the acceleration of free fall on the moon is one sixth of that on earth. what would be the period on the moon of a simple pendulum which has a period of one second on the earth?
period=2PI sqrt(g/length)
so the period on the moon will be
sqrt(1/6) that of the period on Earth.
To find the period of a pendulum on the moon, given that the acceleration due to gravity on the moon is one sixth of that on Earth, we can use the formula for the period of a simple pendulum:
T = 2π√(L/g)
Where:
T = period of the pendulum
L = length of the pendulum
g = acceleration due to gravity
Let's assume the length of the pendulum remains the same on both the moon and Earth.
On Earth, the period is given as 1 second. We need to find the period on the moon.
Given that the acceleration due to gravity on the moon is one sixth of that on Earth, we can say that g_moon = (1/6) * g_earth.
To find the period on the moon, we substitute these values into the formula:
T_moon = 2π√(L/g_moon)
= 2π√(L/((1/6) * g_earth))
= 2π√(6L/g_earth)
Since g_earth cancels out, the final expression for the period on the moon becomes:
T_moon = 2π√(6L/g_earth)
Therefore, the period of the simple pendulum on the moon would be approximately 2π√(6) times longer than on Earth.