At a given point of time the value of displacement of a simple harmonic oscillator is given as $y = A \cos \left(30^{\circ}\right)$. If amplitude is $40\,cm$ and kinetic energy at that time is $200\, J$, the value of force constant is $1.0 \times 10^{ x }\,Nm ^{-1}$. The value of $x$ is ......
  • A$3$
  • B$2$
  • C$4$
  • D$1$
JEE MAIN 2023, Diffcult
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