a) What is the length of a simple pendulum that oscillates with a period of 2.6

Question

a) What is the length of a simple pendulum that oscillates with a period of 2.6 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?

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 2.6 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?

Me=? kg

Mm=? kg

Explanation / Answer

a) T = 2 pi sqrt(L/g)

T2/T1 = sqrt(g1/g2)

T2 = 2.6*sqrt(9.81/3.7)= 4.23 s

b) T = 2 pi sqrt(k/m)

on earth

2.6 = 2*pi*sqrt(20/m)

m=116.8 kg

independent of g

so mmars = 116.8 kg

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