A potential is given by V ( x ) = k ( x + a ) ^ 2 / 2 for x 0 . The...

 Given, potential function for the oscillating particle is

V\left(x\right) = \left\{\begin{cases}\dfrac{k ( x + a{)}^{2}}{2}, & x < 0 \\ \dfrac{k ( x - a{)}^{2}}{2}, & x > 0\end{cases}\right)</p> <p>So, potential energy of the particle (mass<br/>m$) is</p> <p>$U\left(x\right) = \left\{$\begin{cases}\dfrac{km ( x + a{)}^{2}}{2}, & x < 0 \\ \dfrac{km ( x - a{)}^{2}}{2}, & x < 0\end{cases}$\right)

\dfrac{dU}{dx} = \left\{\begin{cases}km ( x + a ) , & x < 0 \\ km ( x - a ) , & x > 0\end{cases}\right)</p> <p>If\dfrac{dU}{dx} = 0,$when$x = \pm a$</p> <p>Now,$\dfrac{{d}^{2}U}{d{x}^{2}} = km > 0$</p> <p>So, particle is in unstable equilibrium at$x = \pm a$</p> <p>Hence, particle is unbounded for$- a > x and $x > a$

In region, $- a \le x \le a$, time period of particle reduces from a maximum.
So, correct graph is (b)..