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Simple Harmonic Motion question

2023 · 13 Apr · Shift 1 · Q69
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  5. /2023 · 13 Apr · Shift 1 · Q69

Simple Harmonic Motion question

2023 · 13 Apr · Shift 1 · Q69

JEE MainPhysicsSimple Harmonic MotionNumerical+4 / −1
At a given point of time the value of displacement of a simple harmonic oscillator is given as y=Acos⁡(30∘)\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}\right)y=Acos(30∘). If amplitude is 40 cm40 \mathrm{~cm}40 cm and kinetic energy at that time is 200 J200 \mathrm{~J}200 J, the value of force constant is 1.0×10x Nm−11.0 \times 10^{x} ~\mathrm{Nm}^{-1}1.0×10x Nm−1. The value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
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