How do you find the ratio of the new/ old world periods of a pendulum if the pendulum wee transported from earth to the moon, where the acceleration due to gravity is 1.63m/s^2?

Im not sure if we use 9.8 at all, I think I'm supposed to use T=2 pi sqrtL/g.

You substitute g of moon for g of earth in the equation; Then do the calculation for g of earth; divide the two to find the ratio.

To find the ratio of the new to the old period of a pendulum when transported from Earth to the Moon, you can use the formula you mentioned: T = 2π√(L/g), where T is the period of the pendulum, L is the length of the pendulum, and g is the acceleration due to gravity.

First, let's find the period of the pendulum on Earth. Assuming you have the length of the pendulum on Earth (L₁) and the acceleration due to gravity on Earth (g₁ = 9.8 m/s²), you can plug these values into the formula to calculate the period (T₁) on Earth.

T₁ = 2π√(L₁/g₁)

Now, to find the period of the pendulum on the Moon, you'll need to use the length of the pendulum on the Moon (L₂) and the acceleration due to gravity on the Moon (g₂ = 1.63 m/s²). These values will differ from those on Earth.

T₂ = 2π√(L₂/g₂)

Finally, to find the ratio of the new (T₂) to the old (T₁) period of the pendulum, you can divide T₂ by T₁:

Ratio = T₂/T₁ = (2π√(L₂/g₂)) / (2π√(L₁/g₁))

Notice that the 2π cancels out, simplifying the equation:

Ratio = √(L₂/g₂) / √(L₁/g₁)

So, to find the ratio, you need to take the square root of (L₂/g₂) and divide it by the square root of (L₁/g₁).