91Ó°ÊÓ

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.

$T=2\mathrm{Ï€}\sqrt{\frac{l}{g}}$

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 $g$.

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

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

The constant part of the formula is:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $9.86m/s^2$.