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i think it is infinite because acceleration due to gravity at the center of the earth is zero and time period of the simple pendulum is given by 2*3.14*sqrt(l/g)....

Q: What is the time period of simple pendulum at center of earth?

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Infinite

At the center of the Earth there would be no effective gravity; a pendulum wouldn't work as a pendulum.

If the plumb point of a pendulum is the center of earth, the pendulum will make diametrical oscillations

Normally the acceleration of gravity is not a factor in the period of a simple pendulum because it does not change on Earth, but if it were to be put on another celestial body the period would change. As gravity increases the period is shorter and as the gravity is less the period is longer.

For small amplitudes, the period can be calculated as 2 x pi x square root of (L / g). Convert the length to meters, and use 9.8 for gravity. The answer will be in seconds. About 1.4 seconds.

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Infinite

At the center of the Earth there would be no effective gravity; a pendulum wouldn't work as a pendulum.

If the plumb point of a pendulum is the center of earth, the pendulum will make diametrical oscillations

at the center of the earth, simple pendulmn has not any gravitational force(if we thought,the earth is an etended object) so at the center the gravitational acceleation is about 'zero' and that's why pendulumn's time period is 'infinite'.

The time period of a pendulum would increases it the pendulum were on the moon instead of the earth. The period of a simple pendulum is equal to 2*pi*√(L/g), where g is acceleration due to gravity. As gravity decreases, g decreases. Since the value of g would be smaller on the moon, the period of the pendulum would increase. The value of g on Earth is 9.8 m/s2, whereas the value of g on the moon is 1.624 m/s2. This makes the period of a pendulum on the moon about 2.47 times longer than the period would be on Earth.

The period of a simple pendulum swinging at a small angle is approximately 2*pi*Sqrt(L/g), where L is the length of the pendulum, and g is acceleration due to gravity. Since gravity on the moon is approximately 1/6 of Earth's gravity, the period of a pendulum on the moon with the same length will be approximately 2.45 times of the same pendulum on the Earth (that's square root of 6).

Yes. The period of the pendulum (the time it takes it swing back and forth once) depends on the length of the pendulum, and also on how strong gravity is. The moon is much smaller and less massive than the earth, and as a result, gravity is considerably weaker. This would make the period of a pendulum longer on the moon than the period of the same pendulum would be on earth.

Normally the acceleration of gravity is not a factor in the period of a simple pendulum because it does not change on Earth, but if it were to be put on another celestial body the period would change. As gravity increases the period is shorter and as the gravity is less the period is longer.

For small amplitudes, the period can be calculated as 2 x pi x square root of (L / g). Convert the length to meters, and use 9.8 for gravity. The answer will be in seconds. About 1.4 seconds.

Approx 80.5 centimetres.

... dependent on the length of the pendulum. ... longer than the period of the same pendulum on Earth. Both of these are correct ways of finishing that sentence.

Nice problem! I get 32.1 centimeters.