MCQ
The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is
- A$\pi$
- B$2\pi $
- ✓$0.5\pi $
- D$0$
✓
Answer
Correct option: C.
$0.5\pi $
c
Let $y=A \sin \omega t$
Let $y=A \sin \omega t$
$\frac{d y}{d t}=A \omega \cos \omega t=A \omega \sin \left(\omega t+\frac{\pi}{2}\right)$
Acceleration $=-A \omega^{2} \sin \omega t$
The phase difference between acceleration and velocity is $\pi / 2$
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