Corrected practice final exam: waves, optics and modern physics (203-SN3-RE, formerly 203-NYC)

$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{100}{0.25}}=\sqrt{400}=20.0\ \mathrm{rad/s}$. Then $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{20}=0.314\ \mathrm{s}$ and $f=\dfrac{1}{T}=3.18\ \mathrm{Hz}$.

$v_{\max}=\omega A=20.0\times 0.080=1.60\ \mathrm{m/s}$, reached when passing through the equilibrium position. And $a_{\max}=\omega^{2}A=400\times 0.080=32.0\ \mathrm{m/s^{2}}$, reached at the extreme positions, where the restoring force is greatest.

$E=\dfrac{1}{2}kA^{2}=0.5\times 100\times 0.080^{2}=0.32\ \mathrm{J}$. The graph does show the horizontal line at $0.32\ \mathrm{J}$: the total energy is constant, which reflects the absence of friction.

At $x=0.040\ \mathrm{m}$: $E_{p}=\dfrac{1}{2}kx^{2}=0.5\times 100\times 0.040^{2}=0.080\ \mathrm{J}$, which is a quarter of the total energy, since $E_{p}$ varies as the square of $x$. So $E_{k}=0.32-0.080=0.24\ \mathrm{J}$, three quarters. The speed follows from $v=\sqrt{\dfrac{2E_{k}}{m}}=\sqrt{\dfrac{0.48}{0.25}}=1.39\ \mathrm{m/s}$, which the formula $v=\omega\sqrt{A^{2}-x^{2}}=20\sqrt{0.0048}=1.39\ \mathrm{m/s}$ confirms. At half amplitude, the block therefore still has $87$ percent of its maximum speed.

On the graph, the two curves cross at the two marked points, at $0.16\ \mathrm{J}$, which is half the total energy. By calculation: $\dfrac{1}{2}kx^{2}=\dfrac{1}{2}kA^{2}-\dfrac{1}{2}kx^{2}$ gives $2x^{2}=A^{2}$, so $x=\pm\dfrac{A}{\sqrt{2}}=\pm 0.0566\ \mathrm{m}$. This is not at half amplitude: the energy is shared equally at about $71$ percent of the amplitude, because the potential energy is quadratic in $x$, not linear.

$T=2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{0.80}{9.81}}=2\pi\times 0.2856=1.79\ \mathrm{s}$. This expression does not contain the mass: the period of a simple pendulum is independent of it, because the driving force and the inertia are both proportional to the mass. It is also independent of the amplitude, but only in the small-oscillation approximation, where $\sin\theta\approx\theta$.

From $T=2\pi\sqrt{\dfrac{L}{g}}$ we get $L=\dfrac{gT^{2}}{4\pi^{2}}=\dfrac{9.81\times 4.0}{4\pi^{2}}=\dfrac{g}{\pi^{2}}=0.994\ \mathrm{m}$. A pendulum about one metre long therefore beats seconds, each swing in one direction lasting one second.

The amplitude is divided by two in $12\ \mathrm{s}$, so $\mathrm{e}^{-12\gamma}=\dfrac{1}{2}$, that is $12\gamma=\ln 2$ and $\gamma=\dfrac{\ln 2}{12}=5.78\times 10^{-2}\ \mathrm{s^{-1}}$.

Solve $\mathrm{e}^{-\gamma t}=0.10$, that is $t=\dfrac{\ln 10}{\gamma}=\dfrac{2.303}{5.78\times 10^{-2}}=39.9\ \mathrm{s}$. The number of oscillations is $\dfrac{39.9}{1.79}=22.2$, so about 22 complete oscillations. Note that the damping hardly changes the period: the pendulum swings at the same rhythm, and only its amplitude shrinks.

The mechanical energy of an oscillator is proportional to the square of the amplitude. If the amplitude is divided by two, the energy is divided by four: a quarter remains, so the pendulum has lost three quarters of its energy, or 75 percent. It is not one half precisely because the relation is quadratic. The same reasoning explains why an amplitude down to 10 percent corresponds to an energy down to 1 percent.

$v=\sqrt{\dfrac{F}{\mu}}=\sqrt{\dfrac{80}{0.020}}=\sqrt{4000}=63.2\ \mathrm{m/s}$. The wavelength is $\lambda=\dfrac{v}{f}=\dfrac{63.2}{120}=0.527\ \mathrm{m}$. The wave speed depends only on the medium, tension and linear mass density, never on the source.

$k=\dfrac{2\pi}{\lambda}=\dfrac{2\pi}{0.527}=11.9\ \mathrm{rad/m}$ and $\omega=2\pi f=2\pi\times 120=754\ \mathrm{rad/s}$. The equation is $y(x, t)=5.0\times 10^{-3}\sin\left(754t-11.9x\right)$, in metres, the minus sign in front of $x$ expressing propagation toward increasing $x$.

The transverse velocity is $\dfrac{\partial y}{\partial t}$, whose amplitude is $v_{\perp\max}=\omega A=754\times 5.0\times 10^{-3}=3.77\ \mathrm{m/s}$. It is well below the wave speed $63.2\ \mathrm{m/s}$, and above all it is of a completely different nature: the wave speed is the speed at which the pattern moves along the string, whereas the transverse velocity is that of a piece of material, perpendicular to the string. No material moves at $63.2\ \mathrm{m/s}$: it is the wave that advances, not the string.

$P=\dfrac{1}{2}\times 0.020\times 63.2\times 754^{2}\times\left(5.0\times 10^{-3}\right)^{2}=0.632\times 5.68\times 10^{5}\times 2.5\times 10^{-5}=8.99\ \mathrm{W}$.

If the tension is doubled, $v=\sqrt{\dfrac{F}{\mu}}$ is multiplied by $\sqrt{2}$. Since $P$ is proportional to $v$, at fixed frequency and amplitude, the power is multiplied by only $\sqrt{2}\approx 1.41$. If the amplitude is doubled, $P$ being proportional to $A^{2}$, it is multiplied by 4. The difference comes from the exponent: the amplitude enters squared, the wave speed to the first power, and the wave speed itself depends on the tension only through a square root. To carry more power, working on the amplitude is therefore far more effective than tightening the string.

From $L=90$ we get $\dfrac{I}{I_{0}}=10^{9}$, so $I=10^{-12}\times 10^{9}=1.0\times 10^{-3}\ \mathrm{W/m^{2}}$. The power crosses a sphere of radius $1.0\ \mathrm{m}$, so $P=I\times 4\pi r^{2}=1.0\times 10^{-3}\times 4\pi=1.26\times 10^{-2}\ \mathrm{W}$. A source of only 13 milliwatts of acoustic power is therefore enough to produce 90 decibels at one metre.

The intensity falls off as the inverse square of the distance: at $10\ \mathrm{m}$ it is divided by $100$, so $I=1.0\times 10^{-5}\ \mathrm{W/m^{2}}$. The level is $L=10\log\left(10^{7}\right)=70\ \mathrm{dB}$. It has therefore dropped by $20\ \mathrm{dB}$: multiplying the distance by ten costs exactly 20 decibels, whatever the source.

Two identical sources double the intensity, so $L=10\log\left(2\times 10^{9}\right)=90+10\log 2=93.0\ \mathrm{dB}$. With ten sources, the intensity is multiplied by ten and $L=90+10=100\ \mathrm{dB}$. Doubling the number of sources adds 3 decibels, multiplying it by ten adds 10.

$120\ \mathrm{dB}$ corresponds to $I=1.0\ \mathrm{W/m^{2}}$. Solve $\dfrac{P}{4\pi r^{2}}=1.0$, that is $r^{2}=\dfrac{1.26\times 10^{-2}}{4\pi}=1.0\times 10^{-3}$, hence $r=3.2\times 10^{-2}\ \mathrm{m}$, about $3.2\ \mathrm{cm}$. You would have to put your ear against the source, which explains the danger of earphones.

Because the decibel scale is logarithmic, not proportional. The level measures the logarithm of the ratio of intensities: doubling the intensity adds $10\log 2\approx 3\ \mathrm{dB}$, a constant amount, not a factor of two. This scale was chosen because the sensation of hearing roughly follows the logarithm of the intensity: a gain of 10 decibels, a factor of ten in intensity, is perceived as a sound about twice as loud.

Source moving toward the observer: $f'=f\dfrac{v}{v-v_{s}}=800\times\dfrac{343}{343-25}=800\times\dfrac{343}{318}=863\ \mathrm{Hz}$. Source moving away: $f'=800\times\dfrac{343}{368}=746\ \mathrm{Hz}$.

The difference is $863-746=117\ \mathrm{Hz}$. It is perfectly audible: it amounts to about two and a half semitones of the musical scale, and it is exactly the jump in pitch you hear when an ambulance passes in front of you.

Observer moving toward a stationary source: $f'=f\dfrac{v+v_{o}}{v}=800\times\dfrac{368}{343}=858\ \mathrm{Hz}$. This is very close to $863\ \mathrm{Hz}$, but not equal. The two situations are therefore not equivalent, contrary to what the relativity of motion might suggest, and the reason is that there is a privileged medium of propagation here, the air: speeds count relative to it. When the source moves, it squeezes the wavelengths; when the observer moves, he simply meets the crests faster. The two effects are different, which gives different formulas.

$M=\dfrac{v_{\mathrm{plane}}}{v_{\mathrm{sound}}}=\dfrac{500}{343}=1.46$. The half-angle satisfies $\sin\mu=\dfrac{1}{M}=\dfrac{343}{500}=0.686$, so $\mu=43.3^{\circ}$, as shown on the figure.

No. Since the aircraft is faster than sound, it constantly stays ahead of the waves it emits: all the sound is confined inside the cone, behind the aircraft. An observer on the ground therefore hears absolutely nothing until the cone has reached them. They see the aircraft pass overhead in silence, then receive the sonic boom all at once as the surface of the cone sweeps over them. The sound does not come from the direction where the aircraft is at that moment, but from the position it occupied earlier.

A pipe open at both ends has a displacement antinode at each end, so $L=\dfrac{\lambda}{2}$ for the fundamental and $f_{1}=\dfrac{v}{2L}=\dfrac{343}{1.20}=286\ \mathrm{Hz}$. All the harmonics exist: $f_{2}=572\ \mathrm{Hz}$ and $f_{3}=858\ \mathrm{Hz}$.

With one closed end, there must be a displacement node at the closed end and an antinode at the open end, so $L=\dfrac{\lambda}{4}$ and $f_{1}=\dfrac{v}{4L}=\dfrac{343}{2.40}=143\ \mathrm{Hz}$, an octave lower than the open pipe of the same length. Only the odd harmonics exist: the next ones are $f_{3}=429\ \mathrm{Hz}$ and $f_{5}=715\ \mathrm{Hz}$. It is this absence of even harmonics that gives the closed pipe its hollower tone.

The beat frequency is the absolute value of the difference: $\left|288.0-285.8\right|=2.17\ \mathrm{Hz}$. You therefore hear the intensity vary about twice per second.

The string must go from $288.0$ to $285.83\ \mathrm{Hz}$. Since $f\propto\sqrt{F}$, we have $\dfrac{F'}{F}=\left(\dfrac{f'}{f}\right)^{2}=\left(\dfrac{285.83}{288.0}\right)^{2}=0.985$. The string must therefore be slackened by about $1.5$ percent. Note that a frequency difference of $0.75$ percent calls for a tension adjustment twice as large, the square root damping the effect.

At $35\ \mathrm{^{\circ}C}$, $v=331+0.6\times 35=352\ \mathrm{m/s}$. The open pipe then rises to $f_{1}=\dfrac{352}{1.20}=293\ \mathrm{Hz}$. The string having stayed at $288.0\ \mathrm{Hz}$, the new beat frequency is $\left|288.0-293.3\right|=5.33\ \mathrm{Hz}$, more than double. This is exactly the problem in orchestras: wind instruments go sharp as the room warms up, since their pitch follows the speed of sound in air, whereas strings tend to go flat as they slacken. The two families therefore drift out of tune in opposite directions, which is why you retune after the instruments have warmed up, not before.

$\dfrac{1}{d_{i}}=\dfrac{1}{15}-\dfrac{1}{20}=\dfrac{4-3}{60}=\dfrac{1}{60}$, so $d_{i}=+60\ \mathrm{cm}$. The magnification is $m=-\dfrac{60}{20}=-3.0$. The image is real since $d_{i}>0$, inverted since $m<0$, and three times larger than the object.

$\dfrac{1}{d_{i}}=\dfrac{1}{15}-\dfrac{1}{10}=\dfrac{2-3}{30}=-\dfrac{1}{30}$, so $d_{i}=-30\ \mathrm{cm}$. Then $m=-\dfrac{-30}{10}=+3.0$. The image is virtual since $d_{i}<0$, upright since $m>0$, and three times larger: this is how a magnifying glass works. The object is here between the lens and its focal point.

$\dfrac{1}{d_{i}}=\dfrac{1}{15}-\dfrac{1}{15}=0$, so $d_{i}$ tends to infinity: the rays emerge parallel and there is no image at a finite distance. This is the collimation position, used in every collimator and at the input of telescopes set to infinity.

Only case a). A real image is formed by rays that actually converge at a point, so a screen can be placed there and the image seen on it. In case b), the emerging rays diverge and cross nowhere: their intersection exists only on their extensions behind the lens. A virtual image can be seen only by an eye or another instrument, never caught on a screen.

The size of the image is $\left|m\right|\times h=3.0\times 2.0=6.0\ \mathrm{cm}$. Since the magnification is negative, the image is inverted: it is $6.0\ \mathrm{cm}$ tall and points down if the object points up.

Total internal reflection occurs when light goes from the more refractive medium to the less refractive one, here from water to air. The critical angle satisfies $n\sin\theta_{c}=1\times\sin 90^{\circ}$, so $\sin\theta_{c}=\dfrac{1}{1.33}=0.752$ and $\theta_{c}=48.8^{\circ}$.

Beyond $48.8^{\circ}$ from the vertical, the surface behaves like a perfect mirror: the diver no longer sees the sky there but the reflection of the bottom. The whole outside world appears squeezed into a cone of half-angle $48.8^{\circ}$ above their head, a phenomenon called Snell's window, and the entire horizon is concentrated on the edge of this circle.

In the prism, the relation between the internal angles is $r+r'=A$. At minimum deviation the path of the ray is symmetric, so $r=r'$, which gives $2r=A$ and $r=\dfrac{A}{2}=30.0^{\circ}$. Snell's law at entry then gives $\sin i=n\sin r=1.52\times\sin 30.0^{\circ}=1.52\times 0.500=0.760$, hence $i=49.5^{\circ}$.

The deviation is $D=i+i'-A$, and by symmetry $i'=i$ at the minimum, so $D_{\min}=2i-A=2\times 49.46-60.0=38.9^{\circ}$.

For blue, $\sin i=1.53\times 0.500=0.765$ gives $i=49.91^{\circ}$ and $D_{\min}=39.8^{\circ}$. For red, $\sin i=1.51\times 0.500=0.755$ gives $i=49.03^{\circ}$ and $D_{\min}=38.1^{\circ}$. The separation is $39.81-38.05=1.76^{\circ}$. Blue is deviated the most, because the index of glass grows as the wavelength decreases. This dispersion is what spreads white light into a spectrum, and it is also what, in raindrops, puts red on the outside of the rainbow and violet on the inside.

The fringe spacing is $i=\dfrac{\lambda D}{d}=\dfrac{589\times 10^{-9}\times 2.0}{0.25\times 10^{-3}}=4.71\times 10^{-3}\ \mathrm{m}$, that is $4.71\ \mathrm{mm}$.

The grating spacing is $d=\dfrac{1\ \mathrm{mm}}{600}=1.667\times 10^{-6}\ \mathrm{m}$. The first order satisfies $\sin\theta_{1}=\dfrac{\lambda}{d}=\dfrac{589\times 10^{-9}}{1.667\times 10^{-6}}=0.353$, so $\theta_{1}=20.7^{\circ}$. Orders exist as long as $\sin\theta=\dfrac{m\lambda}{d}\leq 1$, that is $m\leq\dfrac{d}{\lambda}=2.83$: orders 1 and 2 are therefore observed on each side, and order 3 is out of reach.

The first diffraction minimum satisfies $\sin\theta=\dfrac{\lambda}{a}=\dfrac{589\times 10^{-9}}{0.10\times 10^{-3}}=5.89\times 10^{-3}$, so $\theta=0.337^{\circ}$, a very small angle. The central maximum extends on both sides, and its total width on the screen is $\dfrac{2\lambda D}{a}=2\times 2.0\times 5.89\times 10^{-3}=2.36\times 10^{-2}\ \mathrm{m}$, that is $23.6\ \mathrm{mm}$.

An interference order is missing when it falls exactly on a diffraction minimum, that is when $\dfrac{m\lambda}{d}=\dfrac{n\lambda}{a}$, or $m=n\dfrac{d}{a}=2.5\,n$. For $n=1$ this gives $2.5$, which is not an integer, so no order is lost. For $n=2$ we get $m=5$, and for $n=4$ we get $m=10$. The missing orders are therefore orders 5 and 10: the diffraction envelope erases these interference fringes, which exist in principle but receive zero intensity.

Young's fringes are broad and sinusoidal: the intensity varies gently, so pinpointing the centre of a maximum is imprecise. A grating of 600 lines makes not two but thousands of coherent sources interfere, and all of them must be in phase to give a maximum: the slightest change of angle destroys the agreement and the intensity collapses. The maxima are therefore extremely sharp, which allows their position, and hence $\lambda$, to be measured with far better precision. It is the same principle as the Wheatstone bridge: measuring by looking for a sharp condition is always more precise than reading a slow variation.

$E=\dfrac{1240}{400}=3.10\ \mathrm{eV}$.

Einstein's relation gives $E_{k,\max}=E-W=3.10-2.30=0.80\ \mathrm{eV}$. The stopping potential is the voltage that blocks even the fastest electrons: $eU_{0}=E_{k,\max}$, so $U_{0}=0.80\ \mathrm{V}$. Numerically, the kinetic energy in electronvolts and the stopping potential in volts have the same value, which is the very convenience of the electronvolt.

The threshold corresponds to $E=W$, so $\lambda_{0}=\dfrac{1240}{2.30}=539\ \mathrm{nm}$. Any light of longer wavelength, hence redder than yellow-green, ejects no electron, whatever its intensity.

$E_{k,\max}=0.80\ \mathrm{eV}=0.80\times 1.602\times 10^{-19}=1.28\times 10^{-19}\ \mathrm{J}$. Then $v_{\max}=\sqrt{\dfrac{2E_{k,\max}}{m_{e}}}=\sqrt{\dfrac{2.563\times 10^{-19}}{9.109\times 10^{-31}}}=5.30\times 10^{5}\ \mathrm{m/s}$, about one five-hundredth of the speed of light, which justifies the nonrelativistic calculation.

Doubling the intensity doubles the number of photons per second, hence doubles the number of emitted electrons and so the current. But the maximum kinetic energy does not change at all, because it depends only on the energy of an individual photon, set by the wavelength. This is precisely what the wave model could not explain: in that model, a more intense wave carries more energy and should therefore give more energy to each electron, and a weak wave should eventually eject electrons after a time of accumulation. Experiment shows the opposite: a frequency threshold, no delay, and an energy independent of the intensity. Light had to be admitted to exchange its energy in grains, one photon for one electron.