Question

The length of a pendulum is (1.5 +0.01) m and the acceleration due
gravity is taken into account as (9.8 +0.1) ms'. Calculate the time per
of the pendulum with uncertainty in it?​

Answer 1

Answer:

Since the uncertainty is stated to the nearest hundredth of a meter, I assume the length is given to the nearest hundredth of a meter, so you need to write it as 1.50 m not 1.5 m. Also, the units of the acceleration due to gravity are m/s/s.

The period of a simple pendulum is:

T=2πlg−−√

The uncertainty is gotten from:

T±ΔT=2π(l±Δl)1/2(g±Δg)−1/2

T±ΔT=T(1±Δll)1/2(1±Δgg)−1/2

Taking the first two terms of the series expansion of both factors yields:

T±ΔT≈T(1±Δl2l)(1±Δgg)

Multiplying it out and dropping the much smaller product term gives:

T≈T(1±Δl2l±Δgg)

With the numbers given, this becomes:

T≈2.46(1±1/300±1/200)≈(2.46±0.05) s

So the uncertainty is 0.05 sec. It would not make sense to quote an uncertainty that is more accurate than one significant figure.

Explanation:

Hope it will help you!