Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:

\frac{3\pi}{2} \text{ rad}

\frac{\pi}{2} \text{ rad}

zero

\pi \text{ rad}

Displacement Equation: Let the displacement of a particle executing Simple Harmonic Motion (SHM) be given by: $x = A \sin(\omega t)$ where $A$ is the amplitude and $\omega$ is the angular frequency.
Velocity Equation: Velocity $v$ is the time derivative of displacement: $v = \frac{dx}{dt} = A\omega \cos(\omega t) = A\omega \sin(\omega t + \frac{\pi}{2})$ (Phase difference between displacement and velocity is $\pi/2$ ).
Acceleration Equation: Acceleration $a$ is the time derivative of velocity: $a = \frac{dv}{dt} = \frac{d}{dt}[A\omega \cos(\omega t)] = -A\omega^2 \sin(\omega t)$ Using the trigonometric identity $-\sin \theta = \sin(\theta + \pi)$ , we can write: $a = A\omega^2 \sin(\omega t + \pi)$
Phase Difference: Comparing the phase of displacement $(\omega t)$ and acceleration $(\omega t + \pi)$ , the difference is: $\Delta \phi = (\omega t + \pi) - (\omega t) = \pi \text{ rad}$
Conclusion: The acceleration is always in opposite phase ( $180^\circ$ or $\pi$ radians) to the displacement .

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