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Assertion (A) In S.H.M kinetic and potential energy become equal when the distance is $1 / \sqrt{2}$ times its amplitude. Reason (R) The potential energy of a particle executing S.H.M is periodic with time period being maximum at the extreme displacement.
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Verified Answer
The correct answer is:
$A$ and $R$ are true. $R$ is correct explanation of $A$.
( In SHM,
K. E of oscillator $=\frac{1}{2} m \omega^2\left(a^2-y^2\right)$
P. E of oscillator $=\frac{1}{2} m \omega^2 y^2$
(Here, $a=$ amplitude and $y=$ displacement)
when, $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}$
$$
\begin{aligned}
\Rightarrow & \frac{1}{2} m \omega^2\left(a^2-y^2\right) & =\frac{1}{2} m \omega^2 y^2 \\
\Rightarrow & a^2 & =2 y^2 \\
\text { or } & y & = \pm \frac{a}{\sqrt{2}}
\end{aligned}
$$
And P. E is maximum when $y=a$ or particle is at extreme displacement position. Hence, reason (R) is correct but it does not explain (A).
K. E of oscillator $=\frac{1}{2} m \omega^2\left(a^2-y^2\right)$
P. E of oscillator $=\frac{1}{2} m \omega^2 y^2$
(Here, $a=$ amplitude and $y=$ displacement)
when, $\mathrm{K} . \mathrm{E}=\mathrm{P} . \mathrm{E}$
$$
\begin{aligned}
\Rightarrow & \frac{1}{2} m \omega^2\left(a^2-y^2\right) & =\frac{1}{2} m \omega^2 y^2 \\
\Rightarrow & a^2 & =2 y^2 \\
\text { or } & y & = \pm \frac{a}{\sqrt{2}}
\end{aligned}
$$
And P. E is maximum when $y=a$ or particle is at extreme displacement position. Hence, reason (R) is correct but it does not explain (A).
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