Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of '$B$' to that of '$A$' is

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are:

$A$ solid cylinder of density $\rho_0$,cross-section area $A$ and length $l$ floats in a liquid of density $\rho (\rho > \rho_0)$ with its axis vertical. If it is slightly displaced downward and released,the time period of oscillation will be:

$A$ mass $m = 1.0\,kg$ is placed on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released,the mass executes simple harmonic motion. The spring constant is $k = 500\,N/m$. What is the amplitude $A$ of the motion,so that the mass $m$ tends to get detached from the pan? (Take $g = 10\,m/s^2$).

$A$ mass $m$ is suspended from two coupled springs connected in series. The force constants for the springs are $K_1$ and $K_2$. The time period of the suspended mass will be:

Find the time period of mass $M$ when displaced from its equilibrium position and then released for the system shown in the figure.