In order to study the long-term effects of weightlessness, astronauts in space must be weighed (or at least "massed"). One way in which this is done is to seat them in a chair of known mass attached to a spring of known force constant and measure the period of the oscillations of this system. The 35.4 kg chair alone oscillates with a period of 1.25 s, and the period with the astronaut sitting in the chair is 2.23 s.a. Find the force constant of the spring. (units in N/m)b. Find the mass of the astronaut. (units in kg)

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aristocles aristocles · Answer 1 · 2019-09-09T02:12:02+02:00

Answer:

m = 77.3 kg

Explanation:

Time period of spring block system is given as

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

here we know

m = mass

k = spring constant

so now we have

[tex]1.25 = 2\pi\sqrt{\frac{35.4}{k}}[/tex]

so we have

[tex]k = 894.4 N/m[/tex]

Now time period of chair + astronaut is 2.23 s

so we have

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

[tex]2.23 = 2\pi\sqrt{\frac{35.4 + m}{894.4}}[/tex]

[tex]m = 77.3 kg[/tex]

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-02-17T16:30:43+01:00

The force constant of the spring is 894.4N/m and the mass of the astronaut is 77.3kg

For the chair and spring system, the time period of the simple harmonic motion is given by:

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

where m is the mass = 35.4kg

k is the spring constant

[tex]k=\frac{4\pi^2m}{T^2}\\\\k=\frac{4\pi^2\times 35.4}{1.25\times1.25}\\\\k=894.4N/m[/tex]

Now it is given that the time period of the chair with the astronaut is 2.23s

Let M be the mass of the astronaut and m be the mass of the chair, then the time period is:

[tex]T=2\pi\sqrt{\frac{m+M}{k}}\\\\2.23=2\pi\sqrt{\frac{35.4+M}{894.4} } s\\\\M=77.3kg[/tex]

Learn more about simple harmonic motion:

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