$A$ body of mass $4 \ kg$ attached to a spring of force constant $64 \ N \ m^{-1}$ executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is

Two identical springs of constant $k$ are connected in series and then in parallel. $A$ mass $m$ is suspended from them,the ratio of their frequencies of vertical oscillations will be

$A$ particle of mass $m$ is attached to $3$ springs $A$,$B$ and $C$ of equal force constant $K$ as shown in the figure. If the particle is pushed slightly along the line of spring $C$ and released,find the period of oscillation.

If $b = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \dots$,then $b + \frac{b^2}{2!} + \frac{b^3}{3!} + \frac{b^4}{4!} + \dots \infty = $

Two bodies $A$ and $B$ of equal mass are suspended from two massless springs of spring constant $k_1$ and $k_2$,respectively. If the bodies oscillate vertically such that their amplitudes are equal,the ratio of the maximum velocity of $A$ to the maximum velocity of $B$ is

$A$ light spring is suspended with mass $m_1$ at its lower end and its upper end is fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $T$ seconds. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2} T$. The ratio $\frac{m_1}{m_2}$ is