What is the phase difference between acceleration and velocity of a particle executing simple harmonic?

SHM is represented by $x=A\sin \omega t$

Then, the velocity and acceleration after differentiation will be

$v=A\omega \cos \omega t$

a=−Aω2sinωt

Now, we see displacement is in $\sin \omega t$ & velocity is in $\cos \omega t$

$\Rightarrow$ Phase difference between x & $v=\pi /2$ rad

Similarly, we see velocity in $\cos \omega t$ & acceleration in $\sin \omega t$

$\Rightarrow$ Phase difference between v & $a=\pi /2$rad

Thus, we can conclude that the phase difference between x & $a=\pi$ rad.

The phase difference between the instantaneous velocity and acceleration of a particle executing S.H.M is ____________.

• zero

• 0.5 `pi`

• 0.707 `pi`

• `pi`

The phase difference between the instantaneous velocity and acceleration of a particle executing S.H.M is 0.5 `pi`.

Explanation:

For particle executing S.H.M,

`"x" = "A sin"  omega "t"`

`therefore "v" = "dx"/"dt" = "A"omega  "cos"  omega"t"`

`= "A"omega  "sin" (omega"t" + pi/2)` and `"a" = "dv"/"dt" = -"A"omega^2  "sin"  omega"t" = "A"omega^2  "sin"  (omega"t" + pi)`

Phase difference between acceleration and velocity is,

`pi - pi/2 = pi/2 = 0.5  pi.`

Concept: Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.

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