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(B) The restoring force is given by $F = -kx^3$. The potential energy $U$ is given by $U = \int kx^3 dx = \frac{1}{4} kx^4$.For a particle of mass $m$

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$T/2$

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(B) The restoring force is given by $F = -kx^3$. The potential energy $U$ is given by $U = \int kx^3 dx = \frac{1}{4} kx^4$.For a particle of mass $m$ oscillating with amplitude $A$,the total energy $E$ is $E = \frac{1}{4} kA^4$.The velocity $v$ at any position $x$ is given by $\frac{1}{2} mv^2 + \frac{1}{4} kx^4 = \frac{1}{4} kA^4$.$v = \frac{dx}{dt} = \sqrt{\frac{k}{2m}} \sqrt{A^4 - x^4}$.The time period $T$ is $4 	imes$ time taken to go from $x=0$ to $x=A$:$T = 4 \int_0^A \frac{dx}{\sqrt{\frac{k}{2m}} \sqrt{A^4 - x^4}} = 4 \sqrt{\frac{2m}{k}} \int_0^A \frac{dx}{\sqrt{A^4 - x^4}}$.Let $x = Ay$,then $dx = A dy$. When $x=0, y=0$; when $x=A, y=1$.$T = 4 \sqrt{\frac{2m}{k}} \int_0^1 \frac{A dy}{\sqrt{A^4 - A^4 y^4}} = 4 \sqrt{\frac{2m}{k}} \frac{A}{A^2} \int_0^1 \frac{dy}{\sqrt{1 - y^4}} = \frac{4}{A} \sqrt{\frac{2m}{k}} \int_0^1 \frac{dy}{\sqrt{1 - y^4}}$.Thus,$T \propto \frac{1}{A}$.If the amplitude changes from $A$ to $2A$,the new time period $T'$ is given by $\frac{T'}{T} = \frac{A}{2A} = \frac{1}{2}$.Therefore,$T' = T/2$.

$A$ point particle is acted upon by a restoring force $F = -k x^3$. The time period of oscillation is $T$ when the amplitude is $A$. The time period for an amplitude $2A$ will be

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