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
ali ergin
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")#

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