(a) What is the length of a simple pendulum that oscillates with a period of 3.2 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?



Length (Earth)=?

Length (Mars)=?



(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 3.2 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?



Mass (Earth)=?

Mass (Mars)=?

Respuesta :

Answer:

Explanation:

The expression relating length and time period

T  = 2π [tex]\sqrt{\frac{l}{g} }[/tex]

3.2 = [tex]2\pi \sqrt{\frac{l}{9.8} }[/tex]

l = 2.54 m

On Mars g = 3.7

[tex]3.2 = 2\pi \sqrt{\frac{L}{3.7} }[/tex]

L = .96 m

b )

Expression for elastic constant and time  period is as follows

[tex]T  = 2\pi \sqrt{\frac{m}{k} }[/tex]

[tex]3.2=2\pi \sqrt{\frac{m}{20} }[/tex]

m = 5.19 N/s

Time period of oscillation due to spring is not dependent on g , so same time period will be found on Mars as that on the earth.