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NEET PHYSICSEasy

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:

A

0.5 \pi

B

\pi

C

0.707 \pi

D

zero

Step-by-Step Solution

  1. Analyze Velocity: For a particle in SHM with displacement x=Asin⁡(ωt)x = A \sin(\omega t), the velocity is v=dxdt=Aωcos⁡(ωt)=Aωsin⁡(ωt+π2)v = \frac{dx}{dt} = A\omega \cos(\omega t) = A\omega \sin(\omega t + \frac{\pi}{2}). This shows velocity leads displacement by phase π2\frac{\pi}{2}.
  2. Analyze Acceleration: The acceleration is a=dvdt=−Aω2sin⁡(ωt)=Aω2sin⁡(ωt+π)a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t) = A\omega^2 \sin(\omega t + \pi) . This shows acceleration leads displacement by phase π\pi.
  3. Calculate Phase Difference: The phase difference between acceleration and velocity is (ωt+π)−(ωt+π2)=π2(\omega t + \pi) - (\omega t + \frac{\pi}{2}) = \frac{\pi}{2} radians.
  4. Convert to decimal: π2=0.5π\frac{\pi}{2} = 0.5\pi.
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