Solveeit Logo
Login

Question

Question: In an experiment, the period of oscillation of a simple pendulum was observed to be 2.63 s, 2.56 s, ...

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

Answer

0.11 s

Explanation

Solution

The mean period of oscillations of the determinations of significant figures.

Tmean =∑i=1nTin\mathrm { T } _ { \text {mean } } = \frac { \sum _ { \mathrm { i } = 1 } ^ { \mathrm { n } } \mathrm { T } _ { \mathrm { i } } } { \mathrm { n } }

Tmean =(2.63+2.56+2.42+2.741)5 s\mathrm { T } _ { \text {mean } } = \frac { ( 2.63 + 2.56 + 2.42 + 2.741 ) } { 5 } \mathrm {~s}

=13.125 s=2.624 s=2.62 s= \frac { 13.12 } { 5 } \mathrm {~s} = 2.624 \mathrm {~s} = 2.62 \mathrm {~s}

(Rounded off to two decimal places

The absolute errors in the measurement are

ΔT1=2.62 s−2.63 s−0.01 s\Delta \mathrm { T } _ { 1 } = 2.62 \mathrm {~s} - 2.63 \mathrm {~s} - 0.01 \mathrm {~s} ΔT2=2.62 s−2.56 s=0.06 s\Delta \mathrm { T } _ { 2 } = 2.62 \mathrm {~s} - 2.56 \mathrm {~s} = 0.06 \mathrm {~s} ΔT3=2.62 s−2.42 s=0.20 s\Delta \mathrm { T } _ { 3 } = 2.62 \mathrm {~s} - 2.42 \mathrm {~s} = 0.20 \mathrm {~s} ΔT4=2.62 s−2.71 s=−0.09 s\Delta \mathrm { T } _ { 4 } = 2.62 \mathrm {~s} - 2.71 \mathrm {~s} = - 0.09 \mathrm {~s} ΔT5=2.62 s−2.80=−0.18 s\Delta \mathrm { T } _ { 5 } = 2.62 \mathrm {~s} - 2.80 = - 0.18 \mathrm {~s}

Means absolute error is

ΔTmean =∑i=1n∣ΔTi∣n\Delta T _ { \text {mean } } = \frac { \sum _ { i = 1 } ^ { \mathrm { n } } \left| \Delta T _ { i } \right| } { n } ΔTmena =(0.01+0.06+0.20+0.09+0.18)5\Delta \mathrm { T } _ { \text {mena } } = \frac { ( 0.01 + 0.06 + 0.20 + 0.09 + 0.18 ) } { 5 } =0.545 s=0.11 s= \frac { 0.54 } { 5 } \mathrm {~s} = 0.11 \mathrm {~s}