$A$ 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:

If the period of oscillation of mass $m$ suspended from a spring is $2 \ s$,then the period of oscillation of suspended mass $4m$ with the same spring will be: (in $s$)

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 the unstretched spring. Then $\omega$ is

All the springs in figures $(a)$,$(b)$,and $(c)$ are identical,each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$,and $T_c$ are the time periods of oscillations of the three systems in figures $(a)$,$(b)$,and $(c)$ respectively,then:

$A$ horizontal spring executes $S.H.M.$ with amplitude $A_{1}$,when mass $m_{1}$ is attached to it. When it passes through the mean position,another mass $m_{2}$ is placed on it. Both masses move together with amplitude $A_{2}$. Therefore,the ratio $A_{2}: A_{1}$ is:

What is the condition for a body suspended at the end of a spring to undergo simple harmonic oscillation?