Question

A student measures the time period of 100 oscillation of a simple pendulum four times. He observed 94 s, 95 s, 98 s, 97 s. If the least count of the measuring clock is 1 s, then the reported mean time should be:

• A. (98 ± 1) s
• B. (96 ± 2) s
• C. (98 ± 3) s
• D. (96 ± 4) s

Answer 1

Answer: (B)

(96 ± 2) s

Answer 2

To report the mean time with its associated error, we follow these steps:

1. Calculate the mean (average) of the observations: The given observations are 94 s, 95 s, 98 s, 97 s. Mean time, $\bar{t} = \frac{94 + 95 + 98 + 97}{4} = \frac{384}{4} = 96 \,s$

2. Calculate the absolute deviation for each observation from the mean: $|\Delta t_1| = |94 - 96| = 2 \,s$ $|\Delta t_2| = |95 - 96| = 1 \,s$ $|\Delta t_3| = |98 - 96| = 2 \,s$ $|\Delta t_4| = |97 - 96| = 1 \,s$

3. Calculate the mean absolute error: Mean absolute error, $\Delta \bar{t} = \frac{2 + 1 + 2 + 1}{4} = \frac{6}{4} = 1.5 \,s$

4. Determine the final reported error: The least count of the measuring clock is 1 s. The calculated mean absolute error is 1.5 s. According to standard error reporting conventions, the final uncertainty should be at least the least count of the instrument and is usually rounded to one significant figure. Comparing the mean absolute error (1.5 s) with the least count (1 s), we take the larger value, which is 1.5 s. Rounding 1.5 s to one significant figure gives 2 s.

5. Report the mean time with the error: The reported mean time should be the mean value plus/minus the final error. Reported mean time = $(96 \pm 2)$ s.