Phys 239
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Quantitative Physics
Problem Set 5 Solutions
Waves a’Comin’
Imagine a storm at sea, 2000 km distant. Sustained winds at 10 m/s mostly aligned to point at your distant
coastline whip up waves 10 m peak-to-peak over a region about 50 km across. The waves barrel toward your
coastline, spreading out in a fan roughly 30 degrees in angle. Answer the following questions, assuming
wavelength (crest-to-crest) is not disturbed by the journey:
a. What will be the dominant wavelength?
b. How long will it take for the waves to reach your shore?
c. What will the peak-to-peak amplitude be at your shore before getting into shallow water?
d. What is the wave period before getting to shallow water?
e. What will the period be between breakers on the shore, if the waves break in water one meter deep?
Part a: Wind grabs crests and pushes them along until the phase velocity matches
√ the wind speed (at which
point no more force acts on the wave crest). So vph = 10m/s. Equating to gλ̄ results in λ̄ ≈ 10 m, or
λ ≈ 63 m.
Part b: Wave packets (amplitude envelopes, energy) travel at the group velocity, which is one half the phase
velocity in deep water. So the energy travels at 5 m/s and will cover 2 × 106 m in 4 × 105 s, which is about
4.5 days.
Part c: If the fan of wave energy opens at 30◦ , it will be 1000 km across after traveling 2000 km. Initially,
the storm zone was confined to 50 km. Assuming the envelope thickness is unchanged (for our narrow-band
consideration),
the area of wave energy is reduced by a factor of 20. Since wave energy per area scales as
2
ζ , where ζ measures vertical displacements, we have that ζ 2 is down by a factor of 20, so that ζ is down by
√
a factor of 20 ≈ 4.5. So what were 10 m peak-to-peak waves will now measure just over 2 m peak-to-peak.
Part d: The phase velocity is 10 m/s (speed of crests), while the wavelength is 63 m, so it takes 6.3 sec
between waves—and that’s the period you would record for passing crests.
√
Part e: In shallow water, the phase (and group) velocity is gh ≈ 3 m/s in 1-m-deep water. Assuming
nothing has happened to the wavelength (still 63 m), the period between breaks becomes about 21 sec.
Note that while the problem stipulated an invariant wavelength, this is not really what the water cares
about. As the wave gets into shallow water and the velocity slows, the waves bunch up, leaving period as
a better invariant. In this case, the 6.3 s period would be preserved and the wavelength reduces to 18 m in
1 m water.
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Juice Blob
An astronaut inadvertently spills their juice drink to find a wobbling blob of liquid suspended in mid-air.
Your job is to find the period of oscillation for the lowest (simplest) mode. Model this as a standing wave
traveling around the blob. Be careful to pick a mode that does not violate conservation of momentum for the
blob.
Let’s say the drink was 0.5 L in volume. In spherical form, this is conveniently R = 50 mm in radius.
Modeling as a surface wave, we must satisfy the boundary condition that the wave meet up with itself after
a full circle, so that nλ = 2πR, where n is some integer. We want one wavelength to go halfway around the
blob, or n = 2. Why not the simpler n = 1, going all the way around? Imagine at some snapshot in time
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that the top pole is adding to the mean radius, so that the bottom pole subtracts from the mean radius.
This shifts the center of mass of the blob “up.” Half-a-cycle later the amplitudes are reversed, so that the
blob center of mass is “down.” Without external periodic forcing, this cannot happen. So n = 2 is the
lowest-order mode that does not violate momentum conservation.
We therefore look for a wavelength λ = πR, or λ̄ = R/2. We are dealing with some sort of surface-tension
wave, since gravity clearly has no command over the fluid. A sphere has the lowest surface area for a given
volume, so it is the lowest-energy state given surface tension. Are we in deep water or shallow water? What
does it even mean? I guess we’d say the radius is the depth. Looking at the master equation for waves:
d
γ
g
+ 3 tanh
,
ω2 =
λ̄ ρλ̄
λ̄
we can ignore g (in orbit), set the depth d = R, and λ̄ = R/2. Evaluating the hyperbolic tangent, we learn
that its argument is 2, and tanh 2 ≈ 0.98, so we’re
q well-enough out of the shallow regime, which may come
as a surprise to most. We are left with ω = γ/ρλ̄3 in the deep capillary regime, computing to about
2 rad/s. The period is 2π/ω, or about 3 seconds. So we imagine a pulsation period of 3 seconds, which
seems reasonably well-matched to videos of sloppy astronauts, if not a little slow. Incidentally, we get the
same answer if we ask how long it takes to travel pole-to-pole (a full cycle) at vph .
Now is a spherical geometry an appropriate application of the familiar flat-water dispersion relations? Probably not. But I would not expect radically different answers if done correctly.
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Water as a Gas?
Water acts sort-of like a gas, but more dense and more constrained. Go through a similar development as in
class to determine a) the number density of water molecules; b) the typical linear dimension between adjacent
molecules; c) the mean free path if the typical diameter is 2 Å; d) the approximate diffusion constant for a
molecule of water; and e) how long it would take for a millimeter migration of a molecule in otherwise still
water. Note, this exercise is offensive to the true complexity of liquid molecular interactions, but still offers
some insight (much like a hobo might be offensive to the nose, yet have good advice). The 2 Å figure, for
one, is made up by me and highly suspect.
Part a: Water has a molar mass of 0.018 kg/mol, so that a liter contains 56 mol, and a cubic meter contains
55,555 mol (ignore significant digits: just fun to write that many fives). Multiply by Avogadro’s number to
get a space number density of n ≈ 3.3 × 1028 m−3 .
Part b: Each molecule then occupies a cube volume 1/n ≈ 3 × 10−29 m, which has associated side length of
(30)1/3 Å, or 3 Å.
Part c: We use the result that λ = 1/nσ, and use the notion that the radius of a water molecule is 1 Å, so
that σ = π(2R)2 ≈ 1.3 × 10−19 m2 , making λ ≈ 2 Å. This is comparable to the atomic size (wedged in).
Incidentally, the mean free path, so-calculated, crosses the radius if both are 1.7 Å.
p
Pard d: D = 13 λv, and we have λ so need v = 3kT /m ≈ 640 m/s for water. Incidentally, if you remember
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kT ≈ 40
eV and that a proton/neutron mass as about 940 MeV (water having 18 such nucleons), you can
carry out the computation in eV units, getting a result as a fraction of c. So we would get D ≈ 5×10−8 m2 /s.
Part e: ∆t ∼ ∆R2 /D, which comes to 20 seconds.
This should all be taken with a huge grain of salt. Water molecules are not like billiard balls bouncing off
each other (not interacting until they touch). They are sticky molecules, owing to hydrogen bonds. So it’s
really a jiggling mess.
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Sky Temperature via IR Camera
We saw and messed around with an infrared camera in lecture, operating in the 8–12 micron range. If you
point this particular camera at a clear sky, day or night, it hits its lowest operating readout of −20 C, so it
is clear that the effective IR temperature in this (mostly transparent) atmospheric band is colder than this.
Undeterred, and armed with high-emissivity tape (assume 1.0), you make the following measurements. A
direct sighting of your forehead reads 35 C. You see a reflection of your head in a glass table top on an
outdoor table and note your reflected forehead to measure 22 C, while a piece of tape adhered to the table
reads 20 C. Then you notice that the sky as reflected in the table reads 14 C. From this you conclude the
effective radiative temperature of the sky (in this band) to be X. What is X? Note that reflectivity is the
complement of emissivity (they add to unity).
Central to this discussion is that radiated power goes like
Prad = AεσT 4 ,
where ε characterizes emissivity. Shiny objects have low emissivity. Reflectance and emissivity are complementary, adding to unity. The glass in this case is mostly emissive, but partly reflective. When we have a mix
of sources, like a piece of glass with its own emission but also reflecting something of a different temperature,
we receive a power per area (or flux, F )
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Frecv = εσTsurf
+ (1 − )σTsource
.
Our measuring device does not know how infrared photons came to be present, and just associates the flux
with a temperature, assuming unit emissivity:
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Frecv = σTmeas
.
We can therefore equate these (σ is common to all), and can solve for emissivity, source temperature, etc.
So let’s break down the measurements and what we learn from them. A direct measurement of your forehead
yields a measured temperature of 35 C. A direct measurement of the tape on the glass (a high-emissivity
surface so we can understand the true temperature of the table/glass) yields 20 C. A measurement of your
reflected forehead gives 22 C. So in this reflected measurement scenario, we have Tsource = 35 C, Tsurf = 20 C,
and Tmeas = 22 C. We can now solve for emissivity:
ε=
4
4
− Tsource
Tmeas
,
4
4
Tsurf − Tsource
making sure to use Kelvin for all computations (!) to get ε ≈ 0.875. Armed with this, we can now solve for
Tsource in the sky-reflected case, concluding that Tsource ≈ 227 K, effectively (at 8–12 µm, and assuming unit
emissivity from the sky). This is −46 C, and well below the measurement limit of the camera (−20 C).
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