1. In the equilibrium position.
2. When elastic force = friction force
kx=μmg=0.2* 1. 10=2
X=0.02, if the oscillator speed is exactly 0 at this time,
When the first one reaches the left,
S = kx * x/2μ g =100 * 72 *10 (-4 square) /2*0.2* 10= 12.25 (cm).
12.25-7=5.25
When the vibrator moves 5.25cm to the left of the equilibrium position, the speed is zero and the elastic force is greater than the friction force, so the vibrator will move to the right.
A. Is the English requirement for mechanical design and automation high?
This is what my troops did. First of all, it's easier for mechanical design ma