When a particle of mass m is attached to a vertical spring of spring constant k and released, its motion is described by
y(t) = y₀ sin² $$\omega $$t, where 'y' is measured from the lower end of unstretched spring. Then $$\omega $$ is:

A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of f is :

The displacement time graph of a particle executing S.H.M is given in figure :
(sketch is schematic and not to scale)
Which of the following statements is/are true for this motion?
(A) The force is zero at t = $${{3T} \over 4}$$
(B) The acceleration is maximum at t = T
(C) The speed is maximum at t = $${{T} \over 4}$$
(D) The P.E. is equal to K.E. of the oscillation at t = $${{T} \over 2}$$

A spring mass system (mass m, spring constant k and natural length $$l$$) rest in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity $$\omega $$, (k $$ \gg m{\omega ^2}$$) the relative change in the length of the spring is best given by the option :