Question

A 0.40-kg object connected to a light spring with a force
constant of 19.6 N/m oscillates on a frictionless horizontal
surface. If the spring is compressed 4.0 cm and released
from rest, determine (a) the maximum speed of
the object, (b) the speed of the object when the spring is
compressed 1.5 cm, and (c) the speed of the object when
the spring is stretched 1.5 cm. (d) For what value of x
does the speed equal one-half the maximum speed?

Answer 1

since the spring is compressed by 4cm , amplitude is A = 4cm
as the total energy is constant ,max. KE = max . PE
(a) therefore (.5)*m*v^2 = (.5)*19.6*A^2 this will give max. speed
(b) again use conservation of energy ,the total energy is (.5)*19.6*A^2
total energy = KE + PE at any instant
(C) same as above ,ans will be same too
(D) again by conservation of mechanical energy ,
KE at this instant is (.25) times of maxmimum kinetic energy which you already know

Answer 2

thanks a lot bro!