(a) What is the length of a simple pendulum that oscillates with a period of 3.8
ID: 1496635 • Letter: #
Question
(a) What is the length of a simple pendulum that oscillates with a period of 3.8 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?
LE = m
LM = m
(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 3.8 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?
mE = kg
mM = kg
Explanation / Answer
A)on Earth
Time period =3.8 s
g = 9.8 m/s^2
on Mars
g=3.70m/s^2
length on Earth LE =?
Time period for simple pendulum is
T=2*pi sqrt(L/g)
3.8 = 2*3.14*sqrt(L/9.8)
L=3.588 m
length on Mars
T=2*pi sqrt(L/g)
3.8 = 2*3.14*sqrt(L/3.7)
L=1.35 m
b) k=20N/m
T=3.8s
g=9.8m/s^2 (on earth)
g=3.70m/s^2 (on Mars)
T = 2*pi*sqrt(m/k)
3.8 = 2*3.14*sqrt(m/20)
m=7.322 kg
mass is same every where so mass on Mars also 7.322 kg
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