The force constants of two springs are ${K_1}$ and ${K_2}$. Both are stretched till their elastic energies are equal. If the stretching forces are ${F_1}$ and ${F_2}$, then ${F_1}:{F_2}$ is

A uniform spring of force constant $k$ is cut into two pieces, the lengths of which are in the ratio $1 : 2$. The ratio of the force constants of the shorter and the longer pieces is

Two masses $m_1$ and $m_2$ connected by a spring of spring constant $k$ rest on a frictionless surface. If the masses are pulled apart and let go, the time period of oscillation is

How the period of oscillation depend on the mass of block attached to the end of spring ?

A body of mass $m $ is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass $m$ is slightly pulled down and released , it oscillates with a time period of $3\,s$ . When the mass $m$ is increased by $1\,kg$ , the time period of oscillations becomes $5\,s$ . The value of $m$ in $kg$ is

For the damped oscillator shown in Figure the mass mof the block is $200\; g , k=90 \;N m ^{-1}$ and the damping constant $b$ is $40 \;g s ^{-1} .$ Calculate

$(a)$ the period of oscillation,

$(b)$ time taken for its amplitude of vibrations to drop to half of Its inittal value, and

$(c)$ the time taken for its mechanical energy to drop to half its initial value.