Certainly! To find the acceleration due to gravity using the given length and period of a pendulum, we can use the following fundamental equation for the period of a simple pendulum:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- [tex]\( T \)[/tex] is the period of the pendulum (in seconds),
- [tex]\( L \)[/tex] is the length of the pendulum (in centimeters),
- [tex]\( g \)[/tex] is the acceleration due to gravity (in cm/s[tex]\(^2\)[/tex]).
We are given:
[tex]\[
L = 100 \text{ cm}
\][/tex]
[tex]\[
T = 0.65 \pi \text{ seconds}
\][/tex]
To find [tex]\( g \)[/tex], we need to rearrange the formula to solve for [tex]\( g \)[/tex]:
[tex]\[
T = 2 \pi \sqrt{\frac{L}{g}}
\][/tex]
First, isolate [tex]\(\sqrt{\frac{L}{g}}\)[/tex]:
[tex]\[
\frac{T}{2\pi} = \sqrt{\frac{L}{g}}
\][/tex]
Next, square both sides to eliminate the square root:
[tex]\[
\left(\frac{T}{2\pi}\right)^2 = \frac{L}{g}
\][/tex]
Finally, solve for [tex]\( g \)[/tex]:
[tex]\[
g = \frac{L}{\left(\frac{T}{2\pi}\right)^2}
\][/tex]
Now, substitute the given values into the equation:
[tex]\[
T = 0.65 \pi \text{ seconds}
\][/tex]
[tex]\[
L = 100 \text{ cm}
\][/tex]
Calculate [tex]\(\left(\frac{T}{2\pi}\right)^2\)[/tex]:
[tex]\[
\left(\frac{0.65 \pi}{2 \pi}\right)^2 = \left(\frac{0.65}{2}\right)^2 = \left(0.325\right)^2
\][/tex]
[tex]\[
\left(0.325\right)^2 = 0.105625
\][/tex]
Then, use this result to find [tex]\( g \)[/tex]:
[tex]\[
g = \frac{100}{0.105625}
\][/tex]
[tex]\[
g \approx 946.7455621301774 \text{ cm/s}^2
\][/tex]
Therefore, the acceleration due to gravity for the given pendulum length and period is approximately:
[tex]\[
\boxed{946.7455621301774} \text{ cm/s}^2
\][/tex]
