To find the average absolute error of the recorded periods of oscillation of a simple pendulum, we will follow these steps:
### Step 1: List the recorded periods
The recorded periods of oscillation are:
- \( t_1 = 2.63 \, s \)
- \( t_2 = 2.56 \, s \)
- \( t_3 = 2.42 \, s \)
- \( t_4 = 2.71 \, s \)
- \( t_5 = 2.80 \, s \)
### Step 2: Calculate the average period
To find the average period (\( \bar{t} \)), we sum all the recorded periods and divide by the number of periods (5 in this case).
\[
\bar{t} = \frac{t_1 + t_2 + t_3 + t_4 + t_5}{5}
\]
\[
\bar{t} = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5} = \frac{13.12}{5} = 2.624 \, s
\]
### Step 3: Calculate the absolute errors
Next, we calculate the absolute error for each recorded period by subtracting the average period from each recorded period:
1. \( |t_1 - \bar{t}| = |2.63 - 2.624| = 0.006 \, s \)
2. \( |t_2 - \bar{t}| = |2.56 - 2.624| = 0.064 \, s \)
3. \( |t_3 - \bar{t}| = |2.42 - 2.624| = 0.204 \, s \)
4. \( |t_4 - \bar{t}| = |2.71 - 2.624| = 0.086 \, s \)
5. \( |t_5 - \bar{t}| = |2.80 - 2.624| = 0.176 \, s \)
### Step 4: Calculate the mean absolute error
Now, we sum all the absolute errors and divide by the number of periods (5):
\[
\text{Mean Absolute Error} = \frac{|t_1 - \bar{t}| + |t_2 - \bar{t}| + |t_3 - \bar{t}| + |t_4 - \bar{t}| + |t_5 - \bar{t}|}{5}
\]
\[
= \frac{0.006 + 0.064 + 0.204 + 0.086 + 0.176}{5} = \frac{0.536}{5} = 0.1072 \, s
\]
### Step 5: Round the result
Rounding \( 0.1072 \, s \) gives us approximately \( 0.11 \, s \).
### Final Answer
The average absolute error is approximately \( 0.11 \, s \).
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