The potential energy function for a particle executing linear SHM is given by \(V \left(\right. x \left.\right) = \dfrac{1}{2} \left(kx\right)^{2}\) 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. \(V = 0, K, E\)
(b) Hint: Apply the concept of conservation of energy.
Step 1: Find the potential energy and the kinetic energy at the extreme position.
Total energy is E = PE+ KE
When particle is at x = i.e., at extreme position, returns back. Hence. at x = ; v = 0, K.E. = 0
From Eq. (I)
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