A 50.0 g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 25.0 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
F = -10x, where 10/m is the spring constant, when the object is x distance from the equilibrium point, (or the middle point of the 25 cm amplitude).
Therefore, the acceleration d2x/dt^2 = (F/mass) = --10/(50gm) = 10x/(0.05) = -200x.
Therefore, d/dt (v) = -200x, where v =dx/dt is the velocity of the object.
d/dt(v) = (dv/dx)(dx/dt) = vd/dx= -200x. Or. vdv =-200x*dx
( v^2)/2 =( -200x^2)/2 + C on integration.
But when v=12.5cm= 0.125m, v =0. Therefore,
v^2 = 200*(0.125^2-x^2 ). When x is the position of the particle from the mean position
At the half way of equilibrium position(or mean position), x=0. So v = [200(0.125^2)]^(1/2) = 1.7678 m/s