A student measures the time period of `100` ocillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s, and 92 s. If the minimum division in the measuring clock is `1 s`, then the reported men time should be:

Options

A. `92 pm 5.0s`
B. `92 pm 1.8s`
C. `92 pm 3s`
D. `92 pm 2s`

Answer

To find the reported mean time for the oscillations of a simple pendulum, we will follow these steps:

### Step 1: Calculate the Mean Time
We need to find the mean of the four measurements given: 90 s, 91 s, 95 s, and 92 s.

\[ \text{Mean} = \frac{\text{Sum of all measurements}}{\text{Number of measurements}} = \frac{90 + 91 + 95 + 92}{4} \]

Calculating the sum:

\[ 90 + 91 + 95 + 92 = 368 \]

Now, divide by the number of measurements (which is 4):

\[ \text{Mean} = \frac{368}{4} = 92 \text{ s} \]

### Step 2: Calculate the Absolute Error
Next, we need to calculate the absolute error for the mean time. The absolute error is calculated by finding the difference between each measurement and the mean, then averaging those differences.

\[ \text{Absolute Error} = \frac{|\text{Measurement}_1 - \text{Mean}| + |\text{Measurement}_2 - \text{Mean}| + |\text{Measurement}_3 - \text{Mean}| + |\text{Measurement}_4 - \text{Mean}|}{4} \]

Calculating each difference:

- For 90 s: \( |90 - 92| = 2 \)
- For 91 s: \( |91 - 92| = 1 \)
- For 95 s: \( |95 - 92| = 3 \)
- For 92 s: \( |92 - 92| = 0 \)

Now, sum these absolute differences:

\[ 2 + 1 + 3 + 0 = 6 \]

Now, divide by the number of measurements (4):

\[ \text{Absolute Error} = \frac{6}{4} = 1.5 \text{ s} \]

### Step 3: Round the Absolute Error
Since the minimum division in the measuring clock is 1 s, we round the absolute error to the nearest whole number. Rounding 1.5 gives us 2 s.

### Step 4: Report the Mean Time with Uncertainty
Finally, we report the mean time with the calculated uncertainty:

\[ \text{Reported Mean Time} = 92 \pm 2 \text{ s} \]

### Final Answer
The reported mean time should be \( 92 \pm 2 \text{ s} \).

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