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 67f, what is its length?
L/67
L/8978
L/8.185
L/4489
8.185L
67L
8978L
4489L

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 32f. What is the acceleration due to gravity on this planet? In the answers below, g = 9.8 m/s2.
g/2048
g/32
g/1024
g/5.657
32g
5.657g
1024g
2048g

Explanation / Answer

f = v(g/L)/2p where: f = frequency in cycles per second (Hertz or Hz) = v(g/L)/2p g is the acceleration due to gravity in m/s^ L is the length of the rod or wire in meters A) In this case g is constant and L is varying, f is proportional to v(1/L). B) In this case, L is constant and g is varying, f is proportional to vg

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