Circular Motion and Displacement

[brad sue]

Hello,
I have 2 problems that bother me here:
1-
The maximun speed of a mass attached to a spring is v[sub=max[/sub]=.371m/s, while the maximun acceleration is 1.05m/s².
What is the maximun displacement of the mass?

I tried the equation of v(t)=-A*w*sin(wt+c_o)
and a(t)=-A*w²*cos(wt+c_o)
But I get nothing I was thinking to use vmax=a*w...??

2-
When a certain circular uniform motion is projected on the x and y axis,the projection gives:
x(t)=Rsin(w*t+c_o-pi/2) and y(t)=Rsin(w*t+c_o).
Show that the circular motion is clockwise.
I have no idea of what to do..
Does someone give me some help?
B

 

[mezarashi]

For the first question, I'm going to assume the spring system is horizontal. The maximum velocity happens when the displacement is zero. What does this tell you about the springs total energy? The maximum acceleration occurs when the force acting on the mass is at its max. This occurs when the displacement \Delta x is at its max. The two unknowns are the spring constant and the maximum displacement.

 

[andrevdh]

brad sue,
the maximum values occur when the sin and cos terms are one therefore
v_{max}=-A\omega
and
a_{max}=-A{\omega}^2
use these to solve for \omega
and subsequently for the amplitude.

The second problem can be approached from the standpoint that the tangent of the angle that the rotating vector makes with the x-axis is given by the ratio of the y to the x component.Use the identities
\sin(\frac{\pi}{2}-\theta)=\cos(\theta)
and
\sin(-\theta)=-\sin(\theta)
and observe what happens with this angle as time advances.

 

[brad sue]

brad sue,
the maximum values occur when the sin and cos terms are one therefore
v_{max}=-A\omega
and
a_{max}=-A{\omega}^2
use these to solve for \omega
and subsequently for the amplitude.
The second problem can be approached from the standpoint that the tangent of the angle that the rotating vector makes with the x-axis is given by the ratio of the y to the x component.Use the identities
\sin(\frac{\pi}{2}-\theta)=\cos(\theta)
and
\sin(-\theta)=-\sin(\theta)
and observe what happens with this angle as time advances.

Thank you both( andrevdh and mezarashi).
I will use those info you gave me to finish the problems. If I get stuck I will come back.

B