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We measure the period of oscillation of ...

We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to the 2.63 sec, 2.56 sec, 2.42 se, 2.71 sec, 2.80 sec

Text Solution

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The mean period of oscillation of the pendulum
`T=((2.63+2.56+2.42+2.71+2.80)s)/5`
`" "=(13.22)/5 s =2.624 s=2.62s`
As the periods are measured to a resolution of 0.01 s, all times are to the second decimal, it is proper to put this mean period also to the second decimal.
The errors in the measurements are
2.63 s - 2.62 s = 0.01 s
2.56 s - 2.62 s = -0.06 s
2.42 s - 2.62 s = -0.20 s
2.71 s - 2.62 s = 0.09 s
2.80 s - 2.62 s = 0.18 s
Note that the errors have the same units as the quantity to be measured.
The arithmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes) is
`DeltaT_("mean")=[(0.01+0.06+0.20+0.09+0.18s]//5=0.54 s//5=0.11s`
That means, the period of oscillation of the simple pendulum is `(2.62 pm 0.11)s` i.e. it lies between (2.62 + 0.11) s and (2.62 - 0.11) s or between 2.73 s and 2.51 s. As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a second. Hence there is no point is giving the period to a hundredth. A more correct way will be to write `T=2.6 pm0.1s`.
Note that the last numeral 6 is unreliable, since it may be anything between 5 and 7. We indicate this by saying that the measurements has two significant figures. In this case, the two significant figures are 2, which is reliable and 6, which has an error associated with it. You will learn more about the significant figures in section 2.7.
For this example, the relative error or the percentage error is `deltaa=(0.1)/(2.6) times 100=4.`
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