(a) What is the length of a simple pendulum that oscillates with a period of 1.6
ID: 2265023 • Letter: #
Question
(a) What is the length of a simple pendulum that oscillates with a period of 1.6 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
What physical quantities affect the period of oscillation of a simple pendulum on Earth? m LM =? m What is the length of a simple pendulum that oscillates with a period of 1.6 s on Earth, where the acceleration due to gravity is 9.80 m/s2, and on Mars, where the acceleration due to gravity is 3.70 m/s2?
Explanation / Answer
The period of a pendulum is
T = 2pi*sqrt(L/g)
so, L = T^2*g/(4pi^2)
On Earth,
L = 1.6^2*9.8/(4pi^2)
= 0.6361m
On the moon,
T = 2*pi*sqrt(L/g),
L = 1.6^2*3.70/(4pi^2)
= 0.2401m
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