The acceleration due to gravity on the moon is 1.62 m/s2.
What is the length of a pendulum whose period on the moon matches the period of a 2.00-m-long pendulum on the earth?
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We know that period of a pendulum is given by,
T = 2*pi*sqrt ( L/g)
where T is the period , L is the length of the pendulum and g Is the acceleration due to gravity
In Earth, the length of the pendulum is 2 m,
T (earth)= 2* pi 8sqrt (L/9.81)
And on moon, let the length be L
T(moon)= 2*pi* sqrt (L/1.62)
Since the period of the pendulum are same, we can conclude that,
T(earth) = T (moon)
2 8pi *sqrt (2/9.81) = 2*pi* sqrt (L/1.62)
=sqrt (2/9.81) = sqrt (L/1.62)
2/9.81 = L/ 1.62 [squaring both side]
L = (2/9.81)*1.62 = 0.33 meter
When working with a pendulum, it is important to know the length of the the pendulum and the gravitational pull that is pulling on the mass of the pendulum. The gravitational attraction on mass here on Earth is 9.8 m/s2, while the gravitational attraction of mass on the moon is roughlyu one-sixth of that, 1.6 m/s2. So if you have a 2 meter length of the pendulum on the Earth, and the gravitational attraction is one-sixth that of the Earth, the simplest deduction of the length of the pendulum on the moon would be to divide the Earth length by six, which would give you .33 meters: 2/6 = .33 What this means is the gravitational attraction on the moon is less, so to get the same period swing, or frequency, as the Earth pendulum, you would have to shorten the length of the moon pendulum, which would cause it to accelerate its frequency, which would then equal the frequency of the pendulum on Earth.
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