A particle of mass m is executing oscillation | vedclass

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Question

(A) The potential energy is given by $U(x) = k|x|^3$.The force acting on the particle is $F = -\frac{dU}{dx} = -3k|x|^2 	ext{sgn}(x)$.For a particle

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Proportional to $\frac{1}{\sqrt{a}}$

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(A) The potential energy is given by $U(x) = k|x|^3$.The force acting on the particle is $F = -\frac{dU}{dx} = -3k|x|^2 	ext{sgn}(x)$.For a particle of mass $m$ oscillating with amplitude $a$,the total energy $E$ is conserved and is equal to the potential energy at the extreme position $(x = a)$:$E = U(a) = ka^3$.At any position $x$,the energy is $E = \frac{1}{2}mv^2 + k|x|^3 = ka^3$.Thus,$v = \frac{dx}{dt} = \sqrt{\frac{2k}{m}(a^3 - |x|^3)}$.The time period $T$ is given by $T = 4 \int_{0}^{a} \frac{dx}{v} = 4 \int_{0}^{a} \frac{dx}{\sqrt{\frac{2k}{m}(a^3 - x^3)}}$.Let $x = ay$,then $dx = a dy$. When $x=0, y=0$ and when $x=a, y=1$.$T = 4 \sqrt{\frac{m}{2k}} \int_{0}^{1} \frac{a dy}{\sqrt{a^3(1 - y^3)}} = 4 \sqrt{\frac{m}{2ka}} \int_{0}^{1} \frac{dy}{\sqrt{1 - y^3}}$.Since the integral is a constant,$T \propto \frac{1}{\sqrt{a}}$.

$A$ particle of mass $m$ is executing oscillations about the origin on the $X-$axis. Its potential energy is $U(x) = k|x|^3$,where $k$ is a positive constant. If the amplitude of oscillation is $a$,then its time period $T$ is:

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