說最讓她震撼的一門課是aops 的Physics 1 Mechanics。
我看過aops這門課的講義和作業，感覺除了圖文聲等並茂，似乎也沒有很突出抓人的東西。
她好朋友的家長也說她們的孩子也很喜歡這個課。
這裏有修過這個課孩子的家長嗎。
這是其中的幾個作業。
Problem:
A DVD is a disk used to store digital data. They're often used to hold movies or video games.
By Marcin Sochacki (Wanted) - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=3749407
The digital data is stored in a long sequence of bumps on the bottom of the DVD. The bumps are all about the same size, and run in a spiral around the DVD.
http://www.futurekids.com.sg/doyouknow.html
Computers and video game consoles can use a device called a DVD player to read the sequence of bumps on the DVD.
Greg Krumm, "See Through DVD Drive", https://www.youtube.com/watch?v=8fqr-Fsl9xk
There are roughly
bumps on a DVD. They're too small to see individually with your naked eye, but a DVD player can detect them using a laser.
The DVD player shines a laser onto the DVD from below (red cylinder in the image below). The laser light bounces off the DVD, where it enters the DVD player's detector (blue cylinder). The detector can use the reflected light to read the pattern of bumps on the DVD and send the information to a computer, which can then turn the data into a movie or video game.
The DVD player reads the bumps by following the DVD's spiral pattern. It starts near the center of the DVD and ends at the outside edge.
To follow the spiral, the DVD player has two motions. First, it spins the DVD. This brings new bumps over the laser. You can see the spinning motion in the gif above.
Second, the laser itself moves slowly outward as the DVD spins. This allows it to read tracks further toward the edge of the DVD as it advances along the spiral. The tracks of the spiral are all about the same thickness, so the distance of the laser from the center of the DVD is proportional to the number of rotations the DVD has made.
The DVD player needs to move the bumps past the laser at a constant number of bumps per second.
Which of these is a plot of the angular speed of the DVD over time as the DVD player reads the entire DVD?
By Marcin Sochacki (Wanted) - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=3749407
The digital data is stored in a long sequence of bumps on the bottom of the DVD. The bumps are all about the same size, and run in a spiral around the DVD.
http://www.futurekids.com.sg/doyouknow.html
Computers and video game consoles can use a device called a DVD player to read the sequence of bumps on the DVD.
Greg Krumm, "See Through DVD Drive", https://www.youtube.com/watch?v=8fqr-Fsl9xk
There are roughly
bumps on a DVD. They're too small to see individually with your naked eye, but a DVD player can detect them using a laser.The DVD player shines a laser onto the DVD from below (red cylinder in the image below). The laser light bounces off the DVD, where it enters the DVD player's detector (blue cylinder). The detector can use the reflected light to read the pattern of bumps on the DVD and send the information to a computer, which can then turn the data into a movie or video game.
The DVD player reads the bumps by following the DVD's spiral pattern. It starts near the center of the DVD and ends at the outside edge.
To follow the spiral, the DVD player has two motions. First, it spins the DVD. This brings new bumps over the laser. You can see the spinning motion in the gif above.
Second, the laser itself moves slowly outward as the DVD spins. This allows it to read tracks further toward the edge of the DVD as it advances along the spiral. The tracks of the spiral are all about the same thickness, so the distance of the laser from the center of the DVD is proportional to the number of rotations the DVD has made.
The DVD player needs to move the bumps past the laser at a constant number of bumps per second.
Which of these is a plot of the angular speed of the DVD over time as the DVD player reads the entire DVD?
Problem:
In class, we practiced looking at new situations, finding the energy, and using the coefficients in the equations of kinetic and potential energies to find oscillation periods. Here, we'll practice that same idea in a completely new context.
Radio stations say things like "This is 101.5 FM." This means that the radio station's signal oscillates at frequency of
To listen to this station, your radio needs to find any radio waves with that frequency and ignore all other radio waves around it.
The radio does this with a "tuner". A simple model for a tuner has two components: an inductor and a capacitor. These are devices used in electric circuits. They both store energy. The inductor stores energy when electric charge runs through it, and the capacitor stores energy when there is electric charge on it.
Charge will oscillate between the inductor and capacitor at a certain frequency, which depends on how strong the inductor and capacitor are.
PhET project, "Circuit Construction Kit: DC", https://phet.colorado.edu/en/simulations/circuit-construction-kit-dc
In this animation, the inductor is the coil of wire at the bottom. The capacitor is the two parallel plates at the top. The blue circles are charge. On the left, there is a white circle which can be ignored; in real life the circuit can work without it.
In radios, turning the tuning dial changes the capacitor, affecting the frequency of charge oscillations.
When the frequency of charge oscillations in the tuner is
the radio waves from station 101.5 FM will have the same frequency as the tuner. Then the radio waves will push charge around in the tuner. The radio can detect the precise way that charge is moving and turn it into sound.
Other radio stations with different frequencies will not set a significant amount of charge oscillating in the tuner.
The energy stored by a capacitor can be modeled as
![\[E_C = \dfrac12 \dfrac{1}{C} Q^2\]](https://latex.artofproblemsolving.com/b/a/e/bae53a192b4d2d259a7dcbd50b15fe1bd1d40ea6.png)
where 
is the charge on the capacitor and 
is the capacitance, a measure of how large a capacitor it is. This is a potential energy.
The energy stored by an inductor can be modeled as
![\[ E_I = \dfrac12 L i^2\]](https://latex.artofproblemsolving.com/7/2/f/72f7779699685278473e9788b4f5a717ee7a1e6f.png)
where 
is the inductance, a measure of how strong the inductor is, and 
is how fast charge runs through the inductor, so is similar to a speed, but for charge. You can think of as a rate of change of 
You can think of the energy in the inductor as playing a role similar to kinetic energy.
You change the tuner on a radio from 101.5 FM to 98.0 FM.
As you do, the capacitance of the capacitor gets multiplied by some factor. What is that factor?
Give your answer as a decimal to three decimal places.
Radio stations say things like "This is 101.5 FM." This means that the radio station's signal oscillates at frequency of
To listen to this station, your radio needs to find any radio waves with that frequency and ignore all other radio waves around it.The radio does this with a "tuner". A simple model for a tuner has two components: an inductor and a capacitor. These are devices used in electric circuits. They both store energy. The inductor stores energy when electric charge runs through it, and the capacitor stores energy when there is electric charge on it.
Charge will oscillate between the inductor and capacitor at a certain frequency, which depends on how strong the inductor and capacitor are.
PhET project, "Circuit Construction Kit: DC", https://phet.colorado.edu/en/simulations/circuit-construction-kit-dc
In this animation, the inductor is the coil of wire at the bottom. The capacitor is the two parallel plates at the top. The blue circles are charge. On the left, there is a white circle which can be ignored; in real life the circuit can work without it.
In radios, turning the tuning dial changes the capacitor, affecting the frequency of charge oscillations.
When the frequency of charge oscillations in the tuner is
the radio waves from station 101.5 FM will have the same frequency as the tuner. Then the radio waves will push charge around in the tuner. The radio can detect the precise way that charge is moving and turn it into sound.Other radio stations with different frequencies will not set a significant amount of charge oscillating in the tuner.
The energy stored by a capacitor can be modeled as
where is the charge on the capacitor and is the capacitance, a measure of how large a capacitor it is. This is a potential energy.The energy stored by an inductor can be modeled as
where is the inductance, a measure of how strong the inductor is, and is how fast charge runs through the inductor, so is similar to a speed, but for charge. You can think of as a rate of change of You can think of the energy in the inductor as playing a role similar to kinetic energy.You change the tuner on a radio from 101.5 FM to 98.0 FM.
As you do, the capacitance
of the capacitor gets multiplied by some factor. What is that factor?Give your answer as a decimal to three decimal places.
C98.0FMC101.5FM=????" id="MathJax-Element-357-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size_replace:17.85px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; transform:translateZ(0px); word-spacing:normal" tabindex="0">
Solution:
The capacitor and inductor both have energies that are quadratic, just like the mass on a spring's potential and kinetic energies.
Writing out all the energies, we have
The equations are the same as long as we make the analogies
The question tells us that is like a speed associated with 
so this analogy is valid.
Then the formula for the frequency becomes
![\[f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} \to \dfrac{1}{2\pi} \sqrt{\dfrac{1/C}{L}} = \dfrac{1}{2\pi \sqrt{LC}}\]](https://latex.artofproblemsolving.com/c/9/d/c9d7e155faef618ea64be466fcd3b71be2162de9.png)
The important takeaway is
![\[f \propto \dfrac{1}{\sqrt{C}}\]](https://latex.artofproblemsolving.com/f/0/3/f03cd71c539d8cafa2305005ea7bf2a3c8e0d2cf.png)
This means
![\[ \dfrac{101.5 \times 10^6 \;\mathrm{Hz}}{98.0 \times 10^6 \;\mathrm{Hz}} = \sqrt{\dfrac{C_{\rm 98.0 FM}}{C_{\rm 101.5 FM}}}\]](https://latex.artofproblemsolving.com/5/9/d/59da66f2f9ab45c284b1d8baf4d5a39ff389537f.png)
Solving for
we get
![\[ \dfrac{C_{\rm 98.0 FM}}{C_{\rm 101.5 FM}} = \left(\dfrac{101.5}{98.0}\right)^2 = \boxed{1.073}\]](https://latex.artofproblemsolving.com/5/6/5/565eac349ff9dd8b07754f2f2a96be84ab7a0290.png)
To account for rounding, we accept a small range of nearby answers as well.
Writing out all the energies, we have
The equations are the same as long as we make the analogies
The question tells us that
is like a speed associated with so this analogy is valid.Then the formula for the frequency becomes
The important takeaway is
This means
Solving for
we getTo account for rounding, we accept a small range of nearby answers as well.
Hint(s):
For a mass on a spring, 
Make these analogies:
Find the proportionality between 
and and use it to solve the problem.
Make these analogies:
Find the proportionality between and and use it to solve the problem.Your Response(s):
- $ C98.0FMC101.5FM=1.072" id="MathJax-Element-346-Frame" role="presentation" style="border:0px; box-sizing:border-box; direction:ltr; display:inline-block; float:none; font-size_replace:17.85px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; text-wrap:nowrap; word-spacing:normal" tabindex="0">
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