The potential energy function for a particle executing linear SHM is given by \(V \left(\right. x \left.\right) = \dfrac{1}{2} kx^{2}\) where \(k\) is the force constant of the oscillator (Fig). For \(k = 0.5~\text{N/m},\) the graph of \(V~(x)\) versus \(x\) is shown in the figure. A particle of total energy \(E\) turns back when it reaches \(x = \pm \left(x \right)_{m}.\) If \(V\) and \(K\) indicate the \(PE\) and \(KE,\) respectively of the particle at \(x = + \left(x \right)_{m}, \) then which of the following is correct?
1. \(V = 0, K = E\)
2. \(V = E, K = 0 \)
3. \(V < E, K = 0\)
4. \(V = 0, K, E\)
Step: Find the potential energy and the kinetic energy at the extreme position.
The total energy of the particle is given by; \(E = P.E +K.E\)
When the particle is at \(x= x_m\) i.e., at an extreme position, returns back. Hence. at \(x= x_m; v=0, K.E= 0\)
At the extreme position, \(x= x_m\) the kinetic energy of the particle is zero, \(E = P.E = V(x_m)= \frac{1}{2}kx^2_{m}\)
Therefore, the correct relation is \(V = E, K = 0 \).
Hence, option (2) is the correct answer.
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