
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:
$
{\text{A}}{\text{. 92}} \pm {\text{2s}} \\
{\text{B}}{\text{. 92}} \pm 5.0{\text{s}} \\
{\text{C}}{\text{. 92}} \pm 1.8{\text{s}} \\
{\text{D}}{\text{. 92}} \pm 3{\text{s}} \\
$
Answer
514.2k+ views
Hint: We are given the values in the data set from which we need to calculate the mean value of time. Then the absolute error in the observation is obtained by subtracting the observed value from the mean value. The values obtained give us the required answer.
Complete answer:
We are given that a student measures the time period of 100 oscillations of a simple pendulum four times. The data set is given as 90 s, 91 s, 95 s and 92 s. Therefore, we have
$
{t_1} = 90s \\
{t_2} = 91s \\
{t_3} = 95s \\
{t_4} = 92s \\
$
We can calculate the mean of these values of time obtained for 100 oscillations. It can be calculated as
Mean value ${t_m} = \dfrac{{90 + 91 + 95 + 92}}{4} = 92s$
Now we can calculate the error in calculation for each value of time. It can be obtained for each data value by subtracting it from the mean value of time obtained above.
The absolute errors are given as
$
|\Delta {t_1}| = |{t_m} - {t_1}| = |92 - 90| = 2s \\
|\Delta {t_2}| = |{t_m} - {t_2}| = |92 - 91| = 1s \\
|\Delta {t_3}| = |{t_m} - {t_3}| = |92 - 95| = 3s \\
|\Delta {t_4}| = |{t_m} - {t_4}| = |92 - 92| = 0s \\
$
Now we need to find out the mean of these absolute errors. It is given as
Mean absolute error $\Delta {t_m} = \dfrac{{2 + 1 + 3 + 0}}{4} = 1.5s$
Now we can write the mean time in the following way: $92 \pm 1.5s \simeq 92 \pm 2s$
So, the correct answer is “Option A”.
Note:
The absolute error that we have calculated can be used to calculate the standard deviation of the data. It signifies how much the data is deviated from its mean value. It can be obtained by taking the same of all the squares of the absolute errors obtained in the data points and then dividing it by the number of observations in data and then taking the square root of the obtained value. The square of the standard deviation is known as the variance of the data.
Complete answer:
We are given that a student measures the time period of 100 oscillations of a simple pendulum four times. The data set is given as 90 s, 91 s, 95 s and 92 s. Therefore, we have
$
{t_1} = 90s \\
{t_2} = 91s \\
{t_3} = 95s \\
{t_4} = 92s \\
$
We can calculate the mean of these values of time obtained for 100 oscillations. It can be calculated as
Mean value ${t_m} = \dfrac{{90 + 91 + 95 + 92}}{4} = 92s$
Now we can calculate the error in calculation for each value of time. It can be obtained for each data value by subtracting it from the mean value of time obtained above.
The absolute errors are given as
$
|\Delta {t_1}| = |{t_m} - {t_1}| = |92 - 90| = 2s \\
|\Delta {t_2}| = |{t_m} - {t_2}| = |92 - 91| = 1s \\
|\Delta {t_3}| = |{t_m} - {t_3}| = |92 - 95| = 3s \\
|\Delta {t_4}| = |{t_m} - {t_4}| = |92 - 92| = 0s \\
$
Now we need to find out the mean of these absolute errors. It is given as
Mean absolute error $\Delta {t_m} = \dfrac{{2 + 1 + 3 + 0}}{4} = 1.5s$
Now we can write the mean time in the following way: $92 \pm 1.5s \simeq 92 \pm 2s$
So, the correct answer is “Option A”.
Note:
The absolute error that we have calculated can be used to calculate the standard deviation of the data. It signifies how much the data is deviated from its mean value. It can be obtained by taking the same of all the squares of the absolute errors obtained in the data points and then dividing it by the number of observations in data and then taking the square root of the obtained value. The square of the standard deviation is known as the variance of the data.
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