Question

The free-fall acceleration on Mars is 3.7 m/s2.

Answer 1

Answer:

a) 142.98 cm b) 53.98 cm c) 1.46 kg d) 1.46 kg

Explanation:

Using the equation for the period of a simple pendulum we can answer the first 2 questions.

a) $T=2\pi\sqrt\frac{l}{g}\\T^2=4\pi^2\frac{l}{g}\\l=\frac{T^2g}{4\pi^2}=\frac{(2.4s)^29.8m/s}{4\pi^2}\\l=1.42 m= 142.98 cm$

b) $l=\frac{T^2g}{4\pi^2}=\frac{(2.4s)^23.7m/s}{4\pi^2}\\l=0.53 m= 53.98 cm$

Using the equation for the period of a mass-spring system we can answer the last 2 questions.

c) $T=2\pi\sqrt\frac{m}{k}\\T^2=4\pi^2\frac{m}{k}\\m=\frac{T^2k}{4\pi^2}=\frac{(2.4s)^210N/m}{4\pi^2}\\l=1.46 kg$

d) $T=2\pi\sqrt\frac{m}{k}\\T^2=4\pi^2\frac{m}{k}\\m=\frac{T^2k}{4\pi^2}=\frac{(2.4s)^210N/m}{4\pi^2}\\l=1.46 kg$

we can observe that the period of this system does not depend on the value of gravity.