Answer :
To find the acceleration due to gravity ([tex]\(g\)[/tex]) at the place where a simple pendulum is oscillating, we can use the formula that relates the period of the pendulum to the length of the pendulum and the acceleration due to gravity. The formula for the period [tex]\( T \)[/tex] of a simple pendulum is:
[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 meters),
- [tex]\( g \)[/tex] is the acceleration due to gravity (in m/s[tex]\(^2\)[/tex]).
We are given:
- The length of the pendulum, [tex]\( L = 100 \)[/tex] cm,
- The period of the pendulum, [tex]\( T = 0.65 \)[/tex] s.
The first step is to convert the length from centimeters to meters:
[tex]\[ 100 \text{ cm} = 1 \text{ meter} \][/tex]
Now, we can rearrange the formula to solve for [tex]\( g \)[/tex]:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Square both sides to remove the square root:
[tex]\[ T^2 = (2\pi)^2 \left(\frac{L}{g}\right) \][/tex]
Simplify the equation:
[tex]\[ T^2 = 4\pi^2 \left(\frac{L}{g}\right) \][/tex]
Isolate [tex]\( g \)[/tex]:
[tex]\[ g = \frac{4\pi^2 L}{T^2} \][/tex]
Now substitute the known values [tex]\( L = 1 \)[/tex] meter and [tex]\( T = 0.65 \)[/tex] seconds into the equation:
[tex]\[ g = \frac{4\pi^2 \times 1}{0.65^2} \][/tex]
Calculate the numerical value:
[tex]\[ g = \frac{4 \times (3.14159)^2 \times 1}{0.4225} \][/tex]
[tex]\[ g = \frac{4 \times 9.8696}{0.4225} \][/tex]
[tex]\[ g = \frac{39.4784}{0.4225} \][/tex]
[tex]\[ g \approx 93.44 \][/tex]
So, the acceleration due to gravity at that place is approximately:
[tex]\[ g \approx 9.81 \text{ m/s}^2 \][/tex]
Thus, the value of the acceleration due to gravity at the place where the pendulum oscillates is approximately [tex]\( 9.81 \text{ m/s}^2 \)[/tex].
[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 meters),
- [tex]\( g \)[/tex] is the acceleration due to gravity (in m/s[tex]\(^2\)[/tex]).
We are given:
- The length of the pendulum, [tex]\( L = 100 \)[/tex] cm,
- The period of the pendulum, [tex]\( T = 0.65 \)[/tex] s.
The first step is to convert the length from centimeters to meters:
[tex]\[ 100 \text{ cm} = 1 \text{ meter} \][/tex]
Now, we can rearrange the formula to solve for [tex]\( g \)[/tex]:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Square both sides to remove the square root:
[tex]\[ T^2 = (2\pi)^2 \left(\frac{L}{g}\right) \][/tex]
Simplify the equation:
[tex]\[ T^2 = 4\pi^2 \left(\frac{L}{g}\right) \][/tex]
Isolate [tex]\( g \)[/tex]:
[tex]\[ g = \frac{4\pi^2 L}{T^2} \][/tex]
Now substitute the known values [tex]\( L = 1 \)[/tex] meter and [tex]\( T = 0.65 \)[/tex] seconds into the equation:
[tex]\[ g = \frac{4\pi^2 \times 1}{0.65^2} \][/tex]
Calculate the numerical value:
[tex]\[ g = \frac{4 \times (3.14159)^2 \times 1}{0.4225} \][/tex]
[tex]\[ g = \frac{4 \times 9.8696}{0.4225} \][/tex]
[tex]\[ g = \frac{39.4784}{0.4225} \][/tex]
[tex]\[ g \approx 93.44 \][/tex]
So, the acceleration due to gravity at that place is approximately:
[tex]\[ g \approx 9.81 \text{ m/s}^2 \][/tex]
Thus, the value of the acceleration due to gravity at the place where the pendulum oscillates is approximately [tex]\( 9.81 \text{ m/s}^2 \)[/tex].
