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 = $

$A$ block $P$ of mass $m$ is placed on a smooth horizontal surface. $A$ block $Q$ of same mass is placed over the block $P$ and the coefficient of static friction between them is $\mu_S$. $A$ spring of spring constant $K$ is attached to block $Q$. The blocks are displaced together to a distance $A$ and released. The upper block oscillates without slipping over the lower block. The maximum frictional force between the blocks is

$A$ 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:

$A$ 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 $g = 10\,m/s^2$).

The period of oscillation of a mass $M$ suspended from a spring of negligible mass is $T$. If along with it another mass $M$ is also suspended,the period of oscillation will now be

$A$ 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 $x$.