When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y ( t )= y _{0} \sin ^{2} \omega t ,$ where $'y'$ is measured from the lower end of unstretched spring. Then $\omega$ is

A mass $M$, attached to a horizontal spring, executes S.H.M. with amplitude $A_1$. When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\frac{{{A_1}}}{{{A_2}}}$ is

A block of mass $m$ attached to 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

Two pendulums have time periods $T$ and $\frac{{5T}}{4}.$They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation ..... $^o$

What provides the restoring force in the following cases ?

$(1)$ Compressed spring becomes force for oscillation.

$(2)$ Displacement of water in $U\,-$ tube,

$(3)$ Displacement of pendulum bob from mean position.

Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing SHM with frequency relation $f_{A}=2 f_{B}$, then relation