the world in a different light
I made some refinement to an applet created last year to demonstrate how vector addition can be done either using vector triangle or parallelogram methods.
To access directly in Geogebra, the link is https://www.geogebra.org/m/p2yvdsvs
This new applet is designed for students to practise conversion of common units used in physics on their own. There is a checking algorithm within, which might need some fine-tuning. For full screen view, click here.
The worked solutions given will demonstrate the breakdown of steps that could help students learn the procedure to convert these units.
This little applet is designed to allow students to change the order of magnitude and to use any common prefix to observe how the physical quantities are being written. To view this applet in a new tab, click here.
Standard form (also known as scientific notation) is a way of writing very large or very small numbers that allows for easy comparison of their magnitude by using the powers of ten. Any number that can be expressed as a number, between 1 and 10, multiplied by a power of 10, is said to be in standard form.
For instance, the speed of light in vacuum can be written as 3.00 × 108 m s–1 in standard form.
When a prefix is added to a unit, the unit is multiplied by a numerical value represented by the prefix. e.g. distance = 180 cm = 180 x 10-2 m = 1.80 m
The purpose of using prefixes is to reduce the number of digits used in the expression of values. Hence, students can use the prefix slider to find a user-friendly expression, such as 682 nm instead of 0.000000682 m.
The ten prefixes used are:
| 10-12 | pico |
| 10-9 | nano |
| 10-6 | micro |
| 10-3 | milli |
| 10-2 | centi |
| 10-1 | deci |
| 103 | kilo |
| 106 | mega |
| 109 | giga |
| 1012 | tera |
The internet is abuzz recently with news of ChatGPT (Chat Generative Pre-trained Transformer), an AI chatbot developed by OpenAI that is designed to perform well on tasks that involve generating human-like text.
I was interested to find out how it can help in Physics education and found two main use cases. The first is for students’ own self-directed learning through questioning.
For example, I asked the chatbot what the difference between centrifugal and centripetal force is using the following prompt and was given the response which I pasted here:
Prompt: What is the difference between centripetal and centrifugal force
Centripetal force is a force that acts towards the center of a circular path and is responsible for the change in direction of an object moving in a circular path. It is the force that keeps an object moving in a circular path and is perpendicular to the object’s velocity.
Centrifugal force, on the other hand, is an apparent force that is felt by an observer when an object is moving in a circular path. It is the force that appears to push an object away from the center of a circular path. However, it is not a real force and is only perceived by the observer due to the change in direction of the object.
To illustrate the difference between centripetal and centrifugal force, consider an example of a person swinging a ball on a string in a circular path. The person applies a force on the ball towards the center of the circular path (centripetal force) to keep the ball moving in a circular path. The ball, on the other hand, feels a force pushing it away from the center of the circular path (centrifugal force) due to its change in direction. However, this force is not a real force, but is only perceived by the person holding the ball.
In summary, centripetal force is a real force that acts towards the center of a circular path, while centrifugal force is an apparent force that is only perceived by an observer due to the change in direction of an object moving in a circular path.
I found the responses to the questions to be highly relevant, well-structured and easy to understand. This will serve as a good source of reference for students.
GENERATING LEARNING ACTIVITIES
The second use case for teachers is to generate quick question and activity ideas. I found the responses to be very helpful, even if some modification needs to be done to refine it afterwards.
The first test I did with the AI was to ask it to generate instructions for an experiment and this was the prompt and response:
Prompt: Create an experiment on measuring the speed of sound
Here is an experiment for measuring the speed of sound:
Materials:
- Two meters of string
- One meter stick
- One stopwatch or timer
- Two people
Procedure:
- Tie one end of the string to a stationary object, such as a table leg.
- Have one person hold the other end of the string, and have the second person stand a few meters away.
- The person holding the string should stretch the string taut and make a sharp, loud noise, such as clapping their hands or snapping their fingers.
- The second person should start the stopwatch as soon as they hear the noise, and stop it when they feel the vibration from the string on their hand.
- Measure the distance between the two people using the meter stick.
- Repeat the experiment a few times to ensure accuracy.
To calculate the speed of sound, use the formula: speed = distance / time. The speed of sound is the distance traveled by the sound wave divided by the time it took for the wave to travel that distance.
Note: This experiment is most accurate when performed in a quiet, indoor space with little to no background noise. The string should be stretched taut and not vibrate excessively. It is also important to measure the distance and time accurately to obtain an accurate result.
Not satisfied with the extent of precision and accuracy of the experiment suggested, I decided to give it a more detailed prompt.
Prompt: Create an experiment on measuring the speed of sound in air with the help of a datalogger
Here is an experiment for measuring the speed of sound in air using a datalogger:
Materials:
- A long, straight tube with a diameter of at least 1 inch
- Two clamps
- A datalogger or sound level meter
- A tuning fork with a known frequency (such as 512 Hz)
- A meter stick or ruler
Procedure:
- Clamp one end of the tube to a stationary object, such as a table.
- Clamp the other end of the tube to a stand that is adjustable in height.
- Set up the datalogger or sound level meter at the open end of the tube, making sure it is oriented so that it can accurately measure the sound waves reflected off the other end of the tube.
- Strike the tuning fork and hold it near the open end of the tube.
- Adjust the height of the stand so that the sound waves from the tuning fork are reflected back and forth between the two ends of the tube.
- Use the datalogger or sound level meter to measure the time it takes for the sound waves to travel from one end of the tube to the other and back again.
- Measure the length of the tube using the meter stick or ruler.
- Repeat the experiment a few times to ensure accuracy.
To calculate the speed of sound in air, use the formula: speed = distance / time. The speed of sound in air is the distance traveled by the sound wave (the length of the tube) divided by the time it took for the wave to travel that distance (the time measured by the datalogger or sound level meter).
Note: This experiment is most accurate when performed in a quiet, indoor space with little to no background noise. The tuning fork should be struck firmly and held close to the open end of the tube to ensure that the sound waves are reflected back and forth between the two ends. It is also important to measure the distance and time accurately to obtain an accurate result.
The second test for it was to get it to set a full structured question:
Prompt: Create a data-response question on thermal physics for GCE A-level
Here is a data-response question on thermal physics for GCE A-level:
A student investigates the relationship between the temperature of a substance and its heat capacity. She measures the heat capacity of 50 g of water at several different temperatures, and records her results in the table below:
Temperature (°C) Heat Capacity (J/°C) 20 4.2 40 4.5 60 4.8 80 5.1 Using the data in the table, answer the following questions:
- What is the heat capacity of 50 g of water at a temperature of 60°C?
- Is the heat capacity of 50 g of water a constant value, or does it vary with temperature?
- Can you determine the heat capacity of 50 g of water at a temperature of 0°C from this data? Why or why not?
- The heat capacity of 50 g of water at a temperature of 60°C is 4.8 J/°C.
- The heat capacity of 50 g of water varies with temperature.
- It is not possible to determine the heat capacity of 50 g of water at a temperature of 0°C from this data. This is because the heat capacity of a substance may change at different temperatures, and the data provided only includes measurements at specific temperatures. To determine the heat capacity at a temperature of 0°C, additional measurements would need to be taken at that temperature.
The quality of the questions and corresponding answers were decent, even though some refinement will still have to be done as in the case of the speed of sound experiment. What impressed me was that the formatting, including the tabulation was done for us. We only had to cut and paste the activities into a Word document or Google Doc for further editing before assigning them to students.
LET CHATGPT TELL YOU
After figuring it out for myself, I thought, why not let ChatGPT tell me what else it can do in the context of physics education and this was its response:
ChatGPT could potentially be used to:
It is important to note that while ChatGPT has the potential to be a useful tool for physics education, it is not a replacement for human instructors and should be used in conjunction with other resources and methods of teaching and learning.
This interactive is designed to help students understand the statistical approach underpinning the drawing of a best-fit line for practical work. For context, our national exams have a practical component where students will need to plot their data, often following a linear trend, on graph paper and to draw a best-fit line to determine the gradient and y-intercept.
The instructions to students on how to draw the best-fit line is often procedural without helping students understand the principles behind it. For instance, students are often told to minimise and balance the separation of plots from the best-fit line. However, if there are one or two points that are further from the rest from the best-fit line (but not quite anomalous points that need to be disregarded), students would often neglect that point in an attempt to bring the best fit line as close to the remaining points as possible. This results in a drastic increase in the variance as the differences are squared in order to calculate the “the smallest sum of squares of errors”.
This applet allows students to visualise the changes in the squares, along with the numerical representation of the sum of squares in order to practise “drawing” the best-fit line using a pair of movable dots. A check on how well they have “drawn” the line can be through a comparison with the actual one.
Students can also rearrange the 6 data points to fit any distribution that they have seen before, or teachers can copy and modify the applet in order to provide multiple examples of distribution of points.
Did this simple interactive upon request by a colleague who is teaching the JC1 topic of Oscillations.
Based on the following question, this is used as a quick visual to demonstrate why there must be a minimum depth before the boat approaches harbour.
The rise and fall of water in a harbour is simple harmonic. The depth varies between 1.0 m at low tide and 3.0 m at high tide. The time between successive low tides is 12 hours. A boat, which requires a minimum depth of water of 1.5 m, approaches the harbour at low tide. How long will the boat have to wait before entering?
The equation of the depth of water H based on the amplitude of the tide a can be given by $H = H_o + a \cos \omega t$ where $H_o$ is the average depth of the water.
$H = H_o + a \cos \omega t$
When H = 1.5m,
$1.5 = 2.0 – 1.0 \cos (\dfrac{2 \pi}{12}t)$
$\cos (\dfrac{2 \pi}{12}t) = 0.5$
$t = 2.0 h$