What length of pendulum has a period of 0.9 s on Earth?
Period=2PI sqrt (l/g) right?
T=2π•sqrt(L/g)
g=9.8 m/s²
Solve for ‘L’
To find the length of a pendulum with a period of 0.9 s on Earth, we can use the formula for the period of a simple pendulum:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
First, let's determine the value of g on Earth. The standard value for the acceleration due to gravity on the Earth's surface is approximately 9.8 m/s².
Next, we can rearrange the formula and solve for L:
T = 2π√(L/g)
0.9 = 2π√(L/9.8)
To isolate L, we'll square both sides of the equation:
0.9² = (2π)²(L/9.8)
0.81 = 4π²(L/9.8)
Now, we can solve for L:
L/9.8 = 0.81/(4π²)
L/9.8 ≈ 0.0082
L ≈ 0.0082 * 9.8
L ≈ 0.0804
Therefore, a pendulum with a length of approximately 0.0804 meters (or 8.04 cm) will have a period of 0.9 seconds on Earth.