Oscillations - Result Question 24
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27. The particle executing simple harmonic motion has a kinetic energy $K_0 \cos ^{2} \omega t$. The maximum values of the potential energy and the total energy are respectively
======= ####27. The particle executing simple harmonic motion has a kinetic energy $K_0 \cos ^{2} \omega t$. The maximum values of the potential energy and the total energy are respectively
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/oscillations/oscillations—result-question-24.md (a) $K_0 / 2$ and $K_0$
(c) $K_0$ and $K_0$
(b) $K_0$ and $2 K_0$
(d) 0 and $2 K_0$.
[2007]
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Answer:
Correct Answer: 27. (c)
Solution:
- (c) We have, $U+K=E$ where, $U=$ potential energy, $K=$ Kinetic energy, $E=$ Total energy.
Also, we know that, in S.H.M., when potential energy is maximum, K.E. is zero and vice-versa.
$\therefore U _{\max }+0=E \Rightarrow U _{\max }=E$
Further,
$K . E .=\frac{1}{2} m \omega^{2} a^{2} \cos ^{2} \omega t$
But by question, $K . E .=K_0 \cos ^{2} \omega t$
$\therefore K_0=\frac{1}{2} m \omega^{2} a^{2}$
Hence, total energy, $E=\frac{1}{2} m \omega^{2} a^{2}=K_0$
$\therefore U _{\max }=K_0 & E=K_0$.
