The potential energy of a particle of mass $4\,kg$ in motion along the $x$-axis is given by $U =4(1-\cos 4 x )\,J$. The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{ K }\right)\,s$. The value of $K$ is .......
JEE MAIN 2022, Diffcult
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$U =4(1-\cos 4 x )$
$F =-\frac{ dU }{ dx }=-4(+\sin 4 x ) 4=-16 \sin (4 x )$
For small $\theta$
$\sin \theta \approx \theta$
$F=-64\,x$
$a=-64\,x / m=-16\,x$
$\omega^{2}=16$
$T =\frac{2 \pi}{\omega}=\frac{\pi}{2}$
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