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Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63 $\mathrm{m} / \mathrm{s}^{2}$ .

$T_{2}=2.452 \cdot T_{1}$

Physics 101 Mechanics

Physics 103

Chapter 16

Oscillatory Motion and Waves

Periodic Motion

Wave Optics

University of Michigan - Ann Arbor

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As we know, period off pendulum is equal to soup. I square root El divided by g So we want t smooth, divided by t both. This will be equal to square root g off both divided by G off moon, as you can see to pie and the length are constant. So we know all the values. All we have to do is find the team. So let's substitute team. I'm just writing it now as t subscript m will be equal to t e which is track tee off square voted Geo. Fourth is 9.8 divided by moon's gravity is 1.62 So let's do this calculation when square root 9.8 divided by 1.60 So this is 2.46 t e. So pendulum has frequency. Oh, so time period which is 2.46 times larger than the same pendulum. If it was on coach Thank you

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