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3 years ago

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A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an

additional 5.0 cm and released. Its position as a function of time is approximately what?

1 answer:

Dominik [7]3 years ago

8 0

Answer:

position as a function of time is y = 0.05 × cos(9.9)t

Explanation:

given data

mass = 5 kg

length = 10 cm = 0.1 m

displaced = 5 cm

to find out

position as a function of time

solution

we will apply here equilibrium that is

mass × g = k × length

put here value and find k

k = $\frac{5*9.8}{.01}$

k = 490 N/m

and ω is

ω = $\sqrt{\frac{k}{m} }$

ω = $\sqrt{\frac{490}{5} }$

ω = 9.9

so here position w.r.t time is

y = 0.05 × cosωt

y = 0.05 × cos(9.9)t

so position as a function of time is y = 0.05 × cos(9.9)t

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