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

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$π$

0.707 $π$

zero

0.5 $π$

answer is D.

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Detailed Solution

$L e t y = A \sin ω t v_{i n s} = \frac{d_{v}}{d t} = A ω \cos ω t = A (\sin ω t + \frac{π}{2}) A c c e l e r a t i o n = - A ω^{2} \sin ω t = - A ω^{2} \sin (π + ω t) ∴ ϕ = \frac{π}{2} = 0.5 π$

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