When a particle of mass $m$ moves on the $x$-axis in a potential of the form $V(x)=\mathrm{kx}^2$ it performs simple harmonic motion. The corresponding time period is proportional to $\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$, as can be seen easily using dimensional analysis. However, the motion of a particle can be periodic even when its potential energy increases on both sides of $\mathrm{x}=0$ in a way different from $\mathrm{kx}^2$ and its total energy is such that the particle does not escape to infinity. Consider a particle of mass $\mathrm{m}$ moving on the $x$-axis. Its potential energy is $V(x)=\alpha x^4(\alpha>0)$ for $|x|$ near the origin and becomes a constant equal to $\mathrm{V}_0$ for $|x| \geq X_0$ (see figure). $Image$
$1.$ If the total energy of the particle is $E$, it will perform periodic motion only if
$(A)$ $E$ $<0$ $(B)$ $E$ $>0$ $(C)$ $\mathrm{V}_0 > \mathrm{E}>0$ $(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $\mathrm{A}$, the time period $\mathrm{T}$ of this particle is proportional to
$(A)$ $\mathrm{A} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(B)$ $\frac{1}{\mathrm{~A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(C)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$ $(D)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$
$3.$ The acceleration of this particle for $|\mathrm{x}|>\mathrm{X}_0$ is
$(A)$ proportional to $\mathrm{V}_0$
$(B)$ proportional to $\frac{\mathrm{V}_0}{\mathrm{mX}_0}$
$(C)$ proportional to $\sqrt{\frac{\mathrm{V}_0}{\mathrm{mX}_0}}$
$(D)$ zero
Give the answer qustion $1,2$ and $3.$
IIT 2010, Advanced
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$1.$ Energy must be less than $\mathrm{V}_0$
$2.$ $[\alpha]=\mathrm{ML}^{-2} \mathrm{~T}^{-2}$
Only $(B)$ option has dimension of time Alternatively
$ \frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)^2+\mathrm{kx}^4=\mathrm{kA}^4 $
$ \left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)^2=\frac{2 \mathrm{k}}{\mathrm{m}}\left(\mathrm{A}^4-\mathrm{x}^4\right) $
$ 4 \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \int_0^{\mathrm{A}} \frac{\mathrm{dx}}{\sqrt{\mathrm{A}^4-\mathrm{x}^4}}=\int \mathrm{dt}=\mathrm{T} $
$ 4 \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \frac{1}{\mathrm{~A}} \int_0^1 \frac{\mathrm{du}}{\sqrt{1-\mathrm{u}^4}}=\mathrm{T} \quad \text { Substitute } \mathrm{x}=\mathrm{Au}$
$3.$ As potential energy is constant for $|\mathrm{x}|>\mathrm{X}_0$, the force on the particle is zero hence acceleration is zero.
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