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.