Oscillations Ques 27
27. 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: 27.(b)
Solution:
- (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 $= \pm a$.
BETA
