5.4: Successive approximation - How deep is the well?

The next illustration of taking out the big part emphasizes successive approximation and is disguised as a physics problem.

You drop a stone down a well of unknown depth h and hear the splash 4 s later. Neglecting air resistance, find h to within 5%. Use \(cs = 340 ms^{−1}\) as the speed of sound and \(g = 10 ms^{−2}\) as the strength of gravity.

Approximate and exact solutions give almost the same well depth, but offer significantly different understandings.

Exact depth

The depth is determined by the constraint that the 4 s wait splits into two times: the rock falling freely down the well and the sound traveling up the well. The free-fall time is \(\sqrt{2h/g}\) (Problem 1.3), so the total time is

\[T = \underbrace{\sqrt{\frac{2h}{g}}}_{\text{rock}} +\underbrace{\frac{h}{C_{s}}}_{\text{sound}}. \label{5.32} \]

To solve for \(h\) exactly, either isolate the square root on one side and square both sides to get a quadratic equation in \(h\) (Problem 5.30); or, for a less error-prone method, rewrite the constraint as a quadratic equation in a new variable \(z = \sqrt{h}\)

Problem 5.30 Other quadratic

Solve for \(h\) by isolating the square root on one side and squaring both sides. What are the advantages and disadvantages of this method in comparison with the method of rewriting the constraint as a quadratic in \(z = h\)?

As a quadratic equation in \(z = \sqrt{h}\), the constraint is

\[\frac{1}{c_{s}}Z^{2} + \sqrt{\frac{2}{g}}z - T = 0. \label{5.33} \]

Using the quadratic formula and choosing the positive root yields

\[Z = \frac{-\sqrt{2/g} + \sqrt{2/g + 4T/c_{s}}}{2/c_{s}}. \label{5.34} \]

Because \(z^{2} = h\),

\[h = (\frac{-\sqrt{2/g} + \sqrt{2/g + 4T/c_{s}}}{2/c_{s}})^{2}. \label{5.35} \]

Substituting \(g = 10ms^{−2}\) and \(c_{s} = 340ms^{−1}\) gives \(h ≈ 71.56m\).

Even if the depth is correct, the exact formula for it is a mess. Such high- entropy horrors arise frequently from the quadratic formula; its use often signals the triumph of symbol manipulation over thought. Exact answers, we will find, may be less useful than approximate answers.

Approximate depth

To find a low-entropy, approximate depth, identify the big part—the most important effect. Here, most of the total time is the rock’s free fall: The rock’s maximum speed, even if it fell for the entire 4 s, is only \(gT = 40 ms^{−1}\), which is far below \(c_{s}\). Therefore, the most important effect should arise in the extreme case of infinite sound speed.

Question

If \(c_{s} = \infty\), how deep is the well?

In this zeroth approximation, the free-fall time \(t_{0}\) is the full time \(T = 4s\), so the well depth \(h_{0}\) becomes

\[h_{0} = \frac{1}{2}gt^{2}_{0} = 80m. \label{5.36} \]

Question

Is this approximate depth an overestimate or underestimate? How accurate is it?

This approximation neglects the sound-travel time, so it overestimates the free-fall time and therefore the depth. Compared to the true depth of roughly 71.56 m, it overestimates the depth by only 11% reasonable accuracy for a quick method offering physical insight. Furthermore, this approximation suggests its own refinement.

Question

How can this approximation be improved?

To improve it, use the approximate depth \(h_{0}\) to approximate the sound-travel time.

\[t_{\text{sound}} ≈ \frac{h_{0}}{c_{s}} ≈ 0.24s. \label{5.37} \]

The remaining time is the next approximation to the free-fall time.

\[t_{1} = T - \frac{h_{0}}{c_{s}} ≈ 3.76s. \label{5.38} \]

In that time, the rock falls a distance \(gt^{2}_{1}/2\), so the next approximation to the depth is

\[h_{1} = \frac{1}{2}gt^{2}_{1} ≈ 70.87 m. \label{5.39} \]

Question

Is this approximate depth an overestimate or underestimate? How accurate is it?

The calculation of \(h_{1}\) used \(h_{0}\) to estimate the sound-travel time. Because h0 overestimates the depth, the procedure overestimates the sound-travel time and, by the same amount, underestimates the free-fall time. Thus \(h_{1}\) underestimates the depth. Indeed, \(h_{1}\) is slightly smaller than the true depth of roughly 71.56 m but by only 1.3%.

The method of successive approximation has several advantages over solving the quadratic formula exactly. First, it helps us develop a physical understanding of the system; we realize, for example, that most of the \(T = 4s\) is spent in free fall, so the depth is roughly \(gT^{2}/2\). Second, it has a pictorial explanation (Problem 5.34). Third, it gives a sufficiently accurate answer quickly. If you want to know whether it is safe to jump into the well, why calculate the depth to three decimal places?

Finally, the method can handle small changes in the model. Maybe the speed of sound varies with depth, or air resistance becomes important (Problem 5.32). Then the brute-force, quadratic-formula method fails. The quadratic formula and the even messier cubic and the quartic formulas are rare closed-form solutions to complicated equations. Most equations have no closed-form solution. Therefore, a small change to a solvable model usually produces an intractable model if we demand an exact answer. The method of successive approximation is a robust alternative that produces low-entropy, comprehensible solutions.

Multiple problems

Problem 5.31 Parameter-value inaccuracies

What is \(h_{2}\), the second approximation to the depth? Compare the error in \(h_{1}\) and \(h_{2}\) with the error made by using \(g = 10ms^{-2}\).

Problem 5.32 Effect of air resistance

Roughly what fractional error in the depth is produced by neglecting air resistance (Section 2.4.2)? Compare this error to the error in the first approximation \(h_{1}\) and in the second approximation \(h_{2}\) (Problem 5.31).

Problem 5.33 Dimensionless form of the well-depth analysis

Even the messiest results are cleaner and have lower entropy in dimensionless form. The four quantities \(h, g, T\), and \(c_{s}\) produce two independent dimensionless groups (Section 2.4.1). An intuitively reasonable pair are

\[\bar{h} ≡ \frac{h}{gT^{2}} \text{ and } \bar{T} ≡ \frac{gT}{c_{s}}. \label{5.40} \]

a. What is a physical interpretation of \(\bar{T}\)?

b. With two groups, the general dimensionless form is \(\bar{h}\) = f(\(\bar{T}\)). What is \(\bar{h}\) in the easy case \(\bar{T}\) → 0?

c. Rewrite the quadratic-formula solution

\[h = (\frac{-\sqrt{2/g} + \sqrt{2/g + 4T/c_{s}}}{2/c_{s}})^{2}. \label{5.41} \]

as \(\bar{h} = f(\bar{T}\)). Then check that \(f(\bar{T}\)) behaves correctly in the easy case \(\bar{T} \rightarrow 0\).

Problem 5.34 Spacetime diagram of the well depth

How does the spacetime diagram [44] illustrate the successive approximation of the well depth? On the diagram, mark \(h_{0}\) (the zeroth approximation to the depth), \(h_{1}\), and the exact depth h. Mark \(t_{0}\), the zeroth approximation to the free-fall time. Why are portions of the rock and sound-wavefront curves dotted? How would you redraw the diagram if the speed of sound doubled? If g doubled?