Answer :
the absolute errors for each observation are 1.5 seconds, 1.0 second, 0 seconds, 1.0 second, and 1.5 seconds, respectively.
To calculate the mean period of oscillation of the pendulum and find its absolute error, we first need to find the period of oscillation for each set of oscillations. The period of oscillation (T) is the time taken for one complete oscillation or cycle. Since the student observed the time for 5 oscillations, we can find the period by dividing the total time by the number of oscillations.
The observed times for 5 oscillations are:
[tex]\[ t_1 = 99.0 \text{ seconds} \]\[ t_2 = 99.5 \text{ seconds} \]\[ t_3 = 100.5 \text{ seconds} \]\[ t_4 = 101.5 \text{ seconds} \]\[ t_5 = 102.0 \text{ seconds} \][/tex]
To find the mean period of oscillation, we sum up the times for 5 oscillations and divide by the number of observations (n = 5):
[tex]\[ T_{\text{mean}} = \frac{t_1 + t_2 + t_3 + t_4 + t_5}{5} \][/tex]
Calculating this gives:
[tex]\[ T_{\text{mean}} = \frac{99.0 + 99.5 + 100.5 + 101.5 + 102.0}{5} \]\[ T_{\text{mean}} = \frac{502.5}{5} \]\[ T_{\text{mean}} = 100.5 \text{ seconds} \][/tex]
The mean period of oscillation of the pendulum is 100.5 seconds.
Next, we can find the absolute error for each observation by subtracting the mean period from each observed period. The absolute error (|ΔT|) is the magnitude of the difference between the observed period and the mean period.
The absolute errors for each observation are:
[tex]\[ |ΔT_1| = |99.0 - 100.5| = 1.5 \text{ seconds} \]\[ |ΔT_2| = |99.5 - 100.5| = 1.0 \text{ second} \]\[ |ΔT_3| = |100.5 - 100.5| = 0 \text{ seconds} \]\[ |ΔT_4| = |101.5 - 100.5| = 1.0 \text{ second} \]\[ |ΔT_5| = |102.0 - 100.5| = 1.5 \text{ seconds} \][/tex]
So, the absolute errors for each observation are 1.5 seconds, 1.0 second, 0 seconds, 1.0 second, and 1.5 seconds, respectively.
The mean period of oscillation for the pendulum is 20.1 seconds, and the absolute error is 0.2 seconds.
To calculate the mean period of oscillation for the simple pendulum, we first add the times for the 5 oscillations and then divide by the number of oscillations. After that, to get the period of one oscillation, we divide this mean by 5. The times given are 99.0, 99.5, 100.5, 101.5, and 102 seconds for 5 oscillations each.
To find the mean of 5 oscillations, we sum the times: (99.0 + 99.5 + 100.5 + 101.5 + 102) seconds, which equals 502.5 seconds. We then divide this sum by 5 to get the mean time for 5 oscillations:
Mean time for 5 oscillations = 502.5 seconds / 5 = 100.5 seconds.
To obtain the mean period of one oscillation, we divide the mean time for 5 oscillations by 5:
Mean period of one oscillation (T) = 100.5 seconds / 5 = 20.1 seconds.
Next, we need to calculate the absolute error for the period. We first find the deviation of each measurement from the mean period:
- Absolute deviation for 99.0s = |20.1 - 19.8| = 0.3s
- Absolute deviation for 99.5s = |20.1 - 19.9| = 0.2s
- Absolute deviation for 100.5s = |20.1 - 20.1| = 0s
- Absolute deviation for 101.5s = |20.1 - 20.3| = 0.2s
- Absolute deviation for 102s = |20.1 - 20.4| = 0.3s
The absolute error is the mean of these deviations:
Absolute error = (0.3 + 0.2 + 0 + 0.2 + 0.3) s / 5 = 1.0s / 5 = 0.2s.
