JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

A person of mass

80 \mathrm{~Kg}

sits on a spring and the spring compresses by

2 \mathrm{~m}

. If the person himself is projected at

20 \mathrm{~ms}^{-1}

velocity by the spring then the spring constant of the spring in

\mathrm{Nm}^{-1}

View answer and explanation

Correct option: D

8000

Option D (8000) is correct because the spring constant is calculated using the conservation of energy principle, equating the spring's potential energy to the kinetic energy of the person.

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