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 = $

$A$ body of mass $m$ is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand,so that the spring is neither stretched nor compressed. Suddenly,the support of the hand is removed. The lowest position attained by the mass during oscillation is $4 \, cm$ below the point where it was held in hand.
$(a)$ What is the amplitude of oscillation?
$(b)$ Find the frequency of oscillation.

Two springs of force constants $300 \, N/m$ (Spring $A$) and $400 \, N/m$ (Spring $B$) are joined together in series. The combination is compressed by $8.75 \, cm$. The ratio of energy stored in $A$ and $B$ is $\frac{E_A}{E_B}$. Then $\frac{E_A}{E_B}$ is equal to

$A$ sphere of mass $m$ is attached to the lower end of a light vertical spring with force constant $k$. The upper end of the spring is fixed. The sphere is released from rest at the natural (unstretched) length of the spring and comes to rest again after falling through a distance $x$.

$A$ weightless spring of length $60\, cm$ and force constant $200\, N/m$ is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. $A$ mass of $0.25\, kg$ is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is

$A$ horizontal spring executes $S.H.M.$ with amplitude $A_{1}$,when mass $m_{1}$ is attached to it. When it passes through the mean position,another mass $m_{2}$ is placed on it. Both masses move together with amplitude $A_{2}$. Therefore,the ratio $A_{2}: A_{1}$ is: