Ncert Exercises & Solutions

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. $V = 0, K, E$

Hint: Apply the concept of conservation of energy.

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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