$A$ mass $M$ attached to a horizontal spring executes simple harmonic motion with amplitude $A_1$. When mass $M$ passes the mean position,a smaller mass $m$ is attached to it,and both of them together execute simple harmonic motion with amplitude $A_2$. Then the value of $\frac{A_1}{A_2}$ is

Initially,the system is in equilibrium. The time period of $SHM$ of the block in the vertical direction is

$A$ block of mass $5 \ kg$ moves along the $x$-direction subject to the force $F = (-20x + 10) \ N$,with the value of $x$ in metre. At time $t = 0 \ s$,it is at rest at position $x = 1 \ m$. The position and momentum of the block at $t = (\pi / 4) \ s$ are

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. The amplitude of oscillations is

In a spring-block system as shown in the figure,if the spring constant $K = 9 \pi^2 \ Nm^{-1}$,then the time period of oscillation is (in $s$)

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: