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
$\frac{ K _{2}}{ K _{1}}$
$\frac{ K _{1}}{ K _{2}}$
$\sqrt{\frac{ K _{1}}{ K _{2}}}$
$\sqrt{\frac{ K _{2}}{ K _{1}}}$
Two masses $m_1$ and $m_2$ connected by a spring of spring constant $k$ rest on a frictionless surface. If the masses are pulled apart and let go, the time period of oscillation is
Two bodies $M$ and $N $ of equal masses are suspended from two separate massless springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude $M$ to that of $N$ is
A mass of $2.0\, kg$ is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is $200\, N/m.$ What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take $g = 10 m/s^2$).
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 ratios of time period of oscillation of the two $SHM$ is $\frac{ T _{ b }}{ T _{ a }}=\sqrt{ x },$ where value of $x$ is
(Round off to the Nearest Integer)
A body of mass $0.01 kg$ executes simple harmonic motion $(S.H.M.)$ about $x = 0$ under the influence of a force shown below : The period of the $S.H.M.$ is .... $s$