The moon's gravity is one-sixth that of the earth. What is the period of a 0.25 m long pendulum on the moon?
sqrt (ge/gm) = sqrt 6 times period on earth
No wait I got it!! thankssss
You are welcome :)
To determine the period of a pendulum on the moon, we need to use the formula for the period of a pendulum:
T = 2π√(L/g)
Where:
T = Period of the pendulum
L = Length of the pendulum
g = Acceleration due to gravity
Given that the length of the pendulum is 0.25 m, and the moon's gravity is one-sixth that of the Earth's gravity, we can calculate the value of "g" for the moon.
First, we need to determine the value of "g" on the moon.
Let's denote "g_moon" as the acceleration due to gravity on the moon.
We know that g_moon = (1/6) * g_earth
Assuming that the acceleration due to gravity on Earth is 9.8 m/s², we can calculate the moon's acceleration due to gravity:
g_moon = (1/6) * 9.8 m/s²
g_moon = 1.63 m/s²
Now that we have the value of "g_moon," we can substitute it into the formula for the period of the pendulum to find the answer:
T = 2π√(L/g_moon)
T = 2π√(0.25 m / 1.63 m/s²)
Calculating this further, we get:
T ≈ 2.44 seconds
Therefore, the period of a 0.25 m long pendulum on the moon is approximately 2.44 seconds.