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

Question

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is (A) 0.5 π (B) π (C) 0.707 π (D) zero

Tardigrade Admin · Accepted Answer

Let y=A sin ω t (d y/d t)=A ω cos ω t=A ω sin (ω t+(π/2)) Acceleration =-A ω2 sin ω t The phase difference between acceleration and velocity is π / 2.

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

$0.5 π$

$π$

$0.707 π$

zero

Let $y = A sin ω t$
$\frac{d y}{d t} = A ω cos ω t = A ω sin (ω t + \frac{π}{2})$
Acceleration $= - A ω^{2} sin ω t$
The phase difference between acceleration and velocity is $π /2$ .

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

0.5π

π

0.707π

zero

Let y=Asinωt dtdy​=Aωcosωt=Aωsin(ωt+2π​) Acceleration =−Aω2sinωt The phase difference between acceleration and velocity is π/2.

$0.5 π$

$π$

$0.707 π$

Let $y = A sin ω t$
$\frac{d y}{d t} = A ω cos ω t = A ω sin (ω t + \frac{π}{2})$
Acceleration $= - A ω^{2} sin ω t$
The phase difference between acceleration and velocity is $π /2$ .