Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $k$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system; the amplitude of vibration is

Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and the same mass $m$ is suspended with them. If $T$ is the new time period in this position,then:

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of '$B$' to that of '$A$' is

One end of a spring of force constant $k$ is fixed to a vertical wall and the other to a block of mass $m$ resting on a smooth horizontal surface. There is another wall at a distance $x_0$ from the block. The spring is then compressed by $2 x_0$ and released. The time taken by the block to strike the other wall is

Four massless springs whose force constants are $2K, 2K, K$ and $2K$ respectively are attached to a mass $M$ kept on a frictionless plane as shown in the figure. If mass $M$ is displaced in the horizontal direction,then the frequency of the oscillating system is:

$A$ body of mass $5\; kg$ hangs from a spring and oscillates with a time period of $2\pi\; s$. If the body is removed,the length of the spring will decrease by: