A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be_____ s.
Question
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be_____ s.
Solution
To solve this problem, we need to 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.
However, in this case, the pendulum is not on the surface of the Earth, but at a distance equal to the radius of the Earth. This changes the value of g.
The acceleration due to gravity at a height h from the surface of the Earth is given by:
g' = g * (R/(R+h))^2
where: g' is the new acceleration due to gravity, g is the acceleration due to gravity on the surface of the Earth (approximately 9.81 m/s^2), R is the radius of the Earth (approximately 6.37 x 10^6 m), and h is the height above the surface of the Earth.
In this case, h = R, so the formula simplifies to:
g' = g * (R/(2R))^2 = g/4
Substituting this value into the formula for the period of a simple pendulum, we get:
T = 2π √(L/g') = 2π √(4/ (9.81/4)) = 2π √(16) = 4π s
So, the time period of small oscillations will be 4π seconds.
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