(a) What is the length of a simple pendulum that oscillates with a period of 1.4
ID: 1497601 • Letter: #
Question
(a) What is the length of a simple pendulum that oscillates with a period of 1.4 s on Earth, where the acceleration due to gravity is 9.80 m/s^2, and on Mars, where the acceleration due to gravity is 3.70 m/s^2?
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 1.4 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
LE = mExplanation / Answer
a) on earth,
T = 2*pi*sqrt(Le/ge)
T^2 = 4*pi^2*Le/ge
==> Le = T2*ge/(4*pi^2)
= 1.4^2*9.8/(4*pi^2)
= 0.486 m
similarly,
Lm = T2*gm/(4*pi^2)
= 1.4^2*3.7/(4*pi^2)
= 0.184 m
b) we know, angular frenqcy, w = sqrt(k/m)
we know, T = 2*pi/w
T = 2*pi*sqrt(m/k)
T^2 = 4*pi^2*m/k
==> m = T^2*k/(4*pi^2)
we dont have g term in the above equation. So, m does not depend on g.
mE = mM = 1.4^2*20/(4*pi^2)
= 0.993 kg
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