(a) What is the length of a simple pendulum that oscillates with a period of 3.8

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) L = T^2*g/(4*pi^2) = (3.8^2*9.8)/(4*3.142^2) = 3.58 m...
LE = 3.58 m...

similarly LM = (3.8^2*3.7)/(4*3.142^2) = 1.352 m..

B) m = T^2*k/(4*pi^2)..
her m is independent of g..
so mE =mM..

mE = mM=(3.8^2*20 )/(4*3.142^2) = 7.313 kg

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