Oscillation question: dealing with a maximum velocity

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SUMMARY

The discussion clarifies the relationship between maximum velocity and position in harmonic motion, specifically for a particle attached to a spring. The equations x(t) = Acos(wt + phi) and v(t) = -wAsin(wt + phi) are confirmed, with v(0) calculated as -20, indicating the particle's initial velocity is directed to the left. The convention of positive x being to the right is emphasized, confirming that a leftward initial velocity results in a negative value for v, aligning with the derivative of position over time.

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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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