Corrected practice final exam: waves and modern physics, with heat and climate (203-SN3-RE)

$\omega_{0}=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{1.6\times 10^{4}}{40}}=20$ rad/s.

$f_{0}=\dfrac{\omega_{0}}{2\pi}=3.18$ Hz. Confusing $\omega$ with $f$ loses a factor of $2\pi$, the most common error in the chapter.

The unbalanced laundry drives the tub at the rotation frequency of the drum: resonance occurs when it equals $3.18$ revolutions per second, that is $3.18\times 60=191$ rpm.

$v_{max}=A\omega_{0}=0.0050\times 20=0.10$ m/s and $a_{max}=A\omega_{0}^{2}=0.0050\times 400=2.0$ m/s².

At resonance, the amplitude would be limited only by damping and could become dangerous. By passing quickly through the range around $191$ rpm on the way to $1000$ rpm or more, the drive does not stay at the natural frequency long enough for the amplitude to build up.

Wave number $k=2.5$ rad/m: $\lambda=\dfrac{2\pi}{k}=2.51$ m.

Angular frequency $\omega=40$ rad/s: $f=\dfrac{\omega}{2\pi}=6.37$ Hz.

$v=\dfrac{\omega}{k}=\dfrac{40}{2.5}=16$ m/s. The form $\sin(kx-\omega t)$ travels toward POSITIVE $x$: a constant phase requires $x$ to increase as $t$ increases.

$v_{y}=\dfrac{\partial y}{\partial t}=-0.020\times 40\cos(2.5x-40t)$, with maximum $A\omega=0.80$ m/s, twenty times smaller than the wave speed: two speeds that must never be confused.

$v=\sqrt{\dfrac{T}{\mu}}$, so $T=\mu v^{2}=0.050\times 16^{2}=12.8$ N.

Fundamental of a string fixed at both ends: $L=\dfrac{\lambda_{1}}{2}$, so $\lambda_{1}=0.660$ m and $v=\lambda_{1}f_{1}=0.660\times 440=290$ m/s.

$T=\mu v^{2}=0.60\times 10^{-3}\times 290.4^{2}=50.6$ N. The linear mass density must be in kg/m.

$f_{3}=3f_{1}=1320$ Hz.

The FREQUENCY is conserved in passing from the string to the air, while the wavelength changes with the speed: $\lambda_{air}=\dfrac{343}{440}=0.780$ m, whereas it is $0.660$ m on the string.

The tension and the linear mass density do not change, so neither does $v$; the vibrating length becomes $\dfrac{2}{3}L$: $f=\dfrac{v}{2\cdot\frac{2}{3}L}=\dfrac{3}{2}\times 440=660$ Hz, a perfect fifth above.

$d=\dfrac{1\ \mathrm{mm}}{600}=1.667\times 10^{-6}$ m $=1.667$ µm.

$\sin\theta=\dfrac{486.1\times 10^{-9}}{1.667\times 10^{-6}}=0.2917$, so $\theta=17.0°$.

$\sin\theta=\dfrac{656.3\times 10^{-9}}{1.667\times 10^{-6}}=0.3938$, so $\theta=23.2°$. Red is deviated more than blue: the opposite of a prism.

We need $\sin\theta=\dfrac{m\lambda}{d}\le 1$, that is $m\le\dfrac{d}{\lambda}=2.54$: order $2$ is the last one.

$\sin\theta=\dfrac{2\times 656.3\times 10^{-9}}{1.667\times 10^{-6}}=0.7876$, so $\theta=52.0°$. The higher the order, the farther apart the lines are: the dispersion increases.

$\theta_{min}=1.22\times\dfrac{550\times 10^{-9}}{3.0\times 10^{-3}}=2.24\times 10^{-4}$ rad, or about $0.8$ arcminute.

For small angles, $\theta=\dfrac{s}{L}$: $L=\dfrac{1.5}{2.24\times 10^{-4}}=6.7\times 10^{3}$ m $=6.7$ km. Beyond that, the two headlights merge into a single blob.

$\theta_{min}=1.22\times\dfrac{550\times 10^{-9}}{0.200}=3.36\times 10^{-6}$ rad, sixty-seven times better than the eye.

$s=L\theta=3.84\times 10^{8}\times 3.36\times 10^{-6}=1.29\times 10^{3}$ m $=1.3$ km.

The limit comes from DIFFRACTION by the aperture: each point of the object gives a spot whose angular size is set by $\dfrac{\lambda}{D}$. Magnifying enlarges the spots as much as their separation, without separating them any further. Only a larger diameter, or a shorter wavelength, improves the resolution.

$I=400\cos^{2}60°=400\times 0.25=100$ W/m².

$\cos 90°=0$: no light at all, the screen looks black.

First filter at $45°$: $400\cos^{2}45°=200$ W/m², now polarized at $45°$. The glasses make an angle of $90°-45°=45°$ with this new polarization: $200\cos^{2}45°=100$ W/m².

In unpolarized light, all directions are equally present and the average of $\cos^{2}\theta$ is $\dfrac{1}{2}$: $150$ W/m².

A polarizer does not merely BLOCK part of the light: it REDEFINES the polarization of what it lets through. The intermediate filter rotates the polarization by $45°$, which gives the glasses a nonzero component to transmit. Without it, the polarization of the screen was exactly perpendicular to the glasses.

$f_{1}=\dfrac{3.00\times 10^{8}}{300\times 10^{-9}}=10.0\times 10^{14}$ Hz and $f_{2}=\dfrac{3.00\times 10^{8}}{400\times 10^{-9}}=7.50\times 10^{14}$ Hz.

Subtracting the two relations, $W$ drops out: $e(V_{s1}-V_{s2})=h(f_{1}-f_{2})$, so $h=\dfrac{1.60\times 10^{-19}\times 1.03}{2.50\times 10^{14}}=6.59\times 10^{-34}$ J·s, within $0.5$ % of the accepted value. This is the slope of the line $V_{s}(f)$, multiplied by $e$.

$W=hf_{1}-eV_{s1}=6.59\times 10^{-34}\times 1.00\times 10^{15}-1.60\times 10^{-19}\times 1.87=3.60\times 10^{-19}$ J $=2.25$ eV.

$\lambda_{0}=\dfrac{hc}{W}=\dfrac{6.59\times 10^{-34}\times 3.00\times 10^{8}}{3.60\times 10^{-19}}=5.49\times 10^{-7}$ m $=549$ nm: beyond that, in the yellow and red, no more electrons are emitted.

It does not change: $V_{s}$ measures the MAXIMUM kinetic energy of the electrons, which is set by the frequency. Doubling the intensity doubles the number of photons, and therefore the saturation current, not the energy of each electron.

$E=\dfrac{1240}{97.3}=12.74$ eV.

$E_{n}=-13.6+12.74=-0.86$ eV, so $n^{2}=\dfrac{13.6}{0.86}\approx 16$ and $n=4$. Check: $13.6\left(1-\dfrac{1}{16}\right)=12.75$ eV.

The possible transitions among levels $1$ to $4$ are $4\to 3$, $4\to 2$, $4\to 1$, $3\to 2$, $3\to 1$, $2\to 1$: six lines, $\dfrac{4\times 3}{2}$ pairs of levels.

The longest wavelength corresponds to the smallest energy gap, $4\to 3$: $\Delta E=13.6\left(\dfrac{1}{9}-\dfrac{1}{16}\right)=0.661$ eV and $\lambda=\dfrac{1240}{0.661}=1.88\times 10^{3}$ nm.

$1876$ nm is in the INFRARED: it is the first line of the Paschen series. An atom illuminated with ultraviolet light can re-emit in the infrared and in the visible (the $3\to 2$ transition gives the $656$ nm red line).

$\lambda=\dfrac{\ln 2}{T_{1/2}}=\dfrac{0.693}{12.3\times 3.156\times 10^{7}}=1.79\times 10^{-9}$ s⁻¹. The half-life must be in SECONDS for an activity in becquerels.

$A=\lambda N$, so $N=\dfrac{3.7\times 10^{11}}{1.79\times 10^{-9}}=2.07\times 10^{20}$ nuclei.

$m=2.07\times 10^{20}\times 3.016\times 1.6605\times 10^{-27}=1.04\times 10^{-6}$ kg $=1.04$ mg: about one milligram is enough.

$\dfrac{A}{A_{0}}=2^{-t/T_{1/2}}=2^{-20/12.3}=0.324$, that is $32$ %.

One quarter corresponds to two half-lives: $24.6$ years. It can be read off without a logarithm, because $\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^{2}$.

$\Phi=\dfrac{kA\,\Delta T}{e}=\dfrac{0.033\times 0.80\times 30}{0.030}=26.4$ W.

In one hour, $26.4\times 3600=9.50\times 10^{4}$ J enters, which melts $\dfrac{9.50\times 10^{4}}{3.34\times 10^{5}}=0.285$ kg $=285$ g.

$\dfrac{5.0}{0.285}=17.5$ h.

The thermal resistance doubles and the heat flow is halved: $13.2$ W, and the ice lasts twice as long.

The energy that comes in goes into the CHANGE OF STATE, not into raising the temperature: as long as melting is not complete, the ice-water mixture stays at $0$ °C. This is what makes ice so effective, and it is also what keeps the $30$ K difference constant in the calculation.

The flux varies as $\dfrac{L}{d^{2}}$: $S=1361\times\dfrac{0.5}{0.80^{2}}=1063$ W/m².

$T_{e}=\left(\dfrac{(1-0.35)\times 1063}{4\sigma}\right)^{1/4}=235$ K.

$235-273=-38$ °C.

$T_{s}=1.189\times 235=279$ K, that is $6$ °C.

Yes: between $0$ and $100$ °C under sufficient pressure, water can be liquid. It is the greenhouse effect, and not the distance to the star alone, that decides habitability here: with no atmosphere, the same planet would be frozen.

$\Delta F_{2\times}=5.35\ln 2=3.71$ W/m².

$\Delta T_{eq}=0.80\times 3.71=3.0$ K.

$\Delta F=5.35\ln\dfrac{420}{280}=2.17$ W/m².

$\Delta T_{eq}=0.80\times 2.17=1.7$ K.

Two main reasons. The thermal inertia of the ocean: the climate takes years to centuries to reach equilibrium, and part of the warming due to past emissions is still to come. Pollution aerosols, which raise the albedo and exert a negative forcing, mask part of the effect of the greenhouse gases. Other gases (methane), on the other hand, add their forcing to that of CO₂.