$A$ system consisting of three masses attached to a spring is in equilibrium. When the $700\,g$ mass is removed,the system oscillates with a period of $3\,s$. When the $500\,g$ mass is also removed,what will be the new time period of the system? (in $\text{seconds}$)

$A$ mass $x \ g$ is suspended from a light spring. It is pulled in a downward direction and released so that the mass performs $S.H.M.$ of period $T$. If the mass is increased by $Y \ g$,the period becomes $4T/3$. The ratio of $Y/x$ 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

$A$ body of mass $1 \,kg$ is suspended from a spring of negligible mass. Another body of mass $500 \,g$ moving vertically upwards hits the suspended body with a velocity of $3 \,ms^{-1}$ and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is $\frac{10}{\pi} \,Hz$,the amplitude of the motion and the spring constant are respectively,

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

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: