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It is assumed that the question asks "for the same period".
The period of a simple pendulum, for "very short swings", is 2 pi (L/G)0.5. Since G on the moon is 0.165 that of Earth, L would have to be 0.165, so that 1 m pendulum would have to be about 0.165 m long in order to give the same swing period.
What else can I help you with?
If a simple pendulum with a period of 1 second is set in motion of the moon what is the new period of this pendulum?
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.
What does the frequency of a pendulum depend on?
The frequency of a pendulum depends on the length of the pendulum and the acceleration due to gravity. It is described by the equation f = 1 / (2π) * √(g / L), where f is the frequency, g is the acceleration due to gravity, and L is the length of the pendulum.
What time does the pendulum take for one swing?
The time it takes for a pendulum to complete one full swing is determined by the length of the pendulum and the acceleration due to gravity. The formula for the period of a pendulum is T = 2π√(l/g), where T is the period, l is the length of the pendulum, and g is the acceleration due to gravity. Typically, a pendulum with a length of 1 meter will take about 2 seconds to complete one swing.
What are two factors that alter the oscillation period of a pendulum?
The length of the pendulum and the acceleration due to gravity are two factors that can alter the oscillation period of a pendulum. A longer pendulum will have a longer period, while a stronger gravitational force will result in a shorter period.
What is the length of simple pendulum which ticks seconds?
What you want is a pendulum with a frequency of 1/2 Hz. It swings left for 1 second,then right for 1 second, ticks once in each direction, and completes its cycle in exactly2 seconds.The length of such a pendulum technically depends on the acceleration due to gravityin the place where it's swinging. In fact, pendulum arrangements are used to measurethe local value of gravity.A good representative value for the length of the "seconds pendulum" is 0.994 meter.
The lenth of second's pendulum of the earth is about 1 m What should be the length of second's pendulum on the moon?
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.
What is the difference in period for a pendulum on earth and a pendulum on moon?
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).
What is length of 2nd pendulum on moon?
The length of a pendulum that has a period of 2 seconds is approximately 0.25 meters on Earth. On the moon, where gravity is about 1/6th of Earth's gravity, the length of a 2-second pendulum would be about 0.73 meters.
What is the periodic time of a 0.5m pendulum on the moon?
note: (g(moon)= 1/6g(earth))
If a simple pendulum with a period of 1 second is set in motion of the moon what is the new period of this pendulum?
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.
How does the length of the pendulum effect the pendulum?
The longer the length of the pendulum, the longer the time taken for the pendulum to complete 1 oscillation.
What is the length of rotation of the earth's moon?
the length of the moon's rotation is 27.3 days (about 1 month)
Sketch of graph of pendulum on moon?
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.
Does the period of a pendulum increase or decrease on Moon?
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.
How is the period of the pendulum affected by its length?
the time period of a pendulum is proportional to the square root of length.if the length of the pendulum is increased the time period of the pendulum also gets increased. we know the formula for the time period , from there we can prove that the time period of a pendulum is directly proportional to the effective length of the pendulum. T=2 pi (l\g)^1\2 or, T isproportionalto (l/g)^1/2 or, T is proportional to square root of the effective length.
What does the frequency of a pendulum depend on?
The frequency of a pendulum depends on the length of the pendulum and the acceleration due to gravity. It is described by the equation f = 1 / (2π) * √(g / L), where f is the frequency, g is the acceleration due to gravity, and L is the length of the pendulum.
How would the period of a simple Pendulum be charged if the pendulum were moved from sea level to the moon?
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.
