Q 11-13-083JEE MainJEE Main 2023 (13 Apr, Shift 1)Easy
At a given point of time the value of displacement of a simple harmonic oscillator is given as $y=A\cos(30^\circ)$. If amplitude is $40\ \text{cm}$ and kinetic energy at that time is $200\ \text{J}$, the value of force constant is $1.0\times10^x\ \text{N m}^{-1}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 4
$K=\tfrac12k(A^2-y^2)=\tfrac12kA^2\left(1-\tfrac34\right)=\dfrac{kA^2}{8}$.
$200=\dfrac{k\times0.16}{8}\Rightarrow k=10^4\ \text{N m}^{-1}$, so $x=4$.
Solution by Sreeraj P, M.Sc Physics