Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_{1}$ and $K_{2}$ respectively. If the maximum velocities during oscillations are equal,the ratio of the amplitude of $A$ and $B$ is

$A$ spring-mass system (mass $m$,spring constant $k$,and natural length $\ell_{0}$) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the center of the disc. If the disc,together with the spring-mass system,rotates about its axis with an angular velocity $\omega$ (where $k >> m \omega^{2}$),the relative change in the length of the spring is best given by which option?

$A$ body of mass $m$ is suspended to an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

The vertical extension in a light spring by a weight of $1\, kg$ suspended from it is $9.8\, cm$. What is the period of oscillation?

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. $1$ shows one of them and Fig. $2$ shows their series combination. The ratio of the time period of oscillation of the two $SHM$ is $\frac{T_b}{T_a} = \sqrt{x}$,where the value of $x$ is (Round off to the Nearest Integer).

$A$ $5\, kg$ collar is attached to a spring of spring constant $500\, N/m$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $10\, cm$ and released. The time period of oscillation is