The potential energy of a particle of mass 0. | vedclass

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(D) The force acting on the particle is given by $F = -\frac{dU}{dx}.$Given $U = 5x(x-4) = 5x^2 - 20x.$$F = -\frac{d}{dx}(5x^2 - 20x) = -(10x - 20) =

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all of the above.

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(D) The force acting on the particle is given by $F = -\frac{dU}{dx}.$Given $U = 5x(x-4) = 5x^2 - 20x.$$F = -\frac{d}{dx}(5x^2 - 20x) = -(10x - 20) = -10(x - 2).$Since $F = -k(x - x_0)$ where $x_0 = 2\,m,$ the particle executes simple harmonic motion about the equilibrium position $x = 2\,m.$At the equilibrium position $(x = 2\,m)$,the force is zero,which implies the acceleration is zero and the speed of the particle is maximum.Comparing $F = -10(x - 2)$ with the standard $SHM$ equation $F = -k(x - x_0),$ we get the force constant $k = 10\,N/m.$The time period $T$ is given by $T = 2\pi \sqrt{\frac{m}{k}}.$Substituting $m = 0.1\,kg$ and $k = 10\,N/m,$$T = 2\pi \sqrt{\frac{0.1}{10}} = 2\pi \sqrt{0.01} = 2\pi(0.1) = 0.2\pi = \frac{\pi}{5}\,s.$Since all statements are correct,the correct option is $D$.

The potential energy of a particle of mass $0.1\,kg,$ moving along the $x-$axis,is given by $U = 5x(x-4)\,J,$ where $x$ is in metres. It can be concluded that

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