$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of period $T$. If the mass is increased by $m$,the time period becomes $\frac{5T}{3}$. What is the ratio $\left(\frac{M}{m}\right)$?

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:

$A$ $5\, kg$ 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

$A$ 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

$A$ mass $m$ is vertically suspended from a spring of negligible mass; the system oscillates with a frequency $n$. What will be the frequency of the system if a mass $4m$ is suspended from the same spring?

An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially,the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is: