Answer:
The mass of the astronaut is approximately 119.74 kg
Explanation:
Assuming this problem as a Simple Harmonic Motion of a mass-spring system, the period (T) of the oscillations for a mass (m) and spring constant (k) is:
[tex] T=2\pi\sqrt{\frac{m}{k}} [/tex] (1)
First, we have to calculate the spring constant using equation (1) and the data provided for the oscillations without the astronaut:
(it’s important to note that one complete vibration is the period of the movement)
[tex]1.3=2\pi\sqrt{\frac{42.5}{k}}\Longrightarrow k=42.5(\frac{2\pi}{1.3})^{2} [/tex]
[tex]k\approx992.8\,\frac{N}{m} [/tex]
Now with the value of k, we can use again (1) to find the mass of the astronaut (Ma) that makes the period to be 2.54 seconds
[tex] 2.54=2\pi\sqrt{\frac{42.5+M_{a}}{992.8}}\Longrightarrow M_{a}=992.8(\frac{2.54}{2\pi})^{2}-42.5 [/tex]
[tex] \mathbf{M_{a}\approx119.74\,kg} [/tex]