Two masses $m_1=1 \,kg$ and $m_2=0.5 \,kg$ are suspended together by a massless spring of spring constant $12.5 \,Nm ^{-1}$. When masses are in equilibrium $m_1$ is removed without disturbing the system. New amplitude of oscillation will be ………. $cm$

The length of a spring is $l$ and its force constant is $k$. When a weight $W$ is suspended from it, its length increases by $x$. If the spring is cut into two equal parts and put in parallel and the same weight $W$ is suspended from them, then the extension will be

In the situation as shown in figure time period of vertical oscillation of block for small displacements will be 

A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K _1$ and $K _2$. For the frictionless surface, the time period of oscillation of mass $m$ is

A block with mass $M$ is connected by a massless spring with stiffiess constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: ($i$) when the block is at $x_0$; and ($ii$) when the block is at $x=x_0+A$. In both the cases, a perticle with mass $m$ is placed on the mass $M$ ?

($A$) The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged

($B$) The final time period of oscillation in both the cases is same

($C$) The total energy decreases in both the cases

($D$) The instantaneous speed at $x_0$ of the combined masses decreases in both the cases