Question

The velocity and acceleration of a particle performing simple harmonic motion have a steady phase relationship. The acceleration shows a phase lead over the velocity in radians of

• A. $+\pi$
• B. 0
• C. $+\pi/2$
• D. $-\pi/2$
• E. $-\pi$

Hint:

Memory Trick: In SHM, each derivative introduces a phase lead of $\pi/2$. Displacement $\to$ Velocity ($\pi/2$ lead) $\to$ Acceleration (another $\pi/2$ lead, totaling $\pi$ lead from displacement).

Answer 1

The Correct Option is C

Concept: In Simple Harmonic Motion (SHM), the displacement ($x$), velocity ($v$), and acceleration ($a$) can be represented as:
• $x = A \sin(\omega t)$
• $v = \frac{dx}{dt} = A\omega \cos(\omega t) = A\omega \sin(\omega t + \pi/2)$
• $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi)$

Step 1: {Analyze the phase of velocity relative to displacement.}
Velocity leads displacement by a phase of $\pi/2$.

Step 2: {Analyze the phase of acceleration relative to displacement.}
Acceleration leads displacement by a phase of $\pi$.

Step 3: {Compare acceleration and velocity.}
The phase difference $\Delta \phi$ between acceleration and velocity is: $$\Delta \phi = \text{Phase of } a - \text{Phase of } v = \pi - \frac{\pi}{2} = \frac{\pi}{2}$$ Thus, acceleration leads velocity by $\pi/2$ radians.