The angle between the instantaneous velocity and acceleration of a particle executing SHM is

zero or

π

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π / 2

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zero

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π

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Solution

The correct option is A $π / 2$
$x = A sin (ω t + ϕ) v = A ω cos (ω t + ϕ) a = - A ω^{2} sin (ω t + ϕ) = A ω^{2} cos (ω t + ϕ + \frac{π}{2})$
it is clear from the above two equations that angle between $a$ and $v$ is $\frac{π}{2}$

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The angle between the instantaneous velocity and acceleration of a particle executing SHM is

The correct option is A π/2x=Asin(ωt+ϕ)v=Aωcos(ωt+ϕ)a=−Aω2sin(ωt+ϕ)=Aω2cos(ωt+ϕ+π2)it is clear from the above two equations that angle between a and v is π2

The correct option is A $π / 2$
$x = A sin (ω t + ϕ) v = A ω cos (ω t + ϕ) a = - A ω^{2} sin (ω t + ϕ) = A ω^{2} cos (ω t + ϕ + \frac{π}{2})$
it is clear from the above two equations that angle between $a$ and $v$ is $\frac{π}{2}$