A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K _1$ and $K _2$. For the frictionless surface, the time period of oscillation of mass $m$ is
$\frac{1}{2 \pi} \sqrt{\frac{ K _1+ K _2}{ m }}$
$\frac{1}{2 \pi} \sqrt{\frac{ K _1- K _2}{ m }}$
$2 \pi \sqrt{\frac{ m }{ K _1+ K _2}}$
$2 \pi \sqrt{\frac{m}{K_1-K_2}}$
A spring having a spring constant $‘K’$ is loaded with a mass $‘m’$. The spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant is
A block whose mass is $1 \;kg$ is fastened to a spring. The spring has a spring constant of $50\; N m ^{-1}$. The block is pulled to a distance $x=10\; cm$ from its equilibrlum position at $x=0$ on a frictionless surface from rest at $t=0 .$ Calculate the kinetic, potentlal and total energles of the block when it is $5 \;cm$ away from the mean position.
How the period of oscillation depend on the mass of block attached to the end of spring ?
Two masses $m_1=1 \,kg$ and $m_2=0.5 \,kg$ are suspended together by a massless spring of spring constant $12.5 \,Nm ^{-1}$. When masses are in equilibrium $m_1$ is removed without disturbing the system. New amplitude of oscillation will be .......... $cm$
A $2\, Kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi ^2 N/m$ attached to $2\, Kg$ block at rest kept on a smooth floor. The time for which rear moving block remain in contact with spring will be ... $\sec$