Question

A student measures the time period of $100$ oscillations 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 mean time should be :

• A. $92 \pm 2\, s$
• B. $92 \pm 5.0 \,s$
• C. $92 \pm 1.8 \,s$
• D. $92 \pm 3 \,s$

Answer 1

The Correct Option is A

The problem requires us to calculate the mean time for a set of time period measurements for 100 oscillations of a simple pendulum. Given the data set, the next step is to calculate the mean value and estimate the uncertainty in the measurements.

1. The data set for time periods is: $90 \, s, 91 \, s, 95 \, s, 92 \, s$.
2. First, we calculate the mean time period using the formula for the average:
3. $$ \bar{T} = \frac{T_1 + T_2 + T_3 + T_4}{4} = \frac{90 + 91 + 95 + 92}{4} $$
4. Now, calculating the sum: $90 + 91 + 95 + 92 = 368$
5. Therefore, the mean time period is: $$ \bar{T} = \frac{368}{4} = 92 \, s $$
6. Now, we calculate the uncertainty using the standard deviation, where:
7. $$ \sigma = \sqrt{\frac{\sum(T_i - \bar{T})^2}{n-1}} $$
8. Computing the deviations:
   • $(90 - 92)^2 = 4$
   • $(91 - 92)^2 = 1$
   • $(95 - 92)^2 = 9$
   • $(92 - 92)^2 = 0$
9. Sum of squares: $4 + 1 + 9 + 0 = 14$
10. Calculate the standard deviation: $$ \sigma = \sqrt{\frac{14}{3}} \approx \sqrt{4.67} \approx 2.16 \, s $$
11. Since the minimum division of the clock is $1 \, s$, we round the uncertainty to the nearest significant figure, resulting in $2 \, s$.
12. The reported mean time period with uncertainty is thus: $92 \pm 2 \, s$.

Therefore, the correct answer is $92 \pm 2 \, s$, which matches Option 1.