Oscillations - Result Question 28

31. A particle is executing a simple harmonic motion of amplitude ’ $a$ ‘. Its potential energy is maximum when the displacement from the position of the maximum kinetic energy is

[2002]

(a) 0

(b) $\pm a$

(c) $\pm a / 2$

(d) $-a / 2$

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Answer:

Correct Answer: 31. (b)

Solution:

  1. (b) P.E. of particle executing S.H.M.

$=\frac{1}{2} m \omega^{2} x^{2}$

At $x=a$, P.E. is maximum i.e. $=\frac{1}{2} m \omega^{2} a^{2}$

K.E. $=\frac{1}{2} m \omega^{2}(a^{2}-x^{2})$

At $x=0$, K.E. is maximum. Hence, displacement from position of maximum Kinetic energy $= \pm a$.