is the time period of simple pendulum on the earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:
Question
is the time period of simple pendulum on the earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:
Solution
The time period of a simple pendulum is given by the formula T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity.
On the surface of the Earth, g is approximately 9.8 m/s².
When the pendulum is taken to a height R above the Earth's surface, the acceleration due to gravity changes. It becomes g' = g*(R/(R+h))^2, where R is the Earth's radius and h is the height above the Earth's surface. In this case, h = R, so g' = g*(R/(2R))^2 = g/4.
The time period of the pendulum at this height is T' = 2π√(l/g') = 2π√(l/(g/4)) = 2T.
Therefore, the value of x, which is the ratio of the time period at height R to the time period at the Earth's surface, is x = T'/T = 2.
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