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Physics · 2023

JEE Main · 13 April 2023, Shift 1 · Q57

At a given point of time the value of displacement of a simple harmonic oscillator is given as y=A cos (30^°). If amplitude is 40 cm and kinetic…

At a given point of time the value of displacement of a simple harmonic oscillator is given as $\displaystyle y=A \cos \left(30^{\circ}\right)$. If amplitude is $\displaystyle 40$ cm and kinetic energy at that time is $\displaystyle 200$ J , the value of force constant is $\displaystyle 1.0 \times 10^x \mathrm{Nm}^{-1}$. The value of $\displaystyle x$ is $\displaystyle \_\_\_\_$
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JEE Main 2023 Physics question, with the answer from NTA’s final answer key. Where our answers come from.