The potential energy of a particle of mass 4 | vedclass

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(A) The potential energy is given by $U = 4(1 - \cos 4x)$.For a conservative force,$F = -\frac{dU}{dx}$.$F = -\frac{d}{dx} [4(1 - \cos 4x)] = -4(0 - (

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$2$

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(A) The potential energy is given by $U = 4(1 - \cos 4x)$.For a conservative force,$F = -\frac{dU}{dx}$.$F = -\frac{d}{dx} [4(1 - \cos 4x)] = -4(0 - (-\sin 4x) \cdot 4) = -16 \sin 4x$.For small oscillations,$\sin 	heta \approx 	heta$,so $\sin 4x \approx 4x$.Thus,$F \approx -16(4x) = -64x$.Comparing this with the $SHM$ force equation $F = -m\omega^2 x$,we have $m\omega^2 = 64$.Given mass $m = 4 \, kg$,we get $4 \cdot \omega^2 = 64$,which implies $\omega^2 = 16$,so $\omega = 4 \, rad/s$.The time period $T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \, s$.Comparing this with $\frac{\pi}{K}$,we find $K = 2$.

The potential energy of a particle of mass $4 \, kg$ in motion along the $x$-axis is given by $U = 4(1 - \cos 4x) \, J$. The time period of the particle for small oscillations $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{K}\right) \, s$. The value of $K$ is .......

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