Use a circular pendulum to determine . You can increase the accuracy of the time it takes to go around once by timing$\mathit{N}$ revolutions and then dividing by $\mathit{N}$. This minimizes errors contributed by inaccuracies in starting and stopping the clock. It is wise to start counting from zero $\left(0,1,2,3,4,5\right)$rather than starting from $\left(0,1,2,3,4,5\right)$represents only four revolutions, not five). It also improves accuracy if you start and stop timing at a well-defined event, such as when the mass crosses in front of an easily visible mark. This was the method used by Newton to get an accurate value of $g$. Newton was not only a brilliant theorist but also an excellent experimentalist. For a circular pendulum, he built a large triangular wooden frame mounted on a vertical shaft, and he pushed this around and around while making sure that the string of the circular pendulum stayed parallel to the slanting side of the triangle.

The pendulum's length is $l=1\text{m}$.

Determine how long it took the pendulum to complete 10 oscillations$\left(N=\text{10}\right)$. You know that the pendulum's period is defined by the following formulas.

$T=2\pi \sqrt{\frac{l}{g}}$

Here, $g$is the acceleration due to gravity and $l$is the length.

And,

$T=\frac{t}{N}$

Here,$t$is time and $N$ is the number of oscillations.

According to 10 different time $t$,

$$\begin{gathered} t_1=20.020\text{s,} \\ {t}_2=19.929\text{s,} \\ {t}_3=20.071\text{s,} \\ {t}_4=19.899\text{s,} \\ {t}_5=20.030\text{s,} \\ {t}_6=19.989\text{s,} \\ {t}_7=20.081\text{s,} \\ {t}_8=19.909\text{s,} \\ {t}_9=20.112\text{s,} \\ {t}_{10}=19.949\text{s} \end{gathered}$$

The formula must be inverted above to calculate $g$.

$$\begin{gathered} \frac{t}{N}=2\pi \sqrt{\frac{l}{g}} \\ \sqrt{\frac{l}{g}}=\frac{t}{2\pi N}{/}^2 \\ \frac{l}{g}=\frac{t^2}{4{\pi }^2{N}^2} \end{gathered}$$

$g=\frac{4{\pi }^{2}l{N}^2}{t^2}$ ….. (1)

The constant part of the formula is:

$$\begin{gathered} 4{\pi }^{2}l{N}^2=4\times {\left(3.14\right)}^2\times \left(1\text{m}\right)\times {\left(10\right)}^2 \\ =3943.84\text{m} \end{gathered}$$

Use equation (1) for time $t_1$as below.

$$\begin{gathered} g_1=\frac{4{\pi }^{2}l{N}^2}{{t_1}^2} \\ =\frac{{\displaystyle 3943.84\text{m}}}{{\displaystyle {\left(20.020\text{s}\right)}^2}} \\ =9.84m/s^2 \end{gathered}$$

Use equation (1) for time $t_2$as below.

$$\begin{gathered} g_2=\frac{4{\pi }^{2}l{N}^2}{{t_2}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{1}9.929\text{s}\right)}^2} \\ =9.93m/s^2 \end{gathered}$$

Use equation (1) for time $t_3$as below.

$$\begin{gathered} g_3=\frac{4{\pi }^{2}l{N}^2}{{t_3}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{20.071 s}\right)}^2} \\ =9.79m/s^2 \end{gathered}$$

Use equation (1) for time $t_4$as below.

$$\begin{gathered} g_4=\frac{4{\pi }^{2}l{N}^2}{{t_4}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{19.899 s}\right)}^2} \\ =9.96m/s^2 \end{gathered}$$

Use equation (1) for time $t_5$as below.

$$\begin{gathered} g_5=\frac{4{\pi }^{2}l{N}^2}{{t_5}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{20.03 s}\right)}^2} \\ =9.83m/s^2 \end{gathered}$$

Use equation (1) for time $t_6$as below.

$$\begin{gathered} g_6=\frac{4{\pi }^{2}l{N}^2}{{t_6}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{19.989 s}\right)}^2} \\ =9.87m/s^2 \end{gathered}$$

Use equation (1) for time $t_7$as below.

$$\begin{gathered} g_7=\frac{4{\pi }^{2}l{N}^2}{{t_7}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{20.081 s}\right)}^2} \\ =9.78m/s^2 \end{gathered}$$

Use equation (1) for time $t_8$as below.

$$\begin{gathered} g_8=\frac{4{\pi }^{2}l{N}^2}{{t_8}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{19.909 s}\right)}^2} \\ =9.95m/s^2 \end{gathered}$$

Use equation (1) for time $t_9$as below.

$$\begin{gathered} g_9=\frac{4{\pi }^{2}l{N}^2}{{t_9}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{20.112 s}\right)}^2} \\ =9.75m/s^2 \end{gathered}$$

Use equation (1) for time $t_{10}$as below.

$$\begin{gathered} g_{10}=\frac{4{\pi }^{2}l{N}^2}{{t_{10}}^2} \\ =\frac{3943.84\text{m}}{{\left(\text{19.949 s}\right)}^2} \\ =9.91m/s^2 \end{gathered}$$

The average value of g is:

$g=\frac{g_1+g_2+g_3+g_4+g_5+g_6+g_7+g_8+g_9+g_{10}}{10}$

Substitute the known values in the above equation.

$$\begin{gathered} g=\frac{\left(9.84+9.93+9.79+9.96+9.83+9.87+9.78+9.95+9.75+9.91\right){\displaystyle }{\displaystyle m}{\displaystyle /}{\displaystyle {{\displaystyle s}}^2}}{10} \\ =\frac{{\text{98.61m/s}}^2}{10} \\ =9.86m/s^2 \end{gathered}$$

After ten repetitions of the same measurement, the value of $g$was achieved, which is close to the value of $9.86m/s^2$.