$A$ block of mass $M_1$ is hung by a light spring of force constant $k$ from the top bar of a reverse $U$-frame of mass $M_2$ resting on the floor. The block is pulled down from its equilibrium position by a distance $x$ and then released. Find the minimum value of $x$ such that the reverse $U$-frame will leave the floor momentarily.

$A$ small mass $m$, attached to one end of a spring with a negligible mass and an unstretched length $L$, executes vertical oscillations with angular frequency $\omega_{0}$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 \,cm/s$ and the period is $628 \,ms$. The amplitude of the motion in $cm$ is:

$A$ body of mass $4.9 \, kg$ hangs from a spring and oscillates with a period $0.5 \, s$. On the removal of the body, the spring is shortened by (take $g=10 \, m/s^2, \pi^2=10$)

The scale of a spring balance reading from $0$ to $10 \, kg$ is $0.25 \, m$ long. $A$ body suspended from the balance oscillates vertically with a period of $\pi / 10 \, s$. The mass suspended is ..... $kg$ (neglect the mass of the spring).

$A$ 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):