$A$ force of $6.4 \, N$ stretches a vertical spring by $0.1 \, m$. The mass that must be suspended from the spring so that it oscillates with a period of $\left( \frac{\pi}{4} \right) \, s$ is ... $kg$.

As shown in the figure,$S_1$ and $S_2$ are identical springs with spring constant $K$ each. The oscillation frequency of the mass $m$ is $f$. If the spring $S_2$ is removed,the oscillation frequency will become

In the figure given below,a block of mass $M = 490 \, g$ placed on a frictionless table is connected with two springs having the same spring constant $(K = 2 \, N \, m^{-1})$. If the block is horizontally displaced through '$X$' m,then the number of complete oscillations it will make in $14 \pi$ seconds will be $.........$

$A$ mass on a vertical spring begins its motion at rest at $y = 0 \ cm$. It reaches a maximum height of $y = 10 \ cm$. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy $(KE)$ versus position is given below. Net force on the mass varies with $y$ as

$A$ spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio of $T_2 / T_1$ is

$A$ small block of mass $m$,having charge $q$,is placed on a frictionless inclined plane making an angle $\theta$ with the horizontal. There exists a uniform magnetic field $B$ parallel to the inclined plane but perpendicular to the length of the spring. If $m$ is slightly pulled on the inclined plane in the downward direction and released,the time period of oscillation will be (assume that the block does not leave contact with the plane):