Oscillation question: dealing with a maximum velocity

[insertnamehere]

Hi, I just want to clarify something.
If there is a particle attached to a spring, in which it's maximum velocity is at t=0 towards THE LEFT.
Does that mean that when x(t)= Acos(wt+phi) then v(t)= -wAsin(wt+phi)
therefore v(0)=-wAsin(phi) = - 20
and Vmax= wa= -20 or positive 20.
When it is heading towards the left, should i make the twenty negative when plugging it into the equations?

 

[quasar987]

Mmmh, yes, conventionally, positive x is to the right, and negative x to the left. Hence, if you are told that the initial speed is to the left, it means that the quantity x is decreasing as time goes by. That is to say, the derivative of x wrt t is negative. But dx/dt = v, so yeah, v<0 at this moment.

 

[quicknote]

I would like to clarify that the sign of the velocity depends on the direction you have defined as positive. In the given scenario, if the maximum velocity is towards the left at t=0, then the direction towards the left can be considered as the positive direction. In this case, the maximum velocity would be positive 20. However, if you have defined the direction towards the right as positive, then the maximum velocity would be negative 20. It is important to be consistent in defining the direction as positive or negative in order to accurately interpret the results of your equations. Therefore, the sign of the velocity depends on the direction you have defined as positive and should be consistent throughout your calculations.