$A$ force of $6.4 \, N$ stretches a vertical spring by $0.1 \, m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.

$A$ mass '$m_1$' is suspended from a spring of negligible mass. The spring is pulled slightly in the downward direction and released; the mass performs $S.H.M.$ with a period '$T_1$'. If the mass is increased by '$m_2$',the time period becomes '$T_2$'. The ratio $\frac{m_2}{m_1}$ is

$A$ $5\, kg$ collar is attached to a spring of spring constant $500\, N/m$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $10\, cm$ and released. The time period of oscillation is

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to one-third,then the same mass oscillates with frequency $v_2$. Then:

On a smooth inclined plane,a mass $M$ is attached between two massless springs of force constant $k$ each,as shown in the figure. The other ends of the springs are fixed to firm supports. The period of oscillation of the mass $M$ is

$A$ body of mass $m$ is suspended to an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is