Time period of a simple pendulum is T 1 when on the earth's surface and T 2 when taken to a height '2R' above the earth's surface, where 'R' is the radius of the eart

Question

Time period of a simple pendulum is T 1 ​ when on the earth's surface and T 2 ​ when taken to a height '2R' above the earth's surface, where 'R' is the radius of the earth. The ratio T 1 ​ : T 2 ​ is

TestHub · Accepted Answer

The time period of a simple pendulum is given by $T = 2 π \frac{L}{g}$ .

The acceleration due to gravity at a distance $r$ from the center of the earth is $g = \frac{GM}{r ^{2}}$ .

Substituting this into the time period formula gives $T = 2 π \frac{L r ^{2}}{GM}$ , which implies $T \propto r$ .

On the surface of the earth, the distance from the center is $r_{1} = R$ , so $T_{1} \propto R$ .

At a height $h = 2 R$ above the surface, the distance from the center is $r_{2} = R + 2 R = 3 R$ , so $T_{2} \propto 3 R$ .

The ratio of the time periods is $\frac{T _{1}}{T _{2}} = \frac{R}{3 R} = \frac{1}{3}$ .

Thus, $T_{1} : T_{2} = 1 : 3$ .

Answer: $1 : 3$

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