$A$ mass $M$ is suspended by two springs of force constants $K_1$ and $K_2$ respectively,as shown in the diagram. The total elongation (stretch) of the two springs is

The equivalent spring constant of the system as shown in the figure will be:

$A$ mass $m_1$ performs $S.H.M.$ with amplitude $A$ while connected to a horizontal spring. While mass $m_1$ is passing through the mean position, another mass $m_2$ $(m_2 < m_1)$ is placed on it so that both masses move together with amplitude $A_1$. The ratio of $A/A_1$ is

$A$ body of mass $4.9 \, kg$ hangs from a spring and oscillates with a period $0.5 \, s$. On the removal of the body, the spring is shortened by (take $g=10 \, m/s^2, \pi^2=10$)

Figure $(a)$ shows a spring of force constant $k$ clamped rigidly at one end and a mass $m$ attached to its free end. $A$ force $F$ applied at the free end stretches the spring. Figure $(b)$ shows the same spring with both ends free and attached to a mass $m$ at either end. Each end of the spring in Figure $(b)$ is stretched by the same force $F$.
$(a)$ What is the maximum extension of the spring in the two cases?
$(b)$ If the mass in Figure $(a)$ and the two masses in Figure $(b)$ are released,what is the period of oscillation in each case?

$A$ uniform circular disc of mass $12 \ kg$ is held by two identical springs. When the disc is slightly pressed down and released,it executes $S.H.M.$ of period $2 \ s$. The force constant of each spring is (nearly) (Take $\pi^2=10$) (in $Nm^{-1}$)