A potential is given by V(x) = k(x+a)^2 / 2 f | vedclass

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(B) The potential is given by:$V(x) = \begin{cases} \frac{k(x+a)^2}{2}, & x  0 \end{cases}$This represents two half-pa

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(B) The potential is given by:$V(x) = \begin{cases} \frac{k(x+a)^2}{2}, & x  0 \end{cases}$This represents two half-parabolas joined at $x = 0$. The potential energy $U(x) = mV(x)$ is symmetric about $x = 0$ if we consider the absolute value,but here it is defined piecewise.For a particle of energy $E$,the turning points are found by $U(x) = E$. If $E$ is small,the particle oscillates in one of the two wells (either $x  0$). In this case,the motion is simple harmonic with an effective spring constant $k$,so the period $T = 2\pi \sqrt{m/k}$,which is independent of $E$.As $E$ increases,the particle eventually crosses the barrier at $x = 0$. For $E > V(0) = ka^2/2$,the particle oscillates over both regions. The total period $T$ is the sum of the times spent in each region. As $E$ increases further,the particle spends more time in the flatter parts of the potential,causing the period $T$ to increase with $E$. Thus,the graph shows a constant $T$ for low $E$,followed by a jump and an increasing $T$ for higher $E$,which corresponds to graph $(b)$.

$A$ potential is given by $V(x) = k(x+a)^2 / 2$ for $x < 0$ and $V(x) = k(x-a)^2 / 2$ for $x > 0$. The schematic variation of the oscillation period $T$ for a particle performing periodic motion in this potential as a function of its energy $E$ is:

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