Answer :
Certainly! Let's break down the question and address each part step-by-step.
### Part A: Comparison of Equations
First, let's revisit the pendulum equation:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
This equation describes the period [tex]\(T\)[/tex] of a simple pendulum, where:
- [tex]\(T\)[/tex] is the period (time for one complete oscillation),
- [tex]\(L\)[/tex] is the length of the pendulum,
- [tex]\(g\)[/tex] is the acceleration due to gravity.
To compare with the equation from 1C, we would need that specific equation from part 1C. Since this context isn't provided, we are unable to make a direct comparison. However, the fundamental principles in the pendulum equation remain unchanged and showcase how the period [tex]\(T\)[/tex] depends on both the length [tex]\(L\)[/tex] and the gravitational constant [tex]\(g\)[/tex].
### Part B: Calculating the Period
Given:
- [tex]\( L = 1.50 \, \text{m} \)[/tex] (length of the pendulum),
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] (acceleration due to gravity),
we can use the pendulum equation to calculate the period [tex]\(T\)[/tex]:
[tex]\[ T = 2 \pi \sqrt{\frac{1.50 \, \text{m}}{9.8 \, \text{m/s}^2}} \][/tex]
Let's rewrite the equation with the values plugged in:
[tex]\[ T = 2 \pi \sqrt{\frac{1.50}{9.8}} \][/tex]
By simplifying the fraction within the square root:
[tex]\[ T = 2 \pi \sqrt{0.1530612245} \][/tex]
Taking the square root:
[tex]\[ T \approx 2 \pi \times 0.391739 \approx 2.4582 \][/tex]
So, the calculated period [tex]\(T\)[/tex] for a 1.50-meter pendulum is approximately:
[tex]\[ T \approx 2.458 \, \text{seconds} \][/tex]
### Part C: Comparison and Sources of Error
To compare the calculated period with your data from 2B, you would refer to the graph constructed in that part (which we do not have here). Assuming you've plotted your empirical data points and possibly even a trendline or curve through those points, check if the calculated period ([tex]\(2.458\)[/tex] seconds) matches closely with the value observed in experiments.
Possible sources of error include:
1. Measurement inaccuracies in the length [tex]\(L\)[/tex]: The exact length of the pendulum might have slight errors due to measurement tolerances or misalignments.
2. Variations in gravitational acceleration [tex]\(g\)[/tex]: The value of [tex]\(g\)[/tex] is assumed to be [tex]\(9.8 \, \text{m/s}^2\)[/tex], but it can vary slightly depending on location (e.g., altitude or geophysical factors).
3. Air resistance and friction: These factors can slow the pendulum down slightly, causing the experimental period to be longer than the theoretical period.
4. Human reaction time: When measuring the period [tex]\(T\)[/tex] manually, there can be delays or advances in starting or stopping the timer, leading to less accurate readings.
These sources of error can collectively contribute to differences between the theoretically calculated period and the experimentally observed data. It's important to consider and minimize these errors to achieve more accurate and reliable results.
### Part A: Comparison of Equations
First, let's revisit the pendulum equation:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
This equation describes the period [tex]\(T\)[/tex] of a simple pendulum, where:
- [tex]\(T\)[/tex] is the period (time for one complete oscillation),
- [tex]\(L\)[/tex] is the length of the pendulum,
- [tex]\(g\)[/tex] is the acceleration due to gravity.
To compare with the equation from 1C, we would need that specific equation from part 1C. Since this context isn't provided, we are unable to make a direct comparison. However, the fundamental principles in the pendulum equation remain unchanged and showcase how the period [tex]\(T\)[/tex] depends on both the length [tex]\(L\)[/tex] and the gravitational constant [tex]\(g\)[/tex].
### Part B: Calculating the Period
Given:
- [tex]\( L = 1.50 \, \text{m} \)[/tex] (length of the pendulum),
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] (acceleration due to gravity),
we can use the pendulum equation to calculate the period [tex]\(T\)[/tex]:
[tex]\[ T = 2 \pi \sqrt{\frac{1.50 \, \text{m}}{9.8 \, \text{m/s}^2}} \][/tex]
Let's rewrite the equation with the values plugged in:
[tex]\[ T = 2 \pi \sqrt{\frac{1.50}{9.8}} \][/tex]
By simplifying the fraction within the square root:
[tex]\[ T = 2 \pi \sqrt{0.1530612245} \][/tex]
Taking the square root:
[tex]\[ T \approx 2 \pi \times 0.391739 \approx 2.4582 \][/tex]
So, the calculated period [tex]\(T\)[/tex] for a 1.50-meter pendulum is approximately:
[tex]\[ T \approx 2.458 \, \text{seconds} \][/tex]
### Part C: Comparison and Sources of Error
To compare the calculated period with your data from 2B, you would refer to the graph constructed in that part (which we do not have here). Assuming you've plotted your empirical data points and possibly even a trendline or curve through those points, check if the calculated period ([tex]\(2.458\)[/tex] seconds) matches closely with the value observed in experiments.
Possible sources of error include:
1. Measurement inaccuracies in the length [tex]\(L\)[/tex]: The exact length of the pendulum might have slight errors due to measurement tolerances or misalignments.
2. Variations in gravitational acceleration [tex]\(g\)[/tex]: The value of [tex]\(g\)[/tex] is assumed to be [tex]\(9.8 \, \text{m/s}^2\)[/tex], but it can vary slightly depending on location (e.g., altitude or geophysical factors).
3. Air resistance and friction: These factors can slow the pendulum down slightly, causing the experimental period to be longer than the theoretical period.
4. Human reaction time: When measuring the period [tex]\(T\)[/tex] manually, there can be delays or advances in starting or stopping the timer, leading to less accurate readings.
These sources of error can collectively contribute to differences between the theoretically calculated period and the experimentally observed data. It's important to consider and minimize these errors to achieve more accurate and reliable results.
