You are watching an object that is moving in SHM When the ob

You are watching an object that is moving in SHM. When the object is displaced 0.630 m to the right of its equilibrium position, it has a velocity of 2.35 m/s to the right and an acceleration of 8.45 m/s2 to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

Solution

using law of conservation of energy

Total energy at the 0.6 m = total energy at the right extreme position

(0.5*k*x^2) + (0.5*m*v^2) = (0.5*k*A^2)

F = -k*x

m*a = -k*x

force consatnt k = -m*a/x

then

(-0.5*m*a/x)*x^2 + (0.5*m*v^2) = -(0.5*m*a/x)*A^2

m cancels

(-0.5*8.45*0.630)+(0.5*2.35*2.35) = (-0.5*8.45/0.630)*A^2

A = 0.1217 m is the required distance