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 ......
JEE MAIN 2023, Diffcult
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General equation for displacement is given by
$x=A \sin (\omega t+\phi)$
$\text { at given time }$
$\Rightarrow \omega t+\phi=30^{\circ}$
$\Rightarrow x=40 \times \frac{\sqrt{3}}{2} \Rightarrow 20 \sqrt{3}\,cm$
$\Rightarrow A=40\,cm$
$\Rightarrow K . E=\frac{1}{2} k\left(A^2-x^2\right)=200$
$200=\frac{1}{2} k\left(\frac{1600-1200}{100 \times 100}\right)$
$400 \times 100 \times 100=k \times 400$
$k=10^4$
$x=4$
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