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.
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.
Nice problem! I get 32.1 centimeters.
About 40.7% of that on Earth or about 2.46 times slower.
Increases.
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.
... 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.
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.
Nice problem! I get 32.1 centimeters.
About 40.7% of that on Earth or about 2.46 times slower.
Increases.
The period of a pendulum (for short swings) is about 2 PI (L/g)1/2. The gravity on the moon is less than that on Earth by a factor of six, so the period of the pendulum on the moon would be greater, i.e. slower, by about a factor of 2.5.
This pendulum, which is 2.24m in length, would have a period of 7.36 seconds on the moon.
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
The period is not likely to be charged. However, it would change due to the weaker gravitational force on the moon. Since the surface gravity of the moon is 0.165 that of the earth, the period would increase by a multiple of 1/sqrt(0.165) = 2.462 approx.
The question as asked is tough to answer without some assumptions... The question implies that a comparison is being made to the action of the same pendulum on earth. With that assumption... The graph I assume has time on the x-axis and a form of pendulum oscillating measurement (such as height or back (-1) to forward (+1) ) on the y-axis. The period ( time from peak to peak on the y-axis ) of the pendulum on the moon compared to the same on earth will be 6 times longer assuming that gravity on the moon is 1/6th that of earth. The reason why the period is longer is that the acceleration (gravity) on the moon is much less. This causes the pendulum on the moon to move back and forth less quickly.
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.