$Assertion :$ In simple harmonic motion, the velocity is maximum when the acceleration is minimum.
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$
AIIMS 2014, Medium
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At the middle point velocity of the particle under $SHM$ is maximum but acceleration is zero since displacement is zero. So Assertion is true.
We know that $x=a \sin \omega t$ $...(1)$
Where $x$ is displacement and a is amplitude.
Velocity $=\frac{d x}{d t}=a \omega \cos \omega t$
$=a \omega \cos (-\omega t)=a \omega \sin \left(\frac{\pi}{2}-(-\omega t)\right)$
$=a \omega \sin \left(\omega t+\frac{\pi}{2}\right)$ $...(2)$
From equation $( 1 )$ and $(ii)$ it is clear that
Velocity is ahead of displacement $(x)$ by $\frac{\pi}{2}$ angle.
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