Velocity and Acceleration in SHM – Rankers Physics

Equation of SHM: Practice Problem & Solution

Assertion (A): In SHM the velocity is maximum when the acceleration is minimum. Reason (R): Displacement and velocity in SHM differ in phase by \(\frac{\pi}{2}\) .  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution Explained:

To solve this problem, we apply the core principles of Equation of SHM. Understanding the underlying formula is key to arriving at the correct answer below:

In SHM, velocity is max at equilibrium (where displacement is zero), and acceleration is min (zero) at equilibrium. So A is true.
Displacement `\(x = A\sin(\omega t)\)` and velocity `\(v = A\omega\cos(\omega t)\)` differ in phase by \(\frac{\pi}{2}\). So R is true.
However, R explains phase relation, not why maximum velocity occurs at minimum acceleration. Hence, R does not explain A.

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