The time period of oscillation for the given combination will be:

Frequency of a particle performing $S.H.M.$ is $10 \ Hz$. The particle is suspended from a vertical spring. At the highest point of its oscillation,the spring is unstretched. The maximum speed of the particle is $(g = 10 \ m/s^2)$.

Two small bodies of mass $2 \, kg$ each are attached to each other using a thread of length $10 \, cm$,and hang on a spring whose force constant is $200 \, N/m$,as shown in the figure. We burn the thread. What is the distance between the two bodies when the top body first arrives at its highest position? (Take $\pi^2 = 10$)

$A$ block of mass $100 \,g$ is suspended vertically between two massless springs, each with a spring constant $k=1 \,N/m$. The block is hit from above to impart an impulse of $2 \,Ns$. Calculate the maximum displacement from the equilibrium position of the block. (Take $g=10 \,m/s^2$) (in $\,m$)

$A$ 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

$A$ spring is stretched by $5 \,\,cm$ by a force of $10 \,\,N$. The time period of the oscillations when a mass of $2 \,\,kg$ is suspended by it is: (in $s$)