A student measures the time period of a simple pendulum 4 times. The mean time period is calculated to be 2.00 s . If three of the recorded readings are 1.98 s , 1.99 s , and 2.02 s , what is the percentage relative error of the measurements?

Step-by-step solution

First, find the fourth reading. The mean of 4 readings is 2.00 s , so their sum is 4 2.00 = 8.00 s . The sum of the three given readings is 1.98 + 1.99 + 2.02 = 5.99 s . Therefore, the fourth reading is 8.00 - 5.99 = 2.01 s . Next, calculate the absolute errors for each reading: T₁ = |1.98 - 2.00| = 0.02 s T₂ = |1.99 - 2.00| = 0.01 s T₃ = |2.02 - 2.00| = 0.02 s T₄ = |2.01 - 2.00| = 0.01 s Now, find the mean absolute error: T_m = 0.02 + 0.01 + 0.02 + 0.01 4 = 0.06 4 = 0.015 s Finally, calculate the percentage relati