Question

The time period of oscillation of a simple pendulum in
an experiment is recorded as 2.56 s, 2.62 s, 2.70 s,
2.58 s, 2.45 s respectively. Find the time period,
absolute error in each observation and the percentage
error.
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Answer 1

Answer attached

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Answer 2

Answer:

The time period is 2.582sec

Absolute error in each observation is 0.0624sec

the percentage error is 2.14%

Explanation:

Given

The time period of oscillation of a simple pendulum in

an experiment is recorded as 2.56 s, 2.62 s, 2.70 s,

2.56 s, 2.62 s, 2.70 s,2.58 s, 2.45 s respectively.

let us consider,

• t1=2.56,
• t2 =2.62,
• t3=2.70,
• t4=2.58,
• t5=2.45.

Time period:-

To know time we need to find the mean value mean value = sum of observations/ no.of observation Here number of observations=5

Mean of this values is equal to

=2.56+2.62+2.70+2.58+2.45/5

=2.582sec

Hence Timeperiod is 2.582 sec

Absolute error:-

To find absolute error we need subtract given value from mean value

Absolute error of each observation is

• A1 = T-t1 = 2.582-2.56 =0.022.
• A2 = T-t2 = 2.582-2.62 =-0.038
• A3 = T-t3 = 2.582-2.70 =-0.118
• A4 = T-t4 = 2.582-2.58 =0.002
• A5 = T-t5 = 2.582-2.45 =0.132

Mean Absolute Error is equal to ;;

A1+A2+A3+A4+A5/5 (In magnitude).

So,

=0.022+0.038+0.118+0.002+0.132/5

=0.0624.

Mean Absolute Error is 0.0624 sec

Percentage error:-

Percentage Error = 0.0624/2.582×100

==2.14%.

Hence percentage error is 2.14%

• The time period is 2.582sec
• Mean absolute error is 0.0624sec
• the percentage error is 2.14%