If the length of the second pendulum of the earth is about 1 meter, the length of the second pendulum should be between 0.3 and 0.5 meters.
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).
... 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.
In the eighteenth century, there were two favoured approaches to the definition of the meter. One approach suggested that the metre be defined as the length of a 'seconds pendulum' (pendulum with a half-period of one second). Another suggestion was defining the metre as one ten-millionth of the length of the Earth's meridian along a quadrant (the distance from the Equator to the North Pole).In 1791, the French Academy of Sciences selected the latter definition (the one related to Earth's meridian) over the former (the one with the pendulum) because the force of gravity varies slightly over the surface of the Earth's surface, which affects the period of a pendulum.
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.
Approx 80.5 centimetres.
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.
equation for time in pendulum: t = 2 * pi * ( sq. root (l / g)) key: t = time elapsed ( total, back and forth ) l = length , from pivot to centre of gravity g = acceleration due to gravity say 1 metre length pendulum on earth @ 9.82 (m/s)/s, t = 2.005 seconds same pendulum on neptune @ 11.23 (m/s)/s, t = 1.875 seconds
A 2-meter pendulum at sea level has a period of 2.84 seconds.
The equation is: http://hyperphysics.phy-astr.gsu.edu/HBASE/imgmec/pend.gif T is the period in seconds, L is pendulum length in cm, g is acceleration of gravity in m/s2. We know on earth the period is 1s when the acceleration of gravity is 9.8m/s2, so the pendulum length is 24.824cm. The acceleration of gravity on the moon is 1.6m/s2. Substitute 24.824cm for L and 1.6 for g and you yield 2.475 seconds. The period is 2.475 seconds.
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).
... 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.
This pendulum has a length of 0.45 meters. On the surface of the moon, its period would be 3.31 seconds where g = 1.62m/s^2
Nice problem! I get 32.1 centimeters.
In the eighteenth century, there were two favoured approaches to the definition of the meter. One approach suggested that the metre be defined as the length of a 'seconds pendulum' (pendulum with a half-period of one second). Another suggestion was defining the metre as one ten-millionth of the length of the Earth's meridian along a quadrant (the distance from the Equator to the North Pole).In 1791, the French Academy of Sciences selected the latter definition (the one related to Earth's meridian) over the former (the one with the pendulum) because the force of gravity varies slightly over the surface of the Earth's surface, which affects the period of a pendulum.
Changing the length of a pendulum or the mass of its bob has no effect on g; g is a constant, always equal to 9.8 meters per square second near the surface of Earth.
A pendulum moves not by Earth's rotation, but by gravity pulling down and causing it to swing.