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In an experiment of simple pendulum a student made several observations for the periodof oscillation. His readings turned out to be 2.63s, 2.56s, 2.42s, 2.71s, and 2.80s. Find (i) mean time period of oscillations or most accurate value of time period, (ii) absolute error in each reading (iii) mean absolute error (iv) fractional error and (v) percentage error. |
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Answer» Solution :(i) The MEAN time period of oscillation T=`(2.63 +2.56+2.42+2.71+2.80)/(5)` `=(13.12)/(5)s = 2.624 s~~ 2.62s` (ROUNDED off to 2nd decimal PALCE ) (ii) Taking 2.62s as the true value the absolute errors (true value - measured value ) in the five READINGS are, (2.62 - 2.63) s= -0.01s, (2.62-2.56) s = 0.06s, (2.62 - 2.42)s = 0.20s, (2.62 - 2.71)s = -0.09 s and (2.62-2.80) s =-0.18s (iii) The (maximum) mean absolute error is `(deltaT)_("max")=(0.01+0.06+0.20+0.09+0.18)/(5)`s `=(0.54)/(5)s = 0.108 s ~~ 0.11 `s (iv) The (maximum) fractional error is `((deltaT)/(T))_("max")=(0.11s)/(2.62s)~~0.04` (v) The maximum percentage error is `((deltaT)/(T))_("max")xx100 = 0.04 xx100 =4%` `:.` The value of T should be written as (`2.62 pm 0.11)`s |
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