The period of oscillation of a simple pendulum is T = 2π√(l/g). The value of l is 20cm known to 1mm accuracy, and the time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the accuracy in the determination of g?

Halijoto ya awali, $T_1 = (63.5 \pm 0.5)^\circ C$
Halijoto ya mwisho, $T_2 = (72.6 \pm 0.4)^\circ C$

Ongezeko la halijoto, $\Delta T = T_2 - T_1$.
Kwa thamani za wastani:
$\Delta T_{wastani} = 72.6^\circ C - 63.5^\circ C = 9.1^\circ C$

Wakati wa kutoa au kujumlisha vipimo, uhakika kamili huongezwa.
$\Delta (\Delta T) = \Delta T_1 + \Delta T_2 = 0.5^\circ C + 0.4^\circ C = 0.9^\circ C$

Ongezeko la halijoto ni $\Delta T = (\Delta T_{wastani} \pm \Delta (\Delta T))$.

Urefu, $l = 20 \text{ cm}$. Usahihi ni $1 \text{ mm}$.
Hivyo, $\Delta l = 1 \text{ mm} = 0.1 \text{ cm}$.
$l = (20.0 \pm 0.1) \text{ cm}$

Muda kwa mitetemo 100, $t = 90 \text{ s}$. Azimio la saa ni $1 \text{ s}$.
Hivyo, $\Delta t = 1 \text{ s}$.
$t = (90 \pm 1) \text{ s}$

Kipindi $T = \frac{t}{\text{idadi ya mitetemo}} = \frac{90 \text{ s}}{100} = 0.9 \text{ s}$.
Uhakika katika kipindi $\Delta T = \frac{\Delta t}{\text{idadi ya mitetemo}} = \frac{1 \text{ s}}{100} = 0.01 \text{ s}$.
Hivyo, $T = (0.90 \pm 0.01) \text{ s}$.

Fomula ni $T = 2\pi \sqrt{\frac{l}{g}}$.
Piga mraba pande zote mbili: $T^2 = (2\pi)^2 \frac{l}{g}$
$T^2 = 4\pi^2 \frac{l}{g}$
Tenga $g$: $g = \frac{4\pi^2 l}{T^2}$

Kwa fomula $g = \frac{4\pi^2 l}{T^2}$, uhakika wa asilimia (au uhakika wa sehemu) huhesabiwa kama:

Kokotoa uhakika wa sehemu kwa $l$:

Kokotoa uhakika wa sehemu kwa $T$:

Jumlisha uhakika wa sehemu:

Ili kujumlisha, badilisha kuwa sehemu moja au desimali:

Badilisha kuwa asilimia:

Kuzungusha hadi tarakimu muhimu mbili, tunapata $2.7\%$.