Question

time period is measured by simple pendulum having reading 2.63, 2.56, 2.42, 2.71, 2.80. find the true value of time period ​

Answer 1

Here is your answer

The mean period of oscillation will be

T=2.63+2.56+2.42+2.71+2.8/5=2.62 sec

The errors in the measurement wil be 2.63-2.62=.01sec

2.56-2.62=-.06sec

2.42-2.62=-.2sec

2.71-2.62=.09sec

2.80-2.62=.18 sec

so mean of absolute error will be .11sec

Hence Period of oscillation will be (2.62±0.11)Therefore T=2.6±.1s

relative error or percentage error is given by=0.1/2.6*100

=4%

Answer 2

 ...    

Arithmetic mean,a mean = 2.63+2.56+2.42+2.71+2.80 / 5

= 13.12 / 5

a mean = 2.624 s

Absolute error,

delta a1 = 2.63 - 2.62

= 0.01 s

delta a2 = 2.56 - 2.62

= 0.06 s

delta a3 = 2.42 - 2.62

=0.20 s

delta a4 = 2.71 - 2.62

= 0.09 s

delta a5 = 2.80 - 2.62

= 0.18 s

delta a mean =|delta a1|+|delta a2|+|delta a3|+|delta a4|+|delta a5| / 5

= 0.108

= 0.11 s

a = a mean (+/-) delta a mean

= 2.62 (+/-) 0.11

=2.51 < 2.62 < 2.73

Relative error = 0.11 / 2.6

= 0.04 s

Percentage error = delta a mean / a mean x 100

= 0.11 / 2.6 x 100

= 4 %

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