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  • The potential energy function for a particle executing linear SHM is given by (1/2) kx^2 where k is the force constant of the oscillator.

The potential energy function for a particle executing linear SHM is given by \frac{1}{2}kx^{2} where k is the force constant of the oscillator (Figure). For k = 0.5N/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 xm . If V and K indicate the P.E. and K.E., respectively of the particle at x = \pm xm, then which of the following is correct?

(a) V = O, K = E
(b) V = E, K = O
(c) V < E, K = O
(d) V = O, K < E.

Answers (1)

The answer is the option (b) V = E, K = O

Potential energy at any position,

U = \frac{1}{2} kx^{2}

Total energy of the mass,

E = \frac{1}{2} ka^{2}

Kinetic energy at any position,

K=E-U = \frac{1}{2} ka^{2}+\frac{1}{2} kx^{2}

K= \frac{1}{2} k\left [ a^{2}-x^{2} \right ]

Now,

U_{max}= \frac{1}{2} ka^{2} at extremes. 

U min = 0 {x=0 at mean}

K_{max}= \frac{1}{2} ka^{2} {x=0 at mean}

K min = 0 {extremes}

Hence, E= \frac{1}{2}ka^{2} is constant at all times

When particle is at x = xm (mean), KE =0

So, total energy is E = PE + 0 = \frac{1}{2}kx^{2}

So, option b is correct.

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infoexpert24

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