The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmon

Question

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

Accepted Answer

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

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