(a) What is the length of a simple pendulum that oscillates with a period of 1.2
ID: 1293169 • Letter: #
Question
(a) What is the length of a simple pendulum that oscillates with a period of 1.2 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?
(b) What mass would you need to suspend from a spring with a force constant of 20 N/m in order for the mass-spring system to oscillate with a period of 1.2 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
from
T = 2pi*sqrt(L/g)
lenght on earth = gT^2/(4pi^2) = 9.81*1.2^2/(4*pi^2) = 0.3578 m
Le = 0.3578 m
and
we know Lenght of pendulum is proportional to acceleration due to gravity
from the relation
Lm/Le = gm/ge
Lm = 0.3578* 3.70/(9.80) = 0.135 m
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b)
from the relation
T = 2pi*sqrt(m/K)
1.2 = 2pi*sqrt(m/20)
mass m = 0.73 kg on earth
on mars also mass m = 0.73 kg
since mass is constant it does not change with place
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