Oscillation question: dealing with a maximum velocity

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In oscillatory motion, the maximum velocity of a particle attached to a spring is reached at t=0, moving towards the left. The velocity equation v(t) = -wA sin(wt + phi) indicates that if the initial velocity is -20, then Vmax is also -20 or +20, depending on the direction. Since positive x is defined as rightward, a leftward initial speed means the position x is decreasing over time. Consequently, the velocity v must be negative at this point, confirming that v < 0. Understanding these conventions is crucial for accurately applying the equations of motion.
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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?
 
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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.
 
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