Question:medium

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

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In SHM, acceleration is always directed opposite to displacement ($a = -\omega^2 x$).
Updated On: Jun 19, 2026
  • $\pi$ rad
  • $\pi/2$ rad
  • $\pi/4$ rad
  • 0 rad
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Question:
The question asks for the phase relationship between displacement (\( x \)) and acceleration (\( a \)) for a particle in Simple Harmonic Motion (S.H.M.).

Step 2: Key Formula or Approach:

The standard equation for displacement starting from the extreme position is:
\[ x = A \cos(\omega t) \] The acceleration in S.H.M. is given by the relation:
\[ a = -\omega^2 x \]

Step 3: Detailed Explanation:

Substituting the displacement equation into the acceleration formula:
\[ a = -\omega^2 A \cos(\omega t) \] Using trigonometric identities, \( -\cos(\theta) = \cos(\theta + \pi) \):
\[ a = \omega^2 A \cos(\omega t + \pi) \] Comparing the equations for \( x \) and \( a \):
Phase of \( x \) is \( \omega t \).
Phase of \( a \) is \( \omega t + \pi \).
The phase difference \( \Delta \phi = (\omega t + \pi) - \omega t = \pi \text{ rad} \).

Step 4: Final Answer:

The phase difference between displacement and acceleration in S.H.M. is \( \pi \text{ rad} \).
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