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JEE MAIN 2023
13-04-23 S1
Question
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 $1.0 \times 10^x \mathrm{Nm}^{-1}$. The value of $x$ is $\_\_\_\_$ .
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Question Tags
JEE Main
Physics
Medium
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