Astronauts in space cannot weigh themselves by standing on a bathroom scale. Ins

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.

(a) What is her mass if the spring constant is 190 N/m?

(b) What is her speed when the spring's length is 1.22 m?

Explanation / Answer

You need to figure out where the equilibrium position is, what the amplitude of this motion is, and what displacement a length of 1.22 meters represents relative to equilibrium and maximum.

The lowest point on this curve is 0.6. The highest is 1.4. So the equilibrium position is in the middle, or (0.6 + 1.4)/2 = 1.0. The amplitude of this motion is 0.4, since it goes 1.4-1.0 = 0.4 meters above equilibrium and 1.0 - 0.6 = 0.4 meters below. A value of 1.22 represents a displacement of 0.22 meters relative to equilibrium.

I see that the period from peak to peak is 3 seconds, so f = 1/T = 1/3 Hz.

I need some way to estimate the spring constant, so I can use energy = (1/2)kx^2.

Period = 2*pi*sqrt(m/k) and as I said, the period is 3 seconds. So 3 = 2*pi*sqrt(m/k) or sqrt(m/k) = 3/(2*pi). I have the ratio m/k.
Substituting k =190N/m

m= 43.31kg
The amplitude is 0.4, so the maximum potential energy is (1/2)*k*0.4^2 = E.

Total energy is conserved, so when the displacement is 0.22 I have (1/2)*k*0.4^2 = (1/2)*k*0.22^2 + (1/2)*mv^2. The total PE is converted partly to PE and partly to KE. The (1/2) factors cancel out.

k*0.4^2 - k*0.22^2 = mv^2
190(0.4^2 - 0.22^2) = mv^2

v^2 = (190/43.31)*(0.4^2 - 0.22^2)
v = sqrt(190/43.31) * sqrt(0.4^2 - 0.22^2)
v=0.7m/s

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