Question

Astronauts in space "weigh" themselves by oscillating on a spring. Suppose the position of an oscillating 70 kg astronaut is given by x=(0.34m)sin((πrad/s)⋅t), where t is in s. What force does the spring exert on the astronaut at t = 1.0s. Note that the angle of the sine function is in radians.

Answer 1

Answer:

x = 0 , F= 0

Explanation:

To find the elastic force of a spring we use Hooke's law

      F = - k x

Done k is the Hooke constant and x is the displacement of the spring from its equilibrium point.

They give us the equation of motion of the harmonic oscillator

     x = 0.34 sin (π t)

  The form of the theoretical equation is

     x = A cos (wt + Φ)

To transform the expression the phase must be 90º and the angular velocity is

     w² = k / m

      x = A cos (wt + 90) = A [cos wt sint 90 + cos 90 sint wt]

     Cos 90 = 0

      x = A sin( wt)

Now let's calculate the offset for 1 s

      x = 0.34 sin (π 1)

      x = 0.34 (0)

      x = 0

For this displacement the force is zero