In an experiment, the period of oscillation of a simple pendulum was observed to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. The mean absolute error is

Question

In an experiment, the period of oscillation of a simple pendulum was observed to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. The mean absolute error is (A) 0.11 s (B) 0.12 s (C) 0.13 s (D) 0.14 s

Tardigrade Admin · Accepted Answer

The mean period of oscillation of the pendulum is

T_{m e an} = \frac{i = 1 \sum n T _{i}}{n}

;

T_{m e an} = \frac{( 2.63 + 2.56 + 2.42 + 2.71 + 2.80 )}{5} s

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

(Rounded off to two decimal places)
The absolute errors in the measurement are

Δ T_{1} = 2.62 s - 2.63 s = - 0.01 s

;

Δ T_{2} = 2.62 s - 2.56 s = 0.06 s

Δ T_{3} = 2.62 s - 2.42 s = 0.20 s

;

Δ T_{4} = 2.62 s - 2.71 s = - 0.09 s

Δ T_{5} = 2.62 s - 2.80 s = - 0.18 s

Mean absolute error is

Δ T_{m e an} = \frac{i = 1 \sum n ∣ Δ T _{i} ∣}{n}

Δ T_{m e an} = \frac{( 0.01 + 0.06 + 0.20 + 0.09 + 0.18 )}{5} s

= \frac{0.54}{5} s = 0.11 s

Q. In an experiment, the period of oscillation of a simple pendulum was observed to be $2.63 s$ , $2.56 s$ , $2.42 s$ , $2.71 s$ and $2.80 s$ . The mean absolute error is

$0.11 s$

$0.12 s$

$0.13 s$

$0.14 s$

Q. In an experiment, the period of oscillation of a simple pendulum was observed to be 2.63s, 2.56s, 2.42s, 2.71s and 2.80s. The mean absolute error is

0.11s

0.12s

0.13s

0.14s

Q. In an experiment, the period of oscillation of a simple pendulum was observed to be $2.63 s$ , $2.56 s$ , $2.42 s$ , $2.71 s$ and $2.80 s$ . The mean absolute error is

$0.11 s$

$0.12 s$

$0.13 s$

$0.14 s$