How can I calculate maximum velocity in simple harmonic motion?

I know that the equation for velocity is

V= -omega Asin(omega t +phi)

supposing phi to be zero , cuz if the object is released from the mean position then, at the mean position displacement is zero so,

sin phi = 0 →phi = 0.

I know from the velocity time graph for SHM that max velocity = Aomega.

That is the argument must be -1

How will that be possible ? the argument can only be zero for 2pi/3 right?

Am I missing something?

1 Answer
Dec 26, 2017

"Please see the following explanations."

Explanation:

enter image source here

  • we recommend that you first watch the animation carefully

  • Let P be a point rotating around the O point at a constant linear velocity(u).

  • The projection of P on the x-axis is point B.

  • The movement of point B is limited between A and F.

  • The simple harmonic motion is the action of point B.

  • Please note that the velocity vector changes direction.

  • At points A and F, the velocity of B is zero(green vector).

  • The greatest velocity B has is at O.theta=pi/2 and theta=(3pi)/2

x=r cos theta

theta=omega.t

x=r. cos(omega t)

(d x)/(d t)=v

(d x)/(d t)=v=-r omega sin omega t

"if "omega t=pi/2" or "omega t=(3pi)/2" , "v=-r omega("maximum velocity")

v_("max")=- r omega

"or :"

y=r sin theta

(d x)/(d t)=v=- omega y

x^2+y^2=r^2

y=sqrt(r^2-x^2)

v=- omega sqrt(r^2-x^2)

"if "x=0 " "v=v_("max")