i) Time for $20$ oscillations of a pendulum is measured as

$t_{1} = 39.6 s; t_{2} = 39.9 s a n d t_{3} = 39.5 s$ . What is the precision in the measurements?

ii) Time for $20$ oscillations of a pendulum is measured as
$t_{1} = 39.6 s; t_{2} = 39.9 s a n d t_{3} = 39.5 s$ . The precision in the measurements is $0.005 s$ . What is the accuracy of the measurement?

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Solution

(i) Given,

Time measurement for 20 oscillations of a pendulum.

$t_{1} = 39.6 s$

$t_{2} = 39.9 s$

$t_{3} = 39.5 s$

Since, all the measurements have a maximum of one decimal place, the lease count of the instrument used to measure the time is $0.1 s$ .

Since precision in measurement is equal to the least count of the instrument, therefore
Precision in $20$ oscillations = $0.1 s$

$∴ P r e c i s i o n i n 1 o s c i l l a t i o n = \frac{0.1}{20} = 0.005 s$

Final Answer: $0.005 s$

(ii) Step:1 Find the mean value of time for 20 oscillations.

The mean value of time for $20$ oscillations

$t = \frac{t_{1} + t_{2} + t_{3}}{3}$

= $\frac{39.6 + 39.9 + 39.5}{3}$

= $39.66 s$

Step:2 Find the accuracy of measurement.

Mean time period of one oscillation

= $\frac{39.66}{20} \approx 1.98 s$

Rounding off the time period of second pendulum = $2 s$

Measured time period of the second pendulum = $2 - 0.005 = 1.995 s$

Accuracy of measurement is the maximum observed error, and is given by

= $1.995 - 1.980 = 0.015 s$

Final answer: $0.015 s$

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Similar questions

t_{1} = 39.6 s, t_{2} = 39.9 s

t_{3} = 39.5 s

2.00 s

0.05 s

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