In an experiment, the period of oscillation of a simple pendulum was observed to be $2.63 \mathrm{~s

Question

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

Professor PC Thomas Classes · Answer

(a) : The mean period of oscillation of the pendulum is $ \begin{aligned} T_{	ext {mean }} & =\frac{\sum_{i=1}^n T_i}{n}=\frac{(2.63+2.56+2.42+2.71+2.80)}{5} \mathrm{~s} \ & =\frac{13.12}{5} \mathrm{~s}=2.624 \mathrm{~s}=2.62 \mathrm{~s} \end{aligned} \ $ (Rounded off to two decimal places) The absolute errors in the measurement are $ \begin{aligned} & \Delta T_1=2.62 \mathrm{~s}-2.63 \mathrm{~s}=-0.01 \mathrm{~s} ; \Delta T_2=2.62 \mathrm{~s}-2.56 \mathrm{~s}=0.06 \mathrm{~s} \ & \Delta T_3=2.62 \mathrm{~s}-2.42 \mathrm{~s}=0.20 \mathrm{~s} ; \Delta T_4=2.62 \mathrm{~s}-2.71 \mathrm{~s}=-0.09 \mathrm{~s} \ & \Delta T_5=2.62 \mathrm{~s}-2.80 \mathrm{~s}=-0.18 \mathrm{~s} \end{aligned} $ \  Mean absolute error is \ $ \begin{aligned} \Delta T_{	ext {mean }} & =\frac{\sum_{i=1}^n\left|\Delta T_iight|}{n} \ \Delta T_{	ext {mean }} & =\frac{(0.01+0.06+0.20+0.09+0.18)}{5} \mathrm{~s}=\frac{0.54}{5} \mathrm{~s}=0.11 \mathrm{~s} \end{aligned} $