The frequency of oscillation of a mass $m$ suspended by a spring is $v$. If the mass is reduced to one-fourth,what will be the new frequency of oscillation?

$A$ block of mass $2\,kg$ is attached to two identical springs,each with a spring constant of $20\,N/m$. The block is placed on a frictionless surface,and the outer ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position,it executes simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in $SI$ units. The value of $x$ is $..........$

$A$ simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle, the new time period will be

$A$ sphere of mass $m$ is attached to the lower end of a light vertical spring with force constant $k$. The upper end of the spring is fixed. The sphere is released from rest at the natural (unstretched) length of the spring and comes to rest again after falling through a distance $x$.

$A$ mass $M$ is suspended from a light spring. An additional mass $m$ added displaces the spring further by a distance $x$. Now the combined mass will oscillate on the spring with period

Two bodies $A$ and $B$ of equal mass are suspended from two separate massless springs of spring constants $K_1$ and $K_2$ respectively. The two bodies oscillate vertically such that their maximum velocities are equal. The ratio of the amplitude of $B$ to that of $A$ is