Part A Here on Earth, a simple pendulum of length L is observed to have a freque

Question

Part A Here on Earth, a simple pendulum of length L is observed to have a frequency f. If another pendulum on Earth is observed to have a frequency that is 37f, what is its length?

= 6.083L
= L/2738
= 37L
= L/1369
= L/37
= L/6.083
= 1369L
= 2738L
Part B Here on Earth, a simple pendulum of length L is observed to have a frequency f. On another planet a pendulum of length L is observed to have a frequency of 89f. What is the acceleration due to gravity on this planet? In the answers below, g = 9.8 m/s2.
= g/15842
= g/9.434
= 9.434g
= g/7921
= 89g
= g/89
= 7921g
= 15842g

Explanation / Answer

f proportional to sqrt(g) so g' = 66^2 g = 4356g

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