Question

A particle is performing S.H.M. starting from extreme position. Displacement and acceleration have a phase difference of

• A. $\pi$ rad
• B. $\pi/2$ rad
• C. $\pi/4$ rad
• D. 0 rad

Hint:

In SHM, acceleration is always directed opposite to displacement ($a = -\omega^2 x$).

Answer 1

The Correct Option is A

Step 1: Formula
Displacement $x = A \cos(\omega t)$.
Acceleration $a = \frac{d^2x}{dt^2} = -\omega^2 A \cos(\omega t)$.

Step 2: Analysis
$a = \omega^2 A \cos(\omega t + \pi)$.

Step 3: Calculation
The phase of $x$ is $\omega t$, and the phase of $a$ is $\omega t + \pi$.

Step 4: Conclusion
The phase difference is $\pi$ rad. Final Answer: (A)