di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Mechanics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: A hollow sphere of mass $M$ and radius $R$ is placed under a plank of mass $3 M$ and length $2 R$. The plank is hinged to the floor, and it initially makes an angle $\theta=\frac{\pi}{3}$ rad to the horizontal. Under the weight of the plank, the sphere starts rolling without slipping across the floor. What is the sphere's initial translational acceleration? Assume the plank is frictionless.

<img_4325>

A not-to-scale picture of the sphere-plank setup.

Solution:
Let $N$ be the normal force acting from the ball on the plank. We can use torque at a distance $x=R \cot \frac{\theta}{2}$ from the hinge (by geometry) to write (using $m=3 M$ )
$$
m g \frac{l}{2} \cos \theta-N R \cot \frac{\theta}{2}=\frac{m l^{2}}{3} \alpha
$$
You can also find $N$ from
$$
\begin{aligned}
N \sin \theta-f & =M a_{0} \\
f R & =\frac{2}{3} M R^{2} \alpha \\
a_{0} & =R \alpha
\end{aligned}
$$
You can relate the acceleration at the point of contact $a_{p}=\vec{a}_{0}+\vec{a}_{0} \times \vec{r}=\alpha_{0}(1-\cos \theta)$. Hence combining all equations gives $a=\frac{9}{32} g \approx 2.76 \mathrm{~m} / \mathrm{s}^{2}$.

Final answer: $2.76$ $\mathrm{~m} / \mathrm{s}^{2}$
Subfield: Mechanics


Question: A hollow sphere of mass $M$ and radius $R$ is placed under a plank of mass $3 M$ and length $2 R$. The plank is hinged to the floor, and it initially makes an angle $\theta=\frac{\pi}{3}$ rad to the horizontal. Under the weight of the plank, the sphere starts rolling without slipping across the floor. What is the sphere's initial translational acceleration? Assume the plank is frictionless.

![](https://cdn.mathpix.com/cropped/2023_12_21_7fcf501c85b9e74489f8g-1.jpg?height=263&width=616&top_left_y=188&top_left_x=749)

A not-to-scale picture of the sphere-plank setup.

Solution:
Let $N$ be the normal force acting from the ball on the plank. We can use torque at a distance $x=R \cot \frac{\theta}{2}$ from the hinge (by geometry) to write (using $m=3 M$ )
$$
m g \frac{l}{2} \cos \theta-N R \cot \frac{\theta}{2}=\frac{m l^{2}}{3} \alpha
$$
You can also find $N$ from
$$
\begin{aligned}
N \sin \theta-f & =M a_{0} \\
f R & =\frac{2}{3} M R^{2} \alpha \\
a_{0} & =R \alpha
\end{aligned}
$$
You can relate the acceleration at the point of contact $a_{p}=\vec{a}_{0}+\vec{a}_{0} \times \vec{r}=\alpha_{0}(1-\cos \theta)$. Hence combining all equations gives $a=\frac{9}{32} g \approx 2.76 \mathrm{~m} / \mathrm{s}^{2}$.

Final answer: $2.76$ $\mathrm{~m} / \mathrm{s}^{2}$
Subfield: Optics


Question: Consider a uniform isosceles triangle prism $\mathrm{ABC}$, with the apex angle $\theta=110^{\circ}$ at vertex $\mathrm{A}$. One of the sides, $\mathrm{AC}$, is coated with silver, allowing it to function as a mirror. When a monochrome light-ray of wavelength $\lambda$ approaches side $\mathrm{AB}$ at an angle of incidence $\alpha$, it first refracts, then reaches side AC, reflects, and continues to base BC. After another refraction, the ray eventually exits the prism at the angle of emergence which is also equal to the angle of incidence (see Fig. A).
![](https://cdn.mathpix.com/cropped/2023_12_21_da5107ae3cf532a9d068g-1.jpg?height=520&width=1732&top_left_y=1152&top_left_x=196)



What is the relative refractive index of the prism for that particular wavelength $\lambda$ with respect to the outside environment, given that $\alpha=70^{\circ}$.

Solution:
The light-path refracts on side $A B$ at point $M$, reflects on side $A C$ at point $N$ and refracts on base $\mathrm{BC}$ at point $\mathrm{P}$ (see Fig. A). Define the angle of refraction inside the prism to be $\beta$, then from Snell's law:
$$
\sin \alpha=n \sin \beta .
\tag{29}
$$

From the law of reflection and the $180^{\circ}$-sum of three interior angles inside any triangles:
$$
\begin{aligned}
\widehat{\mathrm{MNA}} & =180^{\circ}-\widehat{\mathrm{NAM}}-\widehat{\mathrm{AMN}}=180^{\circ}-\theta-\left(90^{\circ}-\beta\right) \\
& =\widehat{\mathrm{PNC}}=180^{\circ}-\widehat{\mathrm{NCP}}-\widehat{\mathrm{CPN}}=180^{\circ}-\left(\frac{180^{\circ}-\theta}{2}\right)-\left(90^{\circ}+\beta\right),
\end{aligned}
$$
we obtain the refraction angle $\beta$ to be:
$$
\beta=\frac{3 \theta-180^{\circ}}{4}
$$
Plug this finding into Eq. (29), we get the relative refraction index of the prism with respect to the outside environment:
$$
n=\frac{\sin \alpha}{\sin \beta}=\left.\frac{\sin \alpha}{\sin \left(\frac{3 \theta-180^{\circ}}{4}\right)}\right|_{\alpha=70^{\circ}, \theta=110^{\circ}} \approx 1.5436
$$

Final answer: 1.5436
Subfield: Mechanics


Question: 1. Restore equilibrium 

A wooden plank of length $1 \mathrm{~m}$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height $0.5 \mathrm{~m}$. The specific gravity of the plank (or ratio of plank density to water density) is 0.5 . Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position (exclude the case $\theta=0^{\circ}$ )

<img_4556>

Solution:
<img_4519>

One could find the net buoyant force by integrating about the point of reference. Or one can directly use the fact this upthrust force acts at the center of portion submerged in water.

A sample solution:

Three forces will act on the plank-

1. Weight which will act at centre of plank.


2. Upthrust which will act at centre of submerged portion.
3. Force from the hinge at $\mathrm{O}$.

Notation used: Submerged length $=0.5 \sec \theta, \mathrm{F}=$ Upthrust, $w=$ Weight

Taking moments of all three forces about point O. Moment of hinge force will be zero.

$$
(A \lg \rho) \frac{l}{2} \sin \theta=A(0.5 \sec \theta)\left(\rho_{w}\right)(g)\left(\frac{0.5 \sec \theta}{2}\right) \sin \theta
\tag{1}
$$

Solve to get $\theta=45^{\circ}$

Final answer: $\theta=45$ $^{\circ}$
Subfield: Mechanics


Question: 1. Restore equilibrium 

A wooden plank of length $1 \mathrm{~m}$ and uniform cross-section is hinged at one end to the bottom of a tank as shown in figure. The tank is filled with water up to a height $0.5 \mathrm{~m}$. The specific gravity of the plank (or ratio of plank density to water density) is 0.5 . Find the angle $\theta$ that the plank makes with the vertical in the equilibrium position (exclude the case $\theta=0^{\circ}$ )

![](https://cdn.mathpix.com/cropped/2023_12_21_97ebfc9f8f7a4705c9ebg-1.jpg?height=570&width=799&top_left_y=623&top_left_x=671)

Solution:
![](https://cdn.mathpix.com/cropped/2023_12_21_0c8d119676610ce41c3ag-1.jpg?height=534&width=808&top_left_y=1430&top_left_x=664)

One could find the net buoyant force by integrating about the point of reference. Or one can directly use the fact this upthrust force acts at the center of portion submerged in water.

A sample solution:

Three forces will act on the plank-

1. Weight which will act at centre of plank.


2. Upthrust which will act at centre of submerged portion.
3. Force from the hinge at $\mathrm{O}$.

Notation used: Submerged length $=0.5 \sec \theta, \mathrm{F}=$ Upthrust, $w=$ Weight

Taking moments of all three forces about point O. Moment of hinge force will be zero.

$$
(A \lg \rho) \frac{l}{2} \sin \theta=A(0.5 \sec \theta)\left(\rho_{w}\right)(g)\left(\frac{0.5 \sec \theta}{2}\right) \sin \theta
\tag{1}
$$

Solve to get $\theta=45^{\circ}$

Final answer: $\theta=45$ $^{\circ}$
Subfield: Mechanics

2. State secrets and the stars 

In 1949, the physicist G. I. Taylor used publicly released images of the Trinity nuclear test to estimate the yield of the world's first atomic bomb (i.e., the energy released by the detonation). His estimate was stunningly accurate, despite the fact that the number itself was still classified by the government at the time. In this problem, we will follow a simplified version of his estimate, and use the procedure to estimate the energy released by the supernovae which end the lives of massive stars.

<img_4341>

Figure 1: A snapshot of the Trinity blast wave 15 milliseconds after the explosion, reproduced from Taylor's original paper. A helpful scale bar is included.

Question: (a) After a very short moment, the shock front caused by an explosion sweeps up the material around it, increasing the mass which is blasted outwards. At early times, the energy $E \sim M v^{2} / 2$ can be thought of as roughly conserved.

Let $R$ be the radius of the explosion at some time $t$. Estimate the speed of the shock front as $v \sim R / t$ and the swept up mass $M \sim \rho R^{3}$, where $\rho$ is the density of the surrounding material (don't bother to include factors like $4 \pi / 3$ ).

Estimate the radius of the explosion $R(t)$ as a function of the energy yield $E$, elapsed time $t$, and the density of the surrounding medium $\rho$.

Solution:
Full points should be awarded to a solution as below up to order unity factors.

The energy of the explosion is $E=M v^{2} / 2 \sim M v^{2}$. Using $v \sim R / t$ and $M \sim \rho R^{3}$, the total energy of the explosion is

$$
E \sim \frac{\rho R^{5}}{t^{2}}
\tag{2}
$$

We can then rearrange this expression as

$$
R(t) \sim\left(\frac{E}{\rho}\right)^{1 / 5} t^{2 / 5}
\tag{3}
$$

Final answer: $\left(\frac{E}{\rho}\right)^{1 / 5} t^{2 / 5}$
Subfield: Mechanics

2. State secrets and the stars 

In 1949, the physicist G. I. Taylor used publicly released images of the Trinity nuclear test to estimate the yield of the world's first atomic bomb (i.e., the energy released by the detonation). His estimate was stunningly accurate, despite the fact that the number itself was still classified by the government at the time. In this problem, we will follow a simplified version of his estimate, and use the procedure to estimate the energy released by the supernovae which end the lives of massive stars.

<img_4341>

Figure 1: A snapshot of the Trinity blast wave 15 milliseconds after the explosion, reproduced from Taylor's original paper. A helpful scale bar is included.
Context question:
(a) After a very short moment, the shock front caused by an explosion sweeps up the material around it, increasing the mass which is blasted outwards. At early times, the energy $E \sim M v^{2} / 2$ can be thought of as roughly conserved.

Let $R$ be the radius of the explosion at some time $t$. Estimate the speed of the shock front as $v \sim R / t$ and the swept up mass $M \sim \rho R^{3}$, where $\rho$ is the density of the surrounding material (don't bother to include factors like $4 \pi / 3$ ).

Estimate the radius of the explosion $R(t)$ as a function of the energy yield $E$, elapsed time $t$, and the density of the surrounding medium $\rho$.
Context answer:
\boxed{$\left(\frac{E}{\rho}\right)^{1 / 5} t^{2 / 5}$}


Question: (b) Air has a density $\rho \simeq 1 \mathrm{~kg} \mathrm{~m}^{-3}$. Using the provided image, roughly estimate the yield of the Trinity explosion in kilotons (of TNT).

Note that 1 kiloton is equal to approximately $4.2 \times 10^{12} \mathrm{~J}$.

Solution:
From the included figure, the explosion looks to be approximately $\sim 100 \mathrm{~m}$ in radius at a time $\sim 15 \mathrm{~ms}$. We can use Equation 2 with the given values to obtain

$$
E \sim 11 \text { kilotons of TNT }
\tag{4}
$$

Note that this value may vary significantly depending on the estimated radius from the figure (since $E \propto R^{5}$ depends strongly on $R$ ), but should be in the neighborhood of tens of kilotons of TNT. The official number released by the government at the time was 21 kilotons of TNT, although there has been some uncertainty in this number.

Final answer: $11$ kilotons of TNT
Subfield: Mechanics


Question: (c) In 1054, Chinese astronomers observed and documented a supernova which was bright enough to be visible during the day for around a month. The rubble left behind is a rapidly expanding supernova remnant called the Crab Nebula, which is intensely studied today.


![](https://cdn.mathpix.com/cropped/2023_12_21_f9686d2b2f582fd8d3cfg-1.jpg?height=837&width=898&top_left_y=232&top_left_x=608)

Figure 2: Photographs of the Crab Nebula taken by the Hale Telescope in 1950 (top left) and the Hubble Space Telescope in 2000 (bottom right).

The surrounding interstellar medium contains about 1 proton per cubic meter-protons have a mass $m_{p} \approx 1.7 \times 10^{-27} \mathrm{~kg}$. While not perfectly spherical, Crab Nebula has a radius of roughly $1.7 \mathrm{pc}\left(1 \mathrm{pc}=3.1 \times 10^{16} \mathrm{~m}\right)$. From this information and the given images of the Crab Nebula from 1950 and 2000, estimate the energy released by the initial supernova.

Your answer should be close to the typical energy scales of supernovae, which briefly outshine their host galaxies and can be seen from across the universe.

Solution:
The density of the interstellar medium is

$$
\rho \sim \frac{m_{p}}{\left(1 \mathrm{~cm}^{3}\right)} \sim 1.7 \times 10^{-21} \mathrm{~kg} \mathrm{~cm}^{-3}
\tag{5}
$$

We are given that the Crab Nebula has a radius of approximately 1.7 pc. We see that the explosion seems to expand by $\sim 25 \%$ or so during the fifty years of observation along the long axis, which we take to be a very rough measure of its velocity. In particular,

$$
v \sim \frac{25 \% \times 1.7 \mathrm{pc}}{50 \mathrm{yr}} \sim 8.3 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}
\tag{6}
$$

We can use $E \sim M v^{2}$ and $M \sim \rho R^{3}$ to write

$$
E \sim \rho R^{3} v^{2} \sim 1.7 \times 10^{43} \mathrm{~J}
\tag{7}
$$

where we have used $t \sim 1000 \mathrm{yr}$ (the age of the supernova remnant). This is $E \sim$ $1.7 \times 10^{50} \mathrm{erg}$, which is in the neighborhood of true supernovae which are typically at $\sim 10^{51} \mathrm{erg}$ (a quantity of energy which has been affectionately named the "f.o.e." by astrophysicists because of this fact).

Final answer: $1.7 \times 10^{43}$ J
Subfield: Optics

3. Stick a pin there 

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature $20 \mathrm{~cm}$. The concave surface has a radius of curvature $60 \mathrm{~cm}$. The convex side is silvered and placed on a horizontal surface.

Question: (a) Where should a pin be placed on the optical axis such that its image is formed at the same place?

Solution:
Image of object will coincide with it if ray of light after refraction from the concave surface fall normally on concave mirror so formed by silvering the convex surface. Or image after refraction from concave surface should form at centre of curvature of concave mirror or at a distance of $20 \mathrm{~cm}$ on same side of the combination. Let $x$ be the distance of pin from the given optical system.

Using,

$$
\frac{\mu_{2}}{\nu}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}
\tag{8}
$$

With proper signs,

$$
\frac{1.5}{-20}-\frac{1}{-x}=\frac{1.5-1}{-60}
\tag{9}
$$

Solve to get $x=15 \mathrm{~cm}$

Final answer: 15 cm
Subfield: Optics

3. Stick a pin there 

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature $20 \mathrm{~cm}$. The concave surface has a radius of curvature $60 \mathrm{~cm}$. The convex side is silvered and placed on a horizontal surface.
Context question:
(a) Where should a pin be placed on the optical axis such that its image is formed at the same place?
Context answer:
\boxed{15}


Question: (b) If the concave part is filled with water of refractive index $4 / 3$, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.

<img_4461>

Solution:
There could be different approaches to solve this. A sample solution:

Now, before striking with the concave surface, the ray is first refracted from a plane surface. So, let $\mathrm{x}$ be the distance of pin, then the plane surface will form its image at a distance $\frac{4}{3} x\left(h_{a p p}=\mu h\right)$ from it.

Using,

$$
\frac{\mu_{2}}{\nu}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}
\tag{10}
$$

with proper signs,

$$
\frac{1.5}{-20}-\frac{4 / 3}{-4 x / 3}=\frac{1.5-4 / 3}{-60}
\tag{11}
$$

Solve to get $x=13.84 \mathrm{~cm}$.

Therefore $\Delta x=x_{1}-x_{2}=15 \mathrm{~cm}-13.84 \mathrm{~cm}=1.16 \mathrm{~cm}$ (Downwards)

Final answer: 1.16 cm
Subfield: Optics


Question: (b) If the concave part is filled with water of refractive index $4 / 3$, find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin.

![](https://cdn.mathpix.com/cropped/2023_12_21_c91fdc17266671ace758g-1.jpg?height=341&width=629&top_left_y=949&top_left_x=821)

Solution:
There could be different approaches to solve this. A sample solution:

Now, before striking with the concave surface, the ray is first refracted from a plane surface. So, let $\mathrm{x}$ be the distance of pin, then the plane surface will form its image at a distance $\frac{4}{3} x\left(h_{a p p}=\mu h\right)$ from it.

Using,

$$
\frac{\mu_{2}}{\nu}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}
\tag{10}
$$

with proper signs,

$$
\frac{1.5}{-20}-\frac{4 / 3}{-4 x / 3}=\frac{1.5-4 / 3}{-60}
\tag{11}
$$

Solve to get $x=13.84 \mathrm{~cm}$.

Therefore $\Delta x=x_{1}-x_{2}=15 \mathrm{~cm}-13.84 \mathrm{~cm}=1.16 \mathrm{~cm}$ (Downwards)

Final answer: 1.16 cm
Subfield: Mechanics

4. A complex dance 

In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.

Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary "growth" and "decay" can be translated into oscillations by the Euler identity:

$$
e^{i \theta}=\cos \theta+i \sin \theta
\tag{1}
$$

Question: (a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form

$$
\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
\tag{2}
$$

For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,

$$
\begin{aligned}
& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
\end{aligned}
\tag{3}
$$

where dots represent time derivatives.

Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:

$$
0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
\tag{4}
$$

Solution:
We can write down the equations of motion, multiplying the second one by $i$ :

$$
\begin{aligned}
& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
& 0=\ddot{y}-2 \Omega i \dot{x}-\Omega^{2} i y
\end{aligned}
\tag{19}
$$

We can add these equations together to obtain

$$
0=(\ddot{x}+i \ddot{y})+2 \Omega(\dot{y}-i \dot{x})-\Omega^{2}(x+i y)=\ddot{\eta}+2 \Omega(\dot{y}-i \dot{x})-\Omega^{2} \eta
\tag{20}
$$

We note that

$$
\dot{y}-i \dot{x}=-i(\dot{x}+i \dot{y})=-i \eta
\tag{21}
$$

Then

$$
0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
\tag{22}
$$
Subfield: Mechanics

4. A complex dance 

In this problem, we will solve a number of differential equations corresponding to very different physical phenomena that are unified by the idea of oscillation. Oscillations are captured elegantly by extending our notion of numbers to include the imaginary unit number $i$, strangely defined to obey $i^{2}=-1$. In other words, rather than using real numbers, it is more convenient for us to work in terms of complex numbers.

Exponentials are usually associated with rapid growth or decay. However, with the inclusion of complex numbers, imaginary "growth" and "decay" can be translated into oscillations by the Euler identity:

$$
e^{i \theta}=\cos \theta+i \sin \theta
\tag{1}
$$
Context question:
(a) The usual form of Newton's second law $(\vec{F}=m \vec{a})$ breaks down when we go into a rotating frame, where both the centrifugal and Coriolis forces become important to account for. Newton's second law then takes the form

$$
\vec{F}=m(\vec{a}+2 \vec{v} \times \vec{\Omega}+\vec{\Omega} \times(\vec{\Omega} \times \vec{r}))
\tag{2}
$$

For a particle free of forces confined to the $x-y$ plane in a frame which rotates about the $z$ axis with angular frequency $\Omega$, this becomes the complicated-looking system of differential equations,

$$
\begin{aligned}
& 0=\ddot{x}+2 \Omega \dot{y}-\Omega^{2} x \\
& 0=\ddot{x}-2 \Omega \dot{x}-\Omega^{2} y
\end{aligned}
\tag{3}
$$

where dots represent time derivatives.

Defining $\eta=x+i y$, show that Equations 3 are equivalent to the following single (complex) equation:

$$
0=\ddot{\eta}-2 i \Omega \dot{\eta}-\Omega^{2} \eta
\tag{4}
$$
Context answer:
\boxed{证明题}


Question: (b) Equation 4 is a version of the damped harmonic oscillator, and can be solved by guessing a solution $\eta=\alpha e^{\lambda t}$.

Plugging in this guess, what must $\lambda$ be?

Solution:
We can plug in $\eta=\alpha e^{\lambda t}$ :

$$
0=\lambda^{2} \alpha e^{\lambda t}-2 i \lambda \Omega \alpha e^{\lambda t}-\Omega^{2} \alpha e^{\lambda t}
\tag{23}
$$

Then we can cancel common factors and find that

$$
0=\lambda^{2}-2 i \lambda \Omega-\Omega^{2}=(\lambda-i \Omega)^{2}
\tag{24}
$$

We see that $\lambda=i \Omega$.

Final answer: $\lambda=i \Omega$