Question

Statement 1: In simple harmonic motion, the velocity is maximum when the acceleration is minimum.
Statement 2: Displacement and velocity of $S.H.M$ differ in phase by $\frac{\pi }{2}$

• A. Statement 1 is false, Statement 2 is true
• B. Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation for Statement 1
• C. Statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation for Statement 1
• D. Statement 1 is true, Statement 2 is false

Answer 1

The correct option is B Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation for Statement 1

At the middle point velocity of the particle under $SHM$ is maximum but acceleration is zero since displacement is zero. So Statement 1 is true.
We know that $x=a\sin \omega t...........................\left(i\right)$
Where $x$ is displacement and $a$ is amplitude.
Velocity $=\frac{dx}{dt}=a\omega \cos \omega t$
$=a\omega \cos \left(\omega t\right)=a\omega \sin \left(\frac{\pi }{2}-(-\omega t)\right)$
$=a\omega \sin (\omega t+\frac{\pi }{2})...................................................\left(ii\right)$
From equations $\left(i\right)$ and $\left(ii\right)$ it is clear that
Velocity is ahead of displacement $\left(x\right)$ by $\frac{\pi }{2}$ angle.