt) = Ce^−γt cos (ω_γt + q) and by comparing the real parts and imaginary parts, we have A + B = Ccos q and A − B = i(Csin q). The arbitrary constants can be determined if we know x₀ and v₀.

3. Rationalizing the answer:

“The equation does not know that we are arbitrarily going to take the real part, so it has to present us, so to speak, with a complex conjugate type of solution, so that by putting them together we can make a truly real solution (Feynman et al., 1963, section 24–2 Damped oscillations).”

Feynman says that we must take the real part of the solution, and ask how did the mathematics know that we only wanted the real part? Then, he explains that the physical world has a real solution and expresses the solution as x = e^−γt/2(Ae^iω_γt + A*e^−iω_γt). However, one may visualize complex numbers as equivalent to rotating vectors that have horizontal and vertical components (instead of real and imaginary components). Historically, Steinmetz formalizes the concept of a phasor using rotating vectors and relating the sinusoidal steady-state of a circuit to complex numbers (Araújo & Tonidandel, 2013). He also identifies properties of sinusoids, such as the sum of sinusoids, A₁sin (ωt + q₁) + A₂sin (ωt + q₂), by resolving the horizontal and vertical components, and distinguishes the vertical component using the letter j that means Ö-1.

Feynman ends the section by stating the real solution as x = e^−γt/2(Ae^iω_γt + A*e^−iω_γt) and concludes that our real solution is an oscillation with a phase shift and a damping. Alternatively, some may simplify the solution as Ce^−γtcos (ωt + q) and explain that the phase shift is q as shown in the equation. However, Feynman could have elaborated that the oscillation not only has a phase shift, but there is also a “frequency shift” to ω_γ. One may defend Feynman by explaining the resonance frequency ω_γ is approximately equal to ω₀ because γ is relatively small and this implies that Ö(ω₀² − γ²) » ω₀. Strictly speaking, the phase shift is not equal to q, but we need to consider the phase shift that is due to ω_γt because ω_γ is slightly smaller than ω₀.

Questions for discussion:

1. Feynman guesses that dE/dt = −ωE/Q for a relatively high Q system, but could the equation be dE/dt = −ω₀E/Q or dE/dt = −ω_γE/Q?

2. Would you solve the under-damped oscillation using the equation md²x/dt² + γmdx/dt + mω₀²x = 0 or md²x/dt² + 2γmdx/dt + mω₀²x = 0?

3. Would you explain that we can make a truly real solution because the equation does not know that we are arbitrarily going to take the real part?

The moral of the lesson: the real solution of under-damped oscillation can be expressed as x = e^−γt/2(Ae^iω_γt + A*e^−iω_γt) and it means that our real solution is an oscillation with a phase shift and a light damping.

References:

1. Araújo, A. E. A. D., & Tonidandel, D. A. V. (2013). Steinmetz and the Concept of Phasor: A Forgotten Story. Journal of Control, Automation and Electrical Systems, 24(3), 388-395.

2. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

Section 24–1 The energy of an oscillator

(Work done by a force / Mean stored energy / Q factor)

In this section, Feynman discusses the work done by an external force, the mean energy stored in a forced oscillator, and defines the Q factor.

1. Work done by a force:

“In our problem, of course, F(t) is a cosine function of t. Now let us analyze the situation: how much work is done by the outside force F? (Feynman et al., 1963, section 24–1 The energy of an oscillator).”

Feynman shows that the work done by an external force per second is equal to m[(dx/dt)(d²x/dt²)+ω₀²x(dx/dt)] + γm(dx/dt)². Then, he explains that the average power is equal to < γm(dx/dt)² > because the stored energy does not change in the long run, i.e., < m[(dx/dt)(d²x/dt²) + ω₀²x(dx/dt)] > is effectively zero as t approaches infinity. In defining the Q factor, he uses this result and states that the “work done per cycle” is (γmω²<x²>)(2π/ω). In other words, Feynman has shown that the work done by an external force F₀ cos ωt per second (or average power) is equal to the work done by friction per cycle after a long time. That is, all the energy transferred to an oscillator ultimately ends up in the resistive term γm(dx/dt)².

Feynman elaborates that the rate of electrical energy loss is the electrical resistance times the average square of the current: < P > = R< I² > = ½R I₀². As an analogy, the velocity dx/dt of an object corresponds to the electric current dq/dt, whereas the resistive term mγ corresponds to the electrical resistance R. One may add that the electrical or mechanical energy of an oscillator is constant in the long run, but the total energy is not conserved if we define the oscillator to be an isolated system. However, the work done on the oscillator by a driving force (or electrical source) do not always equal to the work done by the oscillator to overcome a resistive force (or electrical resistance). These two works done are effectively equal in the long run when the system reaches “a steady state” or a stable condition that does not change with time.

2. Mean stored energy:

“At any moment there is a certain amount of stored energy, so we would like to calculate the mean stored energy <E> also. (Feynman et al., 1963, section 24–1 The energy of an oscillator).”

Feynman briefly shows that the mean stored energy <E> = ½m<(dx/dt)²> + ½mω₀²<x²> = ½m(ω²+ω₀²)(½x₀²). Some students may be confused if they consider (dx/dt)² = (iω)²(x²) = -ω²x² because d(x₀e^iωt)/dt = (iω)(x₀e^iωt) = (iω)(x). However, the real part of (iω)(x) = (iω)(x₀)(cos ωt + i sin ωt) is -ωx₀ sin ωt and thus, (dx/dt)² is equal to +ω²x₀² sin² ωt instead of -ω²x². To be clearer, one may include the following steps. For a simple harmonic oscillator, we can express the average kinetic energy stored in an object as <K.E.> = ½mw₀²(½x₀²) and the average potential energy stored in a spring as <P.E.> = ½k(½x₀²). For a forced oscillator with a light damping, we have K.E. = ½mv² = ½mw²x₀²sin² wt and <K.E.> = ½mw²(½x₀²), whereas P.E. = ½kx² = ½mw₀²cos² wt and <P.E.> = ½mω₀²(½x₀²).

Feynman mentions that another interesting feature to discuss is how much energy is stored. However, there are other interesting features, for example, one may consider an oscillator’s stored energy is practically constant only after a sufficiently long time instead of t ® ¥ that seems “unrealistic.” This can be simply achieved by increasing the displacement of the oscillator as a result of an external force. Importantly, the oscillator has a maximum stored energy and maximum displacement at the resonant frequency that is slightly lesser than the natural frequency of oscillation (see section 24–2). When the oscillator is in resonance, the force and the motion of the oscillator are in phase, but the phase difference between the force and the displacement is π/2 radians (see Fig. 23-3).

3. Q factor:

“This is called the Q of the system, and Q is defined as 2π times the mean stored energy, divided by the work done per cycle (Feynman et al., 1963, section 24–1 The energy of an oscillator).”

Feynman defines the Q factor as 2π times the mean stored energy, divided by the work done per cycle. He explains that if we specify the “work done per radian” instead of per cycle, then the 2π disappears. In other words, we can avoid having “2π times the mean stored energy” in the formula by saying the work done per radian such that the term 2π appears in the denominator as “work done/2π.”. Alternatively, we can define Q as a ratio of average energy stored in the oscillator to average energy dissipated during 1 radian of motion (e.g., Kleppner & Kolenkow, 1973). This is because the average work done by an external force is equal to the average work done by friction. In essence, the Q factor is also a comparison between the “energy stored” and the “energy lost.”

According to Feynman, the efficiency of an oscillator can be measured by how much energy is stored as compared to how much work is done by the external force per cycle. Feynman also defines the Q factor as w₀/Dw in which Dw is the frequency width in section 23-2. The frequency width is directly related to the sharpness of resonance depending on the energy stored in the oscillator and the energy lost. The Q factor is sometimes known as the “storage factor” because its numerator is a measure of the energy stored during the oscillation. On the other hand, a system has a higher “quality factor” if its denominator as a measure of the degree of damping has a lower value of friction. Some may prefer the definition of Q factor as w₀/g because the stored energy is proportional to w₀ and the energy loss is dependent on g.

Questions for discussion (Feynman’s explanations were unclear?):

1. How would you explain the work done by an external force F₀ cos ωt per second is equal to the work done by friction per cycle after a long time?

2. How would you explain the mean stored energy of a forced oscillator that has a light damping?

3. Would you explain the Q factor of a system as 2π times the mean stored energy, divided by the work done per cycle?

The moral of the lesson: the Q factor of a system is a measure of the sharpness of resonance that is dependent on the energy stored in the oscillator and the energy lost.

References:

1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

2. Kleppner, D., & Kolenkow, R. (1973). An Introduction to Mechanics. Singapore: McGraw-Hill.

Section 23–4 Resonance in nature

(Mechanical resonance / Nuclear resonance / Resonance particle)

In this section, Feynman discusses mechanical resonance, nuclear resonance, and resonance particle. Specifically, there are six different kinds of resonance, namely, tidal resonance, lattice resonance, nuclear magnetic resonance, nuclear resonance, recoilless nuclear resonance, and resonance particle.

1. Mechanical resonance:

“The first two are from mechanics, the first on a large scale: the atmosphere of the whole earth (Feynman et al., 1963, section 23–4 Resonance in nature).”

Tidal resonance: Feynman says the oscillator (earth’s atmosphere) is driven by the moon, which is effectively revolving about the earth. The resonant frequency corresponding to the rotation of the earth under the moon, which occurs at a period of 12.42 hours - 12 hours due to the earth’s rotation and a little more due to the moon’s rotation. It refers to a tidal resonance in which the largest constituent is the “principal lunar semi-diurnal” tide (M₂ waves). The resonance period is equal to half of a tidal lunar day or the time required for the Earth to rotate once relative to the Moon. This is analogous to the time required for the minute-hand on a watch that starts moving with the hour-hand at 12:00 and they meet again at about 1:05 instead of 1:00.

Feynman explains that we can get the resonant frequency ω₀ and the frequency width γ from the size of the atmospheric tides, and from the phase. He considers it to be an example of poor science if we simply draw a beautiful curve using two numbers instead of measuring something else. However, physicists do not have the liberty to vary the forcing frequencies of the earthquakes. One should recall how Fermi responded to Dyson’s model that uses four arbitrary parameters. Fermi’s reply was “I remember my friend Johnny von Neumann used to say, with four parameters I can fit an elephant, and with five I can make him wiggle his trunk (Dyson, 2015, p. 125).” Nevertheless, we can use the Lorentzian resonance curve or spectral lineshape function (g/2)²/[(w₀-w)² + (g/2)²] to draw the so-called beautiful curve.

Lattice resonance: Feynman discusses another mechanical oscillation that involves a sodium chloride crystal. He suggests that we cannot say whether the resonance width in Fig. 23–7 is natural, or whether it is due to inhomogeneities in the crystal or the finite width of the slit of the spectrometer. To be precise, the absorption of infrared radiation is due to a lattice resonance instead of the so-called mechanical oscillation. Currently, experimental data shows that there is an energy band (or reststrahlen band) in which polar solids such as NaCl absorb and reflect light very strongly. The graph should be more complicated because of the interaction with phonons such that some infrared radiations cannot propagate within a given medium (Fox, 2002).

2. Nuclear resonance:

“Our next example has to do with atomic nuclei. The motions of protons and neutrons in nuclei are oscillatory in certain ways, and we can demonstrate this by the following experiment (Feynman et al., 1963, section 23–4 Resonance in nature).”

Nuclear magnetic resonance: Feynman explains that the nuclear magnetic resonance is about a swinging of atoms that have an angular momentum. Essentially, the frequency of the lateral magnetic field that drives this swinging is kept constant, but it is easier to change the magnetic field strength. One may clarify that the absorption and emission of electromagnetic radiations are related to electrons or atomic nuclei that have spins. The energy of electromagnetic radiations corresponds to the work done against the magnetic fields in turning the “tiny nuclear magnets” from one position to the opposite direction. Nuclear magnetic resonance was developed in 1945 by Felix Bloch and Edward M. Purcell, who were awarded the 1952 Nobel Prize in physics for their research and contributions.

Nuclear resonance: By bombarding a lithium atom with protons, the nuclear reactions produce γ-rays and the graph has a very sharp resonance. Feynman elaborates that the horizontal scale is not a frequency, but it is an energy that is related to the frequency of a wave. This is a nuclear resonance that is caused by the formation of new nuclei (Beryllium) in a particular excited state. On the other hand, the vertical scale corresponds to the intensity of gamma radiations, but it may also be measured in terms of the probability that an incident proton causes an emission of gamma-ray (French, 1971). This probability can be described using the effective target area (or cross section) of a nucleus that is hit by the protons.

Recoilless nuclear resonance: Mössbauer, Feynman’s former colleague, was awarded the Nobel prize in physics for his discovery of recoilless nuclear resonance. Feynman explains that the horizontal scale is velocity and the technique for obtaining different frequencies is related to the Doppler effect (or the relative speed between the source and the absorber). Specifically, a free nucleus (source) recoils after emitting a gamma-ray and the total energy (gamma-ray) absorbed by another nucleus (absorber) is lesser by the recoil energy. Strictly speaking, Mössbauer effect is observed when the nuclei are tightly bound such that the whole crystal recoils after the emission of a gamma-ray. In this case, spectral lines become very sharp because the mass of the crystal is practically infinite and thus, it is an essentially “recoil-less” resonance.

3. Resonance particle:

“We thus determine that there is a resonance at a certain energy for the K⁻ meson (Feynman et al., 1963, section 23–4 Resonance in nature).”

One should be cautioned that Feynman uses two different concepts, resonance states and resonance particles, to explain the resonance in particle physics. According to Feynman, there is a resonance found in a nuclear reaction when a K⁻ meson (kaon) and a proton interact. This results in some kind of a state corresponding to the resonance at a certain energy. In other words, it is a resonance state that corresponds to the energy of a kaon. Experimentally, it appears as a “bump” or “jerk” in a curve, but it could be related to a statistical fluctuation or systematic effect. In essence, resonance states are unstable states that are short-lived and they are similar to atomic energy levels. It is different from the concept of a resonance particle that has an invariant mass.

Feynman elaborates that we do not know whether to call a bump like this a “particle” or simply a resonance. When the resonance is very sharp, it means that it corresponds to a very definite energy as if there were a particle of that energy present in nature. In this view, one may conceptualize the resonance behaves effectively like a particle that has a cross-section and it is able to collide with other particles. In a sense, one may question the existence of resonance particles that have very short lifetimes. However, by using Einstein’s mass-energy relationship, we can deduce the invariant mass of a resonance particle.

Another interesting “bump hunting” is the discovery of a new subatomic particle J/y that is recognized for a Nobel prize in 1976. Lederman gives a nice explanation of the sharpness of resonance that is related to Heisenberg's uncertainty relations: “[t]he shorter the lifetime, the wider the distribution of masses. It is a quantum connection. What we mean by a distribution of masses is that a series of measurements will yield different masses, distributed in a bell-shaped probability curve (Lederman & Teresi, 2006, p. 316).” Lederman was possibly the first person to observe this resonance but he was not awarded for this Nobel prize. In his own words, “I was overjoyed at the breakthrough, a joy tinged, of course, with envy and even just a touch of murderous hatred for the discoverers (Lederman & Teresi, 2006, p. 315).”

Questions for discussion (Feynman’s mistakes?):

1. Should the resonance of sodium chloride crystal be explained as a small scale of mechanical oscillation?

2. Is the nuclear magnetic resonance is really about a swinging of atoms that have an angular momentum? 

3. Should the resonance in particle physics be explained using the concept of resonance states or resonance particles?

The moral of the lesson: the research in resonance has resulted in many Nobel Prizes such as nuclear magnetic resonance in 1952, recoilless nuclear resonance in 1961, and the J/y particle in 1976.

References:

1. Dyson, F. J. (2015). Birds and Frogs: Selected Papers of Freeman Dyson, 1990–2014. Singapore: World Scientific.

2. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

3. Fox, M. (2002). Optical properties of solids. New York: Oxford University Press.

4. French, A. P. (1971). Vibrations and Waves. New York: W.W. Norton.

5. Lederman, L., Teresi, D. (2006). God Particle: If the Universe Is the Answer, What Is the Question?. New York: Dell.