Taylor 5.6-5.8 (today) and 8.1-8.2 (Monday).
The tangent of the phase is the ratio of the imaginary part to the real part (Taylor defines the phase as the negative of this). It is convenient to make the denominator of C real to extract the real and imaginary parts, yielding
The f0 and the quantity in the denominator are common to both the real and imaginary parts, so the phase is simply
Now the particular solution has the form
The amplitude of a sinusoidally driven oscillator varies with the frequency of the applied force. For a weakly damped oscillator, this effect can be dramatic, with small oscillations when the driving frequency is far from the natural frequency of the oscillator, and large oscillations when the driving frequency is close to the natural frequency of the oscillator. As you are to show in a homework problem, the maximum occurs when the driving frequency is ωmax = sqrt(ω0² - 2β²).
The peak in the plot of amplitude versus driving frequency is called the resonance peak. The amplitude at this maximum is approximately the amplitude found when ω≈ω0 which is A = f0/2βω0. The width of the resonance peak is usually given as the full-width at half-maximum (FWHM) and is approximately 2β (as long as β<<ω0.
We discussed the quality factor earlier in the context of the underdamped oscillator. The quality factor, Q, is defined as 2π times the energy stored in the oscillator divided by the energy lost in a single period of oscillation. Therefore we get immediately that, in the case of weak damping,
The quality factor is a measure of the sharpness of the resonance peak, and is also commonly defined as the position of the center of the peak divided by the FWHM. In the weak damping limit, the center of the peak is at approximately ω0, which is approximately the same as ωd, so that the two definitions give the same result.
Let's now look at what happens to the phase of the oscillator near the resonance peak. Recall that the phase is given by tanδ = 2βω / (ω0² - ω²). For ω = ω0, tanδ = ∞, so δ = π/2. For ω < ω0, tanδ > 0, so δ < π/2. For ω > ω0, tanδ < 0, so δ > π/2. For ω --> ∞, tanδ --> 0 from below, so δ --> π/2.
Plot δ versus ω. The important feature is the transition through π/2 near the resonance.
Although it isn't obvious how to solve the problem of a driven oscillator for an arbitrary time-dependent driving force, when the force varies sinusoidally we can completely solve the problem.
The idea behind the Fourier series is that any arbitrary periodic function can be represented by an infinite sum of sines and cosines. "Represented" means that the series converges to the given function in a reasonable way. (Note, if the function is not continuous (has discontinusous jumps), then at a discontinuity, the series converges to a value midway between the function's values on either side of the discontinuity.)
The details are how to find the Fourier series representation of a function, and how to use it in a problem. Assume we have a periodic function of period τ, meaning that the value of the function at time t is the same as the value of the function at time t+τ for any time t, or f(t) = f(t+τ). This function can be represented by an infinite sum of sines and cosines that are integer multiples of the fundamental frequency ω = 2π/τ as follows:
The trick is now to find the Fourier coefficients a0, an and bn (n=1 to ∞).
A most useful fact, that I won't prove here, is that the integral of the product of two cosine and/or sine functions over one or more complete periods of each of the two, is zero unless the product is of the same function, and then it is half the length of the integral.
where τ = 2π/ω. This result gives us a way to find the coefficients by multiplying both sides of the expression for f(t) by the various sines and cosine factors, and integrating over the fundamental period of f(t), so that all the terms except one yield zero on the right hand side, yielding a relation for one of the coefficients. The result is the following integrals for the coefficients
While this seems tedious, usually there are a few tricks that can be used to reduce the amount of work. For instance, if f(t) is even, then all the bn are zero; if f(t) is odd, then all the an (including a0) are zero.
That the Fourier series can be used to solve the problem of an oscillator driven with a periodic force rests on the following observation. The superposition principle says that if x1(t) is the solution to the problem with driving force f1(t), D x1 = f1, and x2(t) is the solution of the problem with driving force f2(t), D x2 = f2, then x(t) = x1(t) + x2(t) is the solution of the problem with driving force f(t) = f1(t) + f2(t) since D x = D(x1+x2) = D x1 + D x2 = f1 + f2 = f. This is related to (relies on) the superposition principle.
Thus, if we can write f(t) as a Fourier series, and we know the solution for each driving frequency, the solution can be written as the infinite sum of discrete solutions. That is, if xn(t) satisfies Dxn = fn, then the solution x(t) = ∑ xn(t).
Consider the case of a weakly damped oscillator with β = 0.2 and ω0 = 2π so that τ0 = 1. Let it be driven by a periodic pulse with Δτ = 0.25, and fmax = 1. Calculate the first six terms for the particular solution x(t) for driving periods τ = 1.0&tau0, 1.5&tau0, 2.0&tau0, and 2.5&tau0.
First we need the Fourier series for the driving force. This depends on the driving period τ. Because the average value of the force is not zero, there is a constant term
We can arrange things so that the force is an even function and all the bn = 0. The other a coefficients are determined with the integral