Assertion : For a particle performing SHM,its | vedclass

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(C) The speed of a particle performing $SHM$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $x$ is the displacement from the

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If the Assertion is correct but Reason is incorrect.

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(C) The speed of a particle performing $SHM$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $x$ is the displacement from the mean position. As the particle moves away from the mean position,the magnitude of displacement $|x|$ increases,which causes the speed $v$ to decrease. Thus,the Assertion is correct.The acceleration in $SHM$ is given by $a = -\omega^2 x$. The acceleration is directed towards the mean position. When the particle moves away from the mean position,the velocity is directed away from the mean position,so the acceleration and velocity are in opposite directions (deceleration). However,when the particle moves towards the mean position,the velocity is directed towards the mean position,so the acceleration and velocity are in the same direction (acceleration). Therefore,the Reason is incorrect because the acceleration is not always opposite to the velocity.

$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.
$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.

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$Assertion :$ For a particle performing $SHM$,its speed decreases as it goes away from the mean position.$Reason :$ In $SHM$,the acceleration is always opposite to the velocity of the particle.