Question

Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Supppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in the figure below.

Answer 1

Given k = 230 N/m

From the fig the the period is T = 3s, so the frequency is f = 1/3 s

   and we know the relation , T = 2 pi sqrt(m/k)

               3 = 2 pi sqrt(m/230)

   solving for mass

               m = 52.434 kgs

b) What is her speed when the spring's length is 1.08 m?

from the fig the average amplitude is 1.4+0.6/2 m = 1.0m
where the amlitude is 0.4 , and our point 1.08 m is 0.08 m above the equilibrium point

by conservation of energy , total energy must be same at any point

   E = 0.5 k*A^2 = 0.5*k*0.4^2

energy when the displacement of 0.08m is 0.5*k*0.08^2+0.5 mv^2

that is 0.5*k*0.4^2 = 0.5*k*0.08^2+0.5 mv^2

   k(0.4^2-0.08^2) = mv^2

   v^2 = k/m(0.4^2-0.08^2)
   v^2 = (230/52.434)(0.4^2-0.08^2)

   v = 0.82083 m/s

the speed is 0.82083 m/s