Answer :
Gravitational field at the pendulum's location is approximately [tex]\( 9.48 \, \text{m/s}^2 \)[/tex] with a period of 1.22 s.
The period (T) of a pendulum can be calculated using the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- [tex]\( T \)[/tex] = period of the pendulum
- [tex]\( L \)[/tex]= length of the pendulum
- [tex]\( g \)[/tex] = gravitational field strength
We can rearrange this formula to solve for [tex]\( g \)[/tex]:
[tex]\[ g = \frac{4\pi^2 L}{T^2} \][/tex]
Given:
- [tex]\( L = 36.9 \)[/tex] cm
- [tex]\( T = 1.22 \)[/tex] s
Let's convert the length from centimeters to meters since the gravitational field strength is usually expressed in meters per second squared (m/s²):
[tex]\[ L = 36.9 \, \text{cm} = 0.369 \, \text{m} \][/tex]
Now, we can plug these values into the formula:
[tex]\[ g = \frac{4\pi^2 \times 0.369}{(1.22)^2} \][/tex]
[tex]\[ g \approx \frac{4 \times (3.14159)^2 \times 0.369}{1.4884} \][/tex]
[tex]\[ g \approx \frac{4 \times 9.8696 \times 0.369}{1.4884} \][/tex]
[tex]\[ g \approx \frac{14.11824}{1.4884} \][/tex]
[tex]\[ g \approx 9.48 \, \text{m/s}^2 \][/tex]