Two springs of constant $k_1$ and $k_2$ are joined in series. The effective spring constant of the combination is given by

The total mechanical energy of a spring-mass system in simple harmonic motion is $E = \frac{1}{2}m\omega^2 A^2$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will:

$A$ $3 \ kg$ block is connected as shown in the figure. The spring constants of the two springs $K_1$ and $K_2$ are $50 \ Nm^{-1}$ and $150 \ Nm^{-1}$ respectively. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is $(g=10 \ ms^{-2})$. (in $ms^{-2}$)

Two springs of force constants $2k$ and $k$ are connected to a mass $m$ as shown. The mass is displaced slightly to one side and released. The frequency of oscillation of the two springs-mass system is

$A$ block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 \, cm$ from its equilibrium position (at $x = 0$) on a frictionless surface from rest. The total energy of the block at $x = 5 \, cm$ is $0.25 \, J$. The spring constant of the spring is $......... \, N \, m^{-1}$.

$A$ block of mass $m$ is attached to two springs of spring constants $k_1$ and $k_2$ as shown in the figure. The block is displaced by $x$ towards the right and released. The velocity of the block when it is at $x/2$ will be