In successive experiments while measuring the period of oscillation of a simple pendulum. The readings turn out to be 2.63 s , 2.56 s , 2.42 s , 2.71 s and 2.80 s. Calculate the mean absolute error.

Question

In successive experiments while measuring the period of oscillation of a simple pendulum. The readings turn out to be 2.63 s , 2.56 s , 2.42 s , 2.71 s and 2.80 s. Calculate the mean absolute error. (A) 0.11 s (B) 0.42 s (C) 0.92 s (D) 0.10 s

Tardigrade Admin · Accepted Answer

The mean period of oscillation of the pendulum,

T_{mean} = \frac{2.63 + 2.56 + 2.42 + 2.71 + 2.80}{5}

= \frac{13.12}{5} = 2.624 = 2.62 s

The absolute errors in the measurements are

Δ T_{1} = 2.63 s - 2.62 s = 0.01 s

Δ T_{2} = 2.56 s - 2.62 s = - 0.06 s

Δ T_{3} = 2.42 s - 2.62 s = - 0.20 s

Δ T_{4} = 2.71 s - 2.62 s = 0.09 s

Δ T_{5} = 2.80 s - 2.62 s = 0.18 s

The arithmetic mean of all the absolute errors is

Δ T_{mean} = i = 1 \sum 5 \frac{∣ Δ T _{i} ∣}{5}

= [(0.01 + 0.06 + 0.20 + 0.09 + 0.18)] /5

= 0.54/5 = 0.108 = 0.11 s

Q. In successive experiments while measuring the period of oscillation of a simple pendulum. The readings turn out to be $2.63 s, 2.56 s, 2.42 s, 2.71 s$ and $2.80 s$ . Calculate the mean absolute error.

$0.11 s$

$0.42 s$

$0.92 s$

$0.10 s$

Q. In successive experiments while measuring the period of oscillation of a simple pendulum. The readings turn out to be 2.63s,2.56s,2.42s,2.71s and 2.80s. Calculate the mean absolute error.

0.11s

0.42s

0.92s

0.10s

Q. In successive experiments while measuring the period of oscillation of a simple pendulum. The readings turn out to be $2.63 s, 2.56 s, 2.42 s, 2.71 s$ and $2.80 s$ . Calculate the mean absolute error.

$0.11 s$

$0.42 s$

$0.92 s$

$0.10 s$